An Explanation for Fictitious Forces
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Fictitious Forces in a Rotating System – Centrifugal and Coriolis
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Exercises with Coriolis and Centrifugal Forces
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[{"Name":"An Explanation for Fictitious Forces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Basic Explanation","Duration":"11m 12s","ChapterTopicVideoID":9003,"CourseChapterTopicPlaylistID":5351,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:02.910","Text":"Hello. In this video,"},{"Start":"00:02.910 ","End":"00:05.505","Text":"I want to talk about fictitious force."},{"Start":"00:05.505 ","End":"00:10.335","Text":"Fictitious force, which is also known as a pseudo force, inertial force,"},{"Start":"00:10.335 ","End":"00:12.915","Text":"or the d\u0027Alembert force is"},{"Start":"00:12.915 ","End":"00:17.835","Text":"an apparent force that is an adjustment to Newton\u0027s second law of motion."},{"Start":"00:17.835 ","End":"00:23.174","Text":"We use it when we\u0027re describing a system in which the observer or frame of reference,"},{"Start":"00:23.174 ","End":"00:24.780","Text":"the place from which we are measuring,"},{"Start":"00:24.780 ","End":"00:27.720","Text":"is in the midst of acceleration."},{"Start":"00:27.720 ","End":"00:33.840","Text":"I want to take the opportunity to describe fictitious forces using an example."},{"Start":"00:33.840 ","End":"00:36.045","Text":"Let\u0027s say we have a road here,"},{"Start":"00:36.045 ","End":"00:41.460","Text":"something like this and on this road, we have a car."},{"Start":"00:41.460 ","End":"00:45.530","Text":"The car, it\u0027s an old Studebaker,"},{"Start":"00:45.530 ","End":"00:48.755","Text":"is going at some velocity to the right,"},{"Start":"00:48.755 ","End":"00:50.855","Text":"and it also has acceleration."},{"Start":"00:50.855 ","End":"00:56.360","Text":"The car gets its acceleration from the force of friction pushing it along the road."},{"Start":"00:56.360 ","End":"00:59.180","Text":"We call this f_s and the force of"},{"Start":"00:59.180 ","End":"01:02.165","Text":"friction is pushing it in the direction of the acceleration."},{"Start":"01:02.165 ","End":"01:06.068","Text":"You could think of this as a person standing behind the car and pushing it,"},{"Start":"01:06.068 ","End":"01:07.925","Text":"but what\u0027s really happening regardless,"},{"Start":"01:07.925 ","End":"01:11.255","Text":"is being pushed along by friction on the road."},{"Start":"01:11.255 ","End":"01:13.595","Text":"For all intents and purposes,"},{"Start":"01:13.595 ","End":"01:18.160","Text":"we\u0027ll assume that f_s is the only force acting upon this car."},{"Start":"01:18.160 ","End":"01:21.635","Text":"The next step is to set up our lab, so to speak,"},{"Start":"01:21.635 ","End":"01:25.460","Text":"the point of observation from which we\u0027re measuring the car\u0027s acceleration,"},{"Start":"01:25.460 ","End":"01:27.960","Text":"position, velocity, or any other measurement."},{"Start":"01:27.960 ","End":"01:30.320","Text":"We make it over here with our x and y-axis,"},{"Start":"01:30.320 ","End":"01:35.130","Text":"that\u0027s our y-axis and this is our x-axis here."},{"Start":"01:35.130 ","End":"01:39.920","Text":"This is the initial point from which we\u0027ll be making observations about the acceleration,"},{"Start":"01:39.920 ","End":"01:42.680","Text":"velocity, and position of our car."},{"Start":"01:42.680 ","End":"01:46.730","Text":"I like to imagine that we have some observer who\u0027s standing here"},{"Start":"01:46.730 ","End":"01:50.705","Text":"at the origin and he\u0027s taking measurements about the car."},{"Start":"01:50.705 ","End":"01:54.980","Text":"For example, if this observer wanted to use Newton\u0027s second law,"},{"Start":"01:54.980 ","End":"02:02.345","Text":"he or she would write out that the sum of the forces equals mass times acceleration."},{"Start":"02:02.345 ","End":"02:05.585","Text":"Now, in this case, remember our only force is f_s."},{"Start":"02:05.585 ","End":"02:08.120","Text":"Now, for all intents and purposes,"},{"Start":"02:08.120 ","End":"02:10.700","Text":"we can assume f_s is any force."},{"Start":"02:10.700 ","End":"02:12.620","Text":"It could be someone pulling on a rope."},{"Start":"02:12.620 ","End":"02:15.140","Text":"It could be someone pushing from behind."},{"Start":"02:15.140 ","End":"02:19.220","Text":"If you don\u0027t understand why the friction is moving the car forward, that\u0027s okay."},{"Start":"02:19.220 ","End":"02:20.375","Text":"We\u0027ll pick it up later."},{"Start":"02:20.375 ","End":"02:23.660","Text":"But again, the point is f_s is our only force"},{"Start":"02:23.660 ","End":"02:27.680","Text":"and so that has to equal ma, mass times acceleration."},{"Start":"02:27.680 ","End":"02:32.000","Text":"Just to reiterate, as long as my measurements are being taken from the laboratory,"},{"Start":"02:32.000 ","End":"02:36.995","Text":"I\u0027ll only have real forces acting upon the object that are accounted for here."},{"Start":"02:36.995 ","End":"02:39.770","Text":"I will not be accounting for any fictitious forces."},{"Start":"02:39.770 ","End":"02:41.980","Text":"There will be no fictitious forces."},{"Start":"02:41.980 ","End":"02:45.610","Text":"As long as we\u0027re measuring from an inertial reference frame,"},{"Start":"02:45.610 ","End":"02:49.917","Text":"that is a reference frame where your observer does not have acceleration,"},{"Start":"02:49.917 ","End":"02:53.285","Text":"you\u0027ll only be measuring real forces."},{"Start":"02:53.285 ","End":"02:55.265","Text":"When I say real forces,"},{"Start":"02:55.265 ","End":"02:59.900","Text":"I mean a force that is carried out by 1 object upon another object."},{"Start":"02:59.900 ","End":"03:06.500","Text":"For example, f_s or force of friction is carried out by the road on the car."},{"Start":"03:06.500 ","End":"03:11.195","Text":"Now, fictitious forces come into play when we have a second observer."},{"Start":"03:11.195 ","End":"03:12.920","Text":"Now this observer, in our case,"},{"Start":"03:12.920 ","End":"03:18.500","Text":"we\u0027ll have him sitting on a skateboard and he\u0027ll have an acceleration as well."},{"Start":"03:18.500 ","End":"03:21.190","Text":"This acceleration will have as a_0."},{"Start":"03:21.190 ","End":"03:25.280","Text":"Now, because our skateboarder is moving with acceleration,"},{"Start":"03:25.280 ","End":"03:28.550","Text":"that means that the acceleration it measures of the car is"},{"Start":"03:28.550 ","End":"03:31.895","Text":"going to be different than the acceleration measured by the laboratory."},{"Start":"03:31.895 ","End":"03:35.345","Text":"We talked about this in our unit on relative motion."},{"Start":"03:35.345 ","End":"03:38.420","Text":"If you recall, in our unit on relative motion,"},{"Start":"03:38.420 ","End":"03:40.565","Text":"we also came up with a formula for this."},{"Start":"03:40.565 ","End":"03:44.810","Text":"If you remember, a tag is going to be our relative acceleration."},{"Start":"03:44.810 ","End":"03:51.640","Text":"This acceleration is the acceleration of the car relative to the skateboarder."},{"Start":"03:51.640 ","End":"03:56.030","Text":"This relative acceleration is going to equal a,"},{"Start":"03:56.030 ","End":"03:59.870","Text":"the acceleration of the car minus a_0,"},{"Start":"03:59.870 ","End":"04:03.270","Text":"the acceleration of the skateboard."},{"Start":"04:04.340 ","End":"04:06.525","Text":"Let\u0027s try an example."},{"Start":"04:06.525 ","End":"04:16.060","Text":"Let\u0027s say that a equals 5m/s^2 and because the skateboard is going slower than the car,"},{"Start":"04:16.060 ","End":"04:21.940","Text":"that the skateboard\u0027s acceleration is 2m/s^2."},{"Start":"04:21.940 ","End":"04:32.115","Text":"That would mean that a tag equals 5 minus 2 or 3m/s^2."},{"Start":"04:32.115 ","End":"04:35.470","Text":"Now let\u0027s assume that we also have a mass for this car."},{"Start":"04:35.470 ","End":"04:38.570","Text":"The mass is 30 kilograms."},{"Start":"04:38.570 ","End":"04:40.630","Text":"If the mass is 30 kilograms,"},{"Start":"04:40.630 ","End":"04:44.200","Text":"we can fill in our sum of forces from before,"},{"Start":"04:44.200 ","End":"04:49.445","Text":"so f_s equals 30kg times"},{"Start":"04:49.445 ","End":"04:57.510","Text":"5m/s^2 or we can"},{"Start":"04:57.510 ","End":"05:02.100","Text":"say it equals 150 Newtons."},{"Start":"05:02.100 ","End":"05:04.720","Text":"Now let\u0027s say we wanted to measure this all from"},{"Start":"05:04.720 ","End":"05:07.510","Text":"the skateboarder\u0027s perspective and do a force equation."},{"Start":"05:07.510 ","End":"05:13.390","Text":"For the skateboarder, the sum of forces equals ma tag,"},{"Start":"05:13.390 ","End":"05:18.069","Text":"because a tag is the acceleration observed by the skateboarder."},{"Start":"05:18.069 ","End":"05:20.620","Text":"The skateboarder doesn\u0027t know that we have a laboratory here."},{"Start":"05:20.620 ","End":"05:23.275","Text":"It doesn\u0027t know about the measurements relative to the ground."},{"Start":"05:23.275 ","End":"05:28.275","Text":"The skateboarder only measures the relative measurements from his or her perspective."},{"Start":"05:28.275 ","End":"05:34.595","Text":"In that case, we would have the sum of forces equals 30 times 3."},{"Start":"05:34.595 ","End":"05:39.790","Text":"But the thing is 150 doesn\u0027t equal 30 times 3,"},{"Start":"05:39.790 ","End":"05:43.960","Text":"so this must be an incorrect equation."},{"Start":"05:43.960 ","End":"05:48.350","Text":"What this reveals to us is that Newton\u0027s second law of motion doesn\u0027t"},{"Start":"05:48.350 ","End":"05:52.599","Text":"work when we\u0027re measuring from the perspective of an observer with acceleration,"},{"Start":"05:52.599 ","End":"05:54.950","Text":"so we need to account for this someway and have an adjustment."},{"Start":"05:54.950 ","End":"05:59.365","Text":"This adjustment is fictitious force or d\u0027Alembert force."},{"Start":"05:59.365 ","End":"06:03.155","Text":"What we do is when we do a force equation,"},{"Start":"06:03.155 ","End":"06:06.013","Text":"the sum of our forces equals ma,"},{"Start":"06:06.013 ","End":"06:08.945","Text":"we need to add some other force here at the beginning,"},{"Start":"06:08.945 ","End":"06:11.590","Text":"and this is our fictitious force."},{"Start":"06:11.590 ","End":"06:16.370","Text":"Now, keep in mind in this equation when we\u0027re talking about fictitious force,"},{"Start":"06:16.370 ","End":"06:20.000","Text":"the sum of forces that we\u0027re talking about here is the same sum of forces,"},{"Start":"06:20.000 ","End":"06:21.470","Text":"the real force that is,"},{"Start":"06:21.470 ","End":"06:22.840","Text":"that we talked about above."},{"Start":"06:22.840 ","End":"06:27.155","Text":"The real force stays the same and we add a fictitious force on top of that."},{"Start":"06:27.155 ","End":"06:29.210","Text":"Now, of course, we have to have a formula for"},{"Start":"06:29.210 ","End":"06:32.635","Text":"calculating the fictitious force and here it is."},{"Start":"06:32.635 ","End":"06:35.045","Text":"This is the formula."},{"Start":"06:35.045 ","End":"06:41.015","Text":"The fictitious force formula is F equals negative ma_0."},{"Start":"06:41.015 ","End":"06:44.195","Text":"Keep in mind, we\u0027re taking the mass of the observed object."},{"Start":"06:44.195 ","End":"06:45.835","Text":"In our case, that\u0027s the car."},{"Start":"06:45.835 ","End":"06:49.235","Text":"The acceleration is the acceleration of the observer."},{"Start":"06:49.235 ","End":"06:51.140","Text":"In our case, that\u0027s the skateboarder."},{"Start":"06:51.140 ","End":"06:54.230","Text":"Usually, the m and the a are from the same object."},{"Start":"06:54.230 ","End":"06:55.880","Text":"In this case, they are not."},{"Start":"06:55.880 ","End":"06:58.430","Text":"Also, keep in mind there\u0027s a negative symbol here."},{"Start":"06:58.430 ","End":"07:00.070","Text":"We\u0027re measuring something negative."},{"Start":"07:00.070 ","End":"07:04.010","Text":"This fictitious force formula gives us that little f we add"},{"Start":"07:04.010 ","End":"07:08.179","Text":"here when we had a non-inertial frame of reference."},{"Start":"07:08.179 ","End":"07:13.235","Text":"Remember, an inertial frame of reference is when we have a stationary observer,"},{"Start":"07:13.235 ","End":"07:14.590","Text":"like in our laboratory."},{"Start":"07:14.590 ","End":"07:17.795","Text":"A non-inertial frame of reference is when we have"},{"Start":"07:17.795 ","End":"07:22.940","Text":"a moving frame of reference with acceleration such as our skateboarder."},{"Start":"07:22.940 ","End":"07:24.905","Text":"Returning to our initial formula,"},{"Start":"07:24.905 ","End":"07:29.130","Text":"remember that we\u0027re measuring a relative acceleration, a tag."},{"Start":"07:29.130 ","End":"07:32.840","Text":"We can insert our formula here and we get"},{"Start":"07:32.840 ","End":"07:41.880","Text":"negative ma_0 plus the sum of forces equals ma tag."},{"Start":"07:41.880 ","End":"07:46.200","Text":"Now, keep in mind a tag we can define as a minus a_0."},{"Start":"07:46.200 ","End":"07:57.250","Text":"We can rewrite this as negative ma_0 plus the sum of forces equals ma minus a_0."},{"Start":"07:58.010 ","End":"08:03.330","Text":"Now, you\u0027ll notice, the negative ma_0 drops out from both sides."},{"Start":"08:03.330 ","End":"08:07.730","Text":"What you\u0027re left with is the sum of forces equals ma."},{"Start":"08:07.730 ","End":"08:10.895","Text":"This formula is the same formula we started with,"},{"Start":"08:10.895 ","End":"08:12.580","Text":"so we know that it\u0027s accurate."},{"Start":"08:12.580 ","End":"08:17.330","Text":"What we\u0027ve essentially done by adding our fictitious force is accounting"},{"Start":"08:17.330 ","End":"08:21.965","Text":"for the perceived change in acceleration due to our observer\u0027s acceleration."},{"Start":"08:21.965 ","End":"08:24.200","Text":"This negative ma_0 accounts for"},{"Start":"08:24.200 ","End":"08:27.770","Text":"the different acceleration that our observer is measuring."},{"Start":"08:27.770 ","End":"08:32.525","Text":"It\u0027s not a real force being acted out from 1 object onto another,"},{"Start":"08:32.525 ","End":"08:34.220","Text":"rather it\u0027s an imaginary force,"},{"Start":"08:34.220 ","End":"08:36.440","Text":"a fictitious force in a true sense,"},{"Start":"08:36.440 ","End":"08:39.440","Text":"and it doesn\u0027t really exist."},{"Start":"08:39.440 ","End":"08:42.680","Text":"What it does do is allow us to measure"},{"Start":"08:42.680 ","End":"08:49.390","Text":"the real forces upon an object using a relative observer."},{"Start":"08:49.390 ","End":"08:53.330","Text":"Let\u0027s add in our numbers from before to check if this works."},{"Start":"08:53.330 ","End":"08:58.741","Text":"Negative ma_0 equals negative"},{"Start":"08:58.741 ","End":"09:04.214","Text":"30 times 2m/s^2 plus the sum of forces is 150,"},{"Start":"09:04.214 ","End":"09:10.060","Text":"and that equals 30 times 3ma tag."},{"Start":"09:10.060 ","End":"09:17.738","Text":"What we end up with is 150 minus 60 equals 90."},{"Start":"09:17.738 ","End":"09:20.388","Text":"This we know to be true,"},{"Start":"09:20.388 ","End":"09:22.180","Text":"so this method works."},{"Start":"09:22.180 ","End":"09:25.720","Text":"The last thing we have to account for is some object"},{"Start":"09:25.720 ","End":"09:30.025","Text":"moving not just along 1 axis but in 2 dimensional space."},{"Start":"09:30.025 ","End":"09:32.710","Text":"Let\u0027s say the car moves instead of along this axis like this,"},{"Start":"09:32.710 ","End":"09:35.275","Text":"the car is moving in this direction,"},{"Start":"09:35.275 ","End":"09:37.350","Text":"maybe in a spiral formation even."},{"Start":"09:37.350 ","End":"09:41.780","Text":"What we need to do in that case is break things down by component."},{"Start":"09:41.780 ","End":"09:46.105","Text":"We can do is first take our x component and say"},{"Start":"09:46.105 ","End":"09:55.145","Text":"negative ma_0x plus Sigma F_x"},{"Start":"09:55.145 ","End":"09:58.860","Text":"equals ma tag x."},{"Start":"09:58.860 ","End":"10:02.680","Text":"Then we can do the same thing for y and we account for each variable."},{"Start":"10:02.680 ","End":"10:05.150","Text":"We can also do this in vector form."},{"Start":"10:05.150 ","End":"10:14.930","Text":"We can say negative ma_0 vector plus the sum of forces in the direction of the vector,"},{"Start":"10:14.930 ","End":"10:19.205","Text":"equals ma tag vector."},{"Start":"10:19.205 ","End":"10:22.085","Text":"It works if we\u0027re doing 1 dimensional motion,"},{"Start":"10:22.085 ","End":"10:25.265","Text":"like we have above along 1 line."},{"Start":"10:25.265 ","End":"10:29.900","Text":"It works in 2 dimensional motion when we have x and y."},{"Start":"10:29.900 ","End":"10:34.024","Text":"We can also do this in terms of vectors and directions."},{"Start":"10:34.024 ","End":"10:37.025","Text":"Now, we can use this in any situation."},{"Start":"10:37.025 ","End":"10:42.110","Text":"One last note is if we\u0027re using relative motion that is,"},{"Start":"10:42.110 ","End":"10:46.205","Text":"and let\u0027s say our skateboarder had an acceleration of 0,"},{"Start":"10:46.205 ","End":"10:48.305","Text":"it was moving without acceleration,"},{"Start":"10:48.305 ","End":"10:51.320","Text":"we would not need to use this whole process."},{"Start":"10:51.320 ","End":"10:53.915","Text":"We would not need to account for fictitious forces."},{"Start":"10:53.915 ","End":"10:55.925","Text":"Fictitious forces, or again,"},{"Start":"10:55.925 ","End":"11:02.405","Text":"d\u0027Alembert force is only necessary when our body has a positive or negative acceleration."},{"Start":"11:02.405 ","End":"11:05.300","Text":"If your observer is moving without acceleration,"},{"Start":"11:05.300 ","End":"11:07.339","Text":"this is not necessary."},{"Start":"11:07.339 ","End":"11:10.025","Text":"That sums up fictitious forces."},{"Start":"11:10.025 ","End":"11:13.500","Text":"I would now suggest moving on to the exercises."}],"ID":9287},{"Watched":false,"Name":"Exercise - Simple Fictitious Force in Elevator","Duration":"13m 5s","ChapterTopicVideoID":10482,"CourseChapterTopicPlaylistID":5351,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"Hello. In this question,"},{"Start":"00:02.160 ","End":"00:08.850","Text":"we\u0027re being told that a man of mass 70 kilograms is standing on a scale in an elevator."},{"Start":"00:08.850 ","End":"00:11.670","Text":"What value will the scale show in"},{"Start":"00:11.670 ","End":"00:16.470","Text":"the following situations: When the elevator is at rest?"},{"Start":"00:16.470 ","End":"00:21.249","Text":"When the elevator is moving at a constant velocity of 5 meters per second?"},{"Start":"00:21.249 ","End":"00:23.345","Text":"When the elevator is accelerating"},{"Start":"00:23.345 ","End":"00:27.455","Text":"upwards with an acceleration of 5 meters per second squared?"},{"Start":"00:27.455 ","End":"00:34.410","Text":"When the elevator is accelerating downwards with an acceleration of 5 meters squared?"},{"Start":"00:35.420 ","End":"00:39.935","Text":"The first thing that we have to realize is that,"},{"Start":"00:39.935 ","End":"00:41.855","Text":"if this is our scale,"},{"Start":"00:41.855 ","End":"00:43.805","Text":"from the man standing on the scale,"},{"Start":"00:43.805 ","End":"00:46.520","Text":"we have 2 forces present."},{"Start":"00:46.520 ","End":"00:53.150","Text":"We have our normal force acting upwards and in the equal and opposite direction,"},{"Start":"00:53.150 ","End":"00:57.600","Text":"we have mass times gravity."},{"Start":"00:59.000 ","End":"01:01.885","Text":"In question number 1,"},{"Start":"01:01.885 ","End":"01:05.300","Text":"we\u0027re being told that the elevator is at rest."},{"Start":"01:05.300 ","End":"01:08.720","Text":"Now what we\u0027re going to see is that the normal force is"},{"Start":"01:08.720 ","End":"01:13.240","Text":"the value that the scale in the elevator is going to show."},{"Start":"01:13.240 ","End":"01:15.425","Text":"Let\u0027s see what this means."},{"Start":"01:15.425 ","End":"01:17.840","Text":"If the elevator is at rest,"},{"Start":"01:17.840 ","End":"01:22.700","Text":"then that means that the sum of all of the forces is equal to 0,"},{"Start":"01:22.700 ","End":"01:24.955","Text":"because we have no acceleration."},{"Start":"01:24.955 ","End":"01:27.015","Text":"The elevator is at rest,"},{"Start":"01:27.015 ","End":"01:29.690","Text":"that means the acceleration is equal to 0,"},{"Start":"01:29.690 ","End":"01:31.850","Text":"so that means that the sum of all of"},{"Start":"01:31.850 ","End":"01:39.260","Text":"our forces if we say that this is our positive direction,"},{"Start":"01:39.260 ","End":"01:45.920","Text":"which is equal to N-mg is equal to mass times acceleration,"},{"Start":"01:45.920 ","End":"01:47.710","Text":"which acceleration is equal to 0."},{"Start":"01:47.710 ","End":"01:53.555","Text":"The sum of all of our forces is equal to 0 and that means that our normal is"},{"Start":"01:53.555 ","End":"02:00.920","Text":"equal to mg. Our normal is what the scale in the elevator is going to show."},{"Start":"02:00.920 ","End":"02:03.785","Text":"What is mg equal to?"},{"Start":"02:03.785 ","End":"02:10.755","Text":"Our mass is 70 kilograms multiplied by our gravity,"},{"Start":"02:10.755 ","End":"02:13.110","Text":"our acceleration due to gravity."},{"Start":"02:13.110 ","End":"02:16.175","Text":"Let\u0027s just round it up to 10 for ease of calculation,"},{"Start":"02:16.175 ","End":"02:19.535","Text":"you can also use 9.8 meters per second squared,"},{"Start":"02:19.535 ","End":"02:22.880","Text":"but we\u0027ll just use it, for now is 10."},{"Start":"02:22.880 ","End":"02:26.030","Text":"Our normal force is going to show"},{"Start":"02:26.030 ","End":"02:32.790","Text":"700 Newtons and this is what our scale is going to show."},{"Start":"02:33.560 ","End":"02:37.270","Text":"Now let\u0027s answer question number 2,"},{"Start":"02:37.270 ","End":"02:43.685","Text":"which is now the elevator is moving at a constant velocity of 5 meters per second."},{"Start":"02:43.685 ","End":"02:47.090","Text":"Again, what is going to be shown on the scale?"},{"Start":"02:47.090 ","End":"02:50.870","Text":"Now a special note, this N over here denotes the normal force,"},{"Start":"02:50.870 ","End":"02:54.260","Text":"and this N over here denotes the unit of force,"},{"Start":"02:54.260 ","End":"02:57.190","Text":"the Newton. Don\u0027t get confused."},{"Start":"02:57.190 ","End":"03:01.805","Text":"Here again, if the elevator is moving at a constant velocity."},{"Start":"03:01.805 ","End":"03:07.760","Text":"A constant velocity means again that our acceleration is equal to 0."},{"Start":"03:07.760 ","End":"03:14.600","Text":"In physics, if some object is at rest or moving at a constant velocity,"},{"Start":"03:14.600 ","End":"03:21.375","Text":"there\u0027s a lot of similarities and not much difference between these 2 physical values."},{"Start":"03:21.375 ","End":"03:23.405","Text":"If the elevator is at rest,"},{"Start":"03:23.405 ","End":"03:27.270","Text":"that\u0027s like saying it\u0027s moving at 0 meters per second."},{"Start":"03:27.800 ","End":"03:31.390","Text":"This is just 5 meters per second."},{"Start":"03:31.390 ","End":"03:33.860","Text":"But either way, the acceleration,"},{"Start":"03:33.860 ","End":"03:36.590","Text":"if we\u0027re at a constant velocity of 0 meters per second,"},{"Start":"03:36.590 ","End":"03:38.000","Text":"or 5 meters per second,"},{"Start":"03:38.000 ","End":"03:40.460","Text":"the acceleration is still equal to 0."},{"Start":"03:40.460 ","End":"03:46.620","Text":"Therefore, we get the same equation over here."},{"Start":"03:46.620 ","End":"03:48.755","Text":"Once we rearrange it,"},{"Start":"03:48.755 ","End":"03:50.435","Text":"we get that again,"},{"Start":"03:50.435 ","End":"03:53.374","Text":"our normal force is equal to mg,"},{"Start":"03:53.374 ","End":"03:56.510","Text":"which is equal to 70 kilograms,"},{"Start":"03:56.510 ","End":"03:59.975","Text":"multiplied by 10 meters per second squared."},{"Start":"03:59.975 ","End":"04:05.120","Text":"Again, we get that the scale is going to show this normal force,"},{"Start":"04:05.120 ","End":"04:09.070","Text":"which is equal to 700 Newtons."},{"Start":"04:09.070 ","End":"04:12.855","Text":"Now let\u0027s answer question number 3."},{"Start":"04:12.855 ","End":"04:16.955","Text":"Question number 3, we\u0027re being told that the elevator is now accelerating"},{"Start":"04:16.955 ","End":"04:21.700","Text":"upwards with an acceleration of 5 meters per second squared."},{"Start":"04:21.700 ","End":"04:25.280","Text":"Here we have our reference frame,"},{"Start":"04:25.280 ","End":"04:30.760","Text":"which is the elevator and we\u0027re told that the elevator has an acceleration."},{"Start":"04:30.760 ","End":"04:36.920","Text":"Whenever you see that there\u0027s some acceleration of a reference frame,"},{"Start":"04:36.920 ","End":"04:39.830","Text":"that means that you have to use fictitious forces."},{"Start":"04:39.830 ","End":"04:44.540","Text":"You can no longer use the normal force equations."},{"Start":"04:44.540 ","End":"04:49.235","Text":"Again you have to add in the fictitious forces in order for us to"},{"Start":"04:49.235 ","End":"04:54.170","Text":"complete the calculation. What does that mean?"},{"Start":"04:54.170 ","End":"05:02.805","Text":"That means that we have to multiply our acceleration by the mass of the object."},{"Start":"05:02.805 ","End":"05:06.375","Text":"Here it\u0027s the mass of the man, so it\u0027s ma."},{"Start":"05:06.375 ","End":"05:10.810","Text":"Then here we\u0027re being told that the acceleration is upwards."},{"Start":"05:10.810 ","End":"05:13.235","Text":"When we\u0027re adding in our fictitious forces,"},{"Start":"05:13.235 ","End":"05:16.625","Text":"we have to add in an equal and opposite force."},{"Start":"05:16.625 ","End":"05:19.730","Text":"Here, if we are accelerating upward,"},{"Start":"05:19.730 ","End":"05:24.440","Text":"our force is mass times acceleration upwards."},{"Start":"05:24.440 ","End":"05:29.375","Text":"Our fictitious force has to be mass times acceleration downwards."},{"Start":"05:29.375 ","End":"05:32.785","Text":"Let\u0027s draw that over here in green."},{"Start":"05:32.785 ","End":"05:35.045","Text":"Here from the same point,"},{"Start":"05:35.045 ","End":"05:40.610","Text":"we\u0027re going to have our mass times our acceleration,"},{"Start":"05:40.610 ","End":"05:48.160","Text":"but in the opposite direction to the reference frames acceleration."},{"Start":"05:49.520 ","End":"05:53.570","Text":"This in green is our fictitious force which is"},{"Start":"05:53.570 ","End":"05:57.125","Text":"in the opposite direction to the acceleration."},{"Start":"05:57.125 ","End":"05:59.135","Text":"Here the acceleration is upwards,"},{"Start":"05:59.135 ","End":"06:02.395","Text":"so our fictitious force is downwards."},{"Start":"06:02.395 ","End":"06:07.335","Text":"Now we can work out the sum of all the forces."},{"Start":"06:07.335 ","End":"06:13.340","Text":"The sum of all of our forces is equal to our normal force which is pointing in"},{"Start":"06:13.340 ","End":"06:21.880","Text":"the positive direction minus our mg minus our ma."},{"Start":"06:21.880 ","End":"06:24.600","Text":"Our normal force, we have to find,"},{"Start":"06:24.600 ","End":"06:28.380","Text":"so we have N minus our mg,"},{"Start":"06:28.380 ","End":"06:31.335","Text":"which we saw was equal to 700."},{"Start":"06:31.335 ","End":"06:38.510","Text":"Mass times gravity where we used our acceleration due to gravity as 10 minus our mass,"},{"Start":"06:38.510 ","End":"06:41.120","Text":"which is 70 times our acceleration,"},{"Start":"06:41.120 ","End":"06:45.440","Text":"which is 5, so minus 350."},{"Start":"06:45.440 ","End":"06:50.790","Text":"All of this is equal to 0."},{"Start":"06:51.740 ","End":"06:54.140","Text":"Once we rearrange this,"},{"Start":"06:54.140 ","End":"06:55.775","Text":"we\u0027ll get that our normal force,"},{"Start":"06:55.775 ","End":"06:57.830","Text":"which is what the scale will show,"},{"Start":"06:57.830 ","End":"07:03.355","Text":"is equal to 1,050 Newtons."},{"Start":"07:03.355 ","End":"07:06.860","Text":"Now, how come I said that the sum of all of my forces is"},{"Start":"07:06.860 ","End":"07:10.915","Text":"equal to mass times acceleration which is equal to 0?"},{"Start":"07:10.915 ","End":"07:12.780","Text":"Why did I put a 0 here?"},{"Start":"07:12.780 ","End":"07:17.260","Text":"Let\u0027s write that this is equal to our mass times our acceleration."},{"Start":"07:17.260 ","End":"07:22.120","Text":"We have an acceleration of 5 meters per second squared."},{"Start":"07:22.120 ","End":"07:27.290","Text":"However, I can say now that my acceleration is equal to"},{"Start":"07:27.290 ","End":"07:32.315","Text":"0 because I substitute it in my fictitious force."},{"Start":"07:32.315 ","End":"07:37.250","Text":"My fictitious force means that I can cancel out this acceleration over"},{"Start":"07:37.250 ","End":"07:43.355","Text":"here and say therefore that now I can consider my reference frame as inertial,"},{"Start":"07:43.355 ","End":"07:46.250","Text":"ie, my reference frame isn\u0027t accelerating,"},{"Start":"07:46.250 ","End":"07:50.165","Text":"because I added this equal and opposite force."},{"Start":"07:50.165 ","End":"07:52.670","Text":"This is my fictitious force."},{"Start":"07:52.670 ","End":"07:59.550","Text":"Therefore, I could say that this is equal to 0 and then come to this answer."},{"Start":"08:00.640 ","End":"08:04.715","Text":"What\u0027s important to remember here is that when you have"},{"Start":"08:04.715 ","End":"08:09.210","Text":"a reference frame which is not inertial, ie,"},{"Start":"08:09.210 ","End":"08:12.200","Text":"it\u0027s accelerating, then you work out the sum of"},{"Start":"08:12.200 ","End":"08:17.735","Text":"all the forces that you have naturally and then you add in the fictitious force."},{"Start":"08:17.735 ","End":"08:20.750","Text":"Here, it was a minus just because we decided"},{"Start":"08:20.750 ","End":"08:24.645","Text":"that this upwards direction is the positive direction."},{"Start":"08:24.645 ","End":"08:28.790","Text":"Here we subtracted our fictitious force, but it\u0027s the same thing."},{"Start":"08:28.790 ","End":"08:31.640","Text":"It\u0027s just how we defined our axes."},{"Start":"08:31.640 ","End":"08:35.930","Text":"Then you subtract or add the fictitious force."},{"Start":"08:35.930 ","End":"08:41.435","Text":"Then that means that you can consider the elevator or the reference frame as inertial."},{"Start":"08:41.435 ","End":"08:42.860","Text":"Once again, ie,"},{"Start":"08:42.860 ","End":"08:47.225","Text":"it\u0027s not accelerating anymore and that\u0027s why you could say that this,"},{"Start":"08:47.225 ","End":"08:50.240","Text":"the sum of all the forces is again equal to 0,"},{"Start":"08:50.240 ","End":"08:52.690","Text":"as we did in questions 1 and 2."},{"Start":"08:52.690 ","End":"08:55.675","Text":"Now let\u0027s answer question number 4."},{"Start":"08:55.675 ","End":"08:59.285","Text":"This time the elevator is accelerating downwards"},{"Start":"08:59.285 ","End":"09:03.700","Text":"with an acceleration of 5 meters per second squared."},{"Start":"09:03.700 ","End":"09:05.630","Text":"Again, what we\u0027re going to do,"},{"Start":"09:05.630 ","End":"09:09.420","Text":"so we\u0027re going to add in our fictitious force."},{"Start":"09:09.730 ","End":"09:15.185","Text":"Our fictitious force is simply the mass of our man over here"},{"Start":"09:15.185 ","End":"09:20.665","Text":"multiplied by the acceleration of the non-inertial reference frame."},{"Start":"09:20.665 ","End":"09:23.195","Text":"Which we know is 5 meters per second squared,"},{"Start":"09:23.195 ","End":"09:26.940","Text":"but we\u0027re told that it\u0027s accelerating downwards."},{"Start":"09:28.100 ","End":"09:30.645","Text":"If our acceleration is downwards,"},{"Start":"09:30.645 ","End":"09:34.875","Text":"our fictitious force has to be the opposite, so upwards."},{"Start":"09:34.875 ","End":"09:39.260","Text":"This time, our fictitious force is up in this direction and again,"},{"Start":"09:39.260 ","End":"09:43.390","Text":"it\u0027s equal to the mass times the acceleration."},{"Start":"09:44.420 ","End":"09:50.020","Text":"Now we can say that the sum of all of our forces is equal to,"},{"Start":"09:50.020 ","End":"09:52.970","Text":"so in the positive direction as we defined it,"},{"Start":"09:52.970 ","End":"09:56.930","Text":"we have our normal force plus this time,"},{"Start":"09:56.930 ","End":"10:00.050","Text":"our fictitious force, mass times acceleration."},{"Start":"10:00.050 ","End":"10:04.610","Text":"It\u0027s now in the positive direction because it\u0027s in"},{"Start":"10:04.610 ","End":"10:11.375","Text":"the opposite direction to the direction of acceleration of the non-inertial frame."},{"Start":"10:11.375 ","End":"10:13.640","Text":"Then in our negative direction,"},{"Start":"10:13.640 ","End":"10:18.870","Text":"we have mg. Now we\u0027re trying to find the normal."},{"Start":"10:18.870 ","End":"10:24.365","Text":"We have that the normal plus our mass times acceleration,"},{"Start":"10:24.365 ","End":"10:27.545","Text":"which is a mass of 70 kilograms,"},{"Start":"10:27.545 ","End":"10:31.280","Text":"multiplied by an acceleration of 5 meters per second squared,"},{"Start":"10:31.280 ","End":"10:38.430","Text":"so it\u0027s 350 Newtons minus mg,"},{"Start":"10:38.430 ","End":"10:39.900","Text":"which is 70 times,"},{"Start":"10:39.900 ","End":"10:43.140","Text":"we said that g is equal to 10 here,"},{"Start":"10:43.140 ","End":"10:46.540","Text":"so again, it\u0027s minus 700."},{"Start":"10:46.760 ","End":"10:49.085","Text":"Then all of this is,"},{"Start":"10:49.085 ","End":"10:50.720","Text":"again, for the same reason,"},{"Start":"10:50.720 ","End":"10:53.555","Text":"is question 3 equal to 0,"},{"Start":"10:53.555 ","End":"10:56.435","Text":"because we\u0027ve added in our fictitious force,"},{"Start":"10:56.435 ","End":"11:00.650","Text":"we can consider the elevator to be not accelerating,"},{"Start":"11:00.650 ","End":"11:03.350","Text":"to be traveling at some constant velocity,"},{"Start":"11:03.350 ","End":"11:07.410","Text":"which means that the sum of all of the forces is equal to 0."},{"Start":"11:08.120 ","End":"11:11.555","Text":"Now, once we rearrange everything,"},{"Start":"11:11.555 ","End":"11:16.760","Text":"we\u0027ll get that our normal force or the value that the scale will show is"},{"Start":"11:16.760 ","End":"11:22.960","Text":"going to be equal to 350 Newtons."},{"Start":"11:24.350 ","End":"11:26.570","Text":"That\u0027s the end of this question."},{"Start":"11:26.570 ","End":"11:30.440","Text":"Quickly though, before we finish this lesson,"},{"Start":"11:30.440 ","End":"11:35.750","Text":"let\u0027s imagine that the elevator was accelerating downwards with an acceleration of"},{"Start":"11:35.750 ","End":"11:41.955","Text":"10 meters per second squared over the value of g. Our acceleration due to gravity."},{"Start":"11:41.955 ","End":"11:45.195","Text":"In that case, if here we have our scale,"},{"Start":"11:45.195 ","End":"11:47.460","Text":"the forces acting would be,"},{"Start":"11:47.460 ","End":"11:49.910","Text":"again we would have our normal force."},{"Start":"11:49.910 ","End":"11:53.930","Text":"Then if it\u0027s accelerating downwards with an acceleration of g,"},{"Start":"11:53.930 ","End":"11:58.010","Text":"so our fictitious force would be mass times the acceleration,"},{"Start":"11:58.010 ","End":"12:02.015","Text":"which here is g, and then of course we have our force"},{"Start":"12:02.015 ","End":"12:07.110","Text":"due to gravity of mg pointing downwards."},{"Start":"12:07.110 ","End":"12:12.390","Text":"Now we can see that we\u0027ve taken into account our fictitious force."},{"Start":"12:12.430 ","End":"12:16.685","Text":"The sum of all of our forces would be equal to"},{"Start":"12:16.685 ","End":"12:21.015","Text":"our normal force plus our fictitious force mg,"},{"Start":"12:21.015 ","End":"12:25.620","Text":"minus our force due to gravity mg and this would,"},{"Start":"12:25.620 ","End":"12:28.230","Text":"of course, all equal to 0."},{"Start":"12:28.230 ","End":"12:30.980","Text":"Because we\u0027ve taken into account our fictitious force,"},{"Start":"12:30.980 ","End":"12:34.165","Text":"which means that our acceleration is now equal to 0."},{"Start":"12:34.165 ","End":"12:38.855","Text":"Now we can see that mg-mg is of course 0."},{"Start":"12:38.855 ","End":"12:42.440","Text":"We\u0027d get that our normal force would be equal to 0,"},{"Start":"12:42.440 ","End":"12:46.625","Text":"ie, our scale will show that our man in the elevator,"},{"Start":"12:46.625 ","End":"12:51.400","Text":"which is falling down to earth with an acceleration due to gravity,"},{"Start":"12:51.400 ","End":"12:54.585","Text":"will not be standing on the scale."},{"Start":"12:54.585 ","End":"12:58.025","Text":"The man will be floating inside the elevator,"},{"Start":"12:58.025 ","End":"13:02.010","Text":"which is exactly what we would\u0027ve expected to see."},{"Start":"13:02.260 ","End":"13:05.790","Text":"That\u0027s the end of this lesson."}],"ID":10844},{"Watched":false,"Name":"Exercise - Triangular Car","Duration":"28m 55s","ChapterTopicVideoID":10415,"CourseChapterTopicPlaylistID":5351,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"Hello. In this question,"},{"Start":"00:02.280 ","End":"00:07.635","Text":"we\u0027re given a triangular car which has an angle Theta at its front."},{"Start":"00:07.635 ","End":"00:14.085","Text":"On the car, a mass M is placed and we\u0027re being told that between the car and the mass,"},{"Start":"00:14.085 ","End":"00:15.975","Text":"there is a frictional force."},{"Start":"00:15.975 ","End":"00:22.560","Text":"We\u0027re told that sine of the angle Theta is equal to 0.6 and the coefficient of"},{"Start":"00:22.560 ","End":"00:25.650","Text":"kinetic friction is equal to the coefficient of"},{"Start":"00:25.650 ","End":"00:30.540","Text":"static friction and that that is equal to 0.2."},{"Start":"00:30.540 ","End":"00:32.295","Text":"The first question is,"},{"Start":"00:32.295 ","End":"00:35.445","Text":"what is the condition on the acceleration,"},{"Start":"00:35.445 ","End":"00:43.225","Text":"a, the acceleration of the car such that our mass M won\u0027t slip downwards?"},{"Start":"00:43.225 ","End":"00:47.660","Text":"The first thing we\u0027re being told that our coefficient of"},{"Start":"00:47.660 ","End":"00:52.180","Text":"static friction is equal to our coefficient of kinetic friction,"},{"Start":"00:52.180 ","End":"00:55.115","Text":"that, that is equal to 0.2."},{"Start":"00:55.115 ","End":"01:00.910","Text":"Let\u0027s just call this Mu coefficient of friction and"},{"Start":"01:00.910 ","End":"01:07.055","Text":"we\u0027re told that our car over here is accelerating with some acceleration a,"},{"Start":"01:07.055 ","End":"01:09.260","Text":"which is what we want to find out."},{"Start":"01:09.260 ","End":"01:13.055","Text":"We want to see that our mass M doesn\u0027t slip downwards."},{"Start":"01:13.055 ","End":"01:15.920","Text":"Now we\u0027re not going to speak about the case where a car is"},{"Start":"01:15.920 ","End":"01:20.030","Text":"accelerating such that our mass slips upwards."},{"Start":"01:20.030 ","End":"01:21.830","Text":"If you want, you can do that on your own,"},{"Start":"01:21.830 ","End":"01:27.020","Text":"but we\u0027re specifically dealing with stopping the mass from slipping downwards."},{"Start":"01:27.020 ","End":"01:29.960","Text":"Now, this question can be solved either in"},{"Start":"01:29.960 ","End":"01:34.235","Text":"the labs frame of reference or in the car\u0027s frame of reference."},{"Start":"01:34.235 ","End":"01:38.840","Text":"We\u0027re going to solve it in the car\u0027s frame of reference because it\u0027s slightly less"},{"Start":"01:38.840 ","End":"01:43.760","Text":"complicated as we will see in the later questions."},{"Start":"01:43.760 ","End":"01:46.474","Text":"If I\u0027m in the car\u0027s frame of reference,"},{"Start":"01:46.474 ","End":"01:53.500","Text":"that means that I can imagine that I\u0027m standing in the car and I\u0027m moving with it."},{"Start":"01:53.500 ","End":"01:55.635","Text":"If I look around me,"},{"Start":"01:55.635 ","End":"01:57.479","Text":"I\u0027m not accelerating,"},{"Start":"01:57.479 ","End":"02:01.325","Text":"and we want that our mass over here at the top,"},{"Start":"02:01.325 ","End":"02:04.905","Text":"that it won\u0027t move downwards."},{"Start":"02:04.905 ","End":"02:09.590","Text":"What does that mean? That means that the acceleration of the mass on top,"},{"Start":"02:09.590 ","End":"02:15.085","Text":"so let\u0027s call that a tag is going to be equal to 0."},{"Start":"02:15.085 ","End":"02:18.500","Text":"That means it\u0027s moving at"},{"Start":"02:18.500 ","End":"02:24.490","Text":"a constant velocity or it\u0027s not accelerating or its velocity is equal to 0."},{"Start":"02:24.490 ","End":"02:28.790","Text":"If my acceleration of the mass is equal to 0,"},{"Start":"02:28.790 ","End":"02:32.555","Text":"so that means that the sum of all of my forces acting"},{"Start":"02:32.555 ","End":"02:36.830","Text":"on this mass must also be equal to 0."},{"Start":"02:36.830 ","End":"02:43.400","Text":"Now what we\u0027re going to do is we\u0027re going to draw a free body diagram for this mass."},{"Start":"02:43.400 ","End":"02:46.925","Text":"Now we can choose our axes to be however we want."},{"Start":"02:46.925 ","End":"02:54.920","Text":"I\u0027m specifically going to choose them to be according to this slope over here."},{"Start":"02:54.920 ","End":"02:57.830","Text":"Let\u0027s draw the free body diagram."},{"Start":"02:57.830 ","End":"02:59.195","Text":"So first of all,"},{"Start":"02:59.195 ","End":"03:06.949","Text":"we have our Mg pointing downwards and then we have our normal force,"},{"Start":"03:06.949 ","End":"03:12.605","Text":"which has obviously normal or perpendicular to our surface,"},{"Start":"03:12.605 ","End":"03:15.620","Text":"where this is the y-axis and this is the x-axis."},{"Start":"03:15.620 ","End":"03:18.980","Text":"Then we have our frictional force."},{"Start":"03:18.980 ","End":"03:22.550","Text":"We have some static frictional force pointing in"},{"Start":"03:22.550 ","End":"03:27.529","Text":"this direction and now we have one last fourth force,"},{"Start":"03:27.529 ","End":"03:30.185","Text":"and that is our fictitious force."},{"Start":"03:30.185 ","End":"03:35.675","Text":"If we have our car traveling in this direction with some acceleration a,"},{"Start":"03:35.675 ","End":"03:37.880","Text":"so that means that we\u0027re going to have"},{"Start":"03:37.880 ","End":"03:46.560","Text":"some fictitious force pointing in the opposite direction with magnitude Ma."},{"Start":"03:46.900 ","End":"03:50.315","Text":"Now we\u0027ve drawn our free body diagram,"},{"Start":"03:50.315 ","End":"03:54.350","Text":"let\u0027s work out the sum of all of the forces."},{"Start":"03:54.350 ","End":"03:58.055","Text":"Again, we\u0027re going to break this up into the different axis."},{"Start":"03:58.055 ","End":"04:03.760","Text":"Let\u0027s work out first the sum of all of the forces on the y-axis."},{"Start":"04:03.760 ","End":"04:05.450","Text":"In the y direction,"},{"Start":"04:05.450 ","End":"04:09.335","Text":"we have our normal force in the positive y direction."},{"Start":"04:09.335 ","End":"04:12.200","Text":"Then in the negative y direction,"},{"Start":"04:12.200 ","End":"04:15.455","Text":"we have our fictitious force."},{"Start":"04:15.455 ","End":"04:18.575","Text":"Now, if we look at the geometry of the problem,"},{"Start":"04:18.575 ","End":"04:22.670","Text":"we can see that our fictitious force is going to be first of all that\u0027s in"},{"Start":"04:22.670 ","End":"04:27.395","Text":"the negative y direction, so negative Ma."},{"Start":"04:27.395 ","End":"04:34.235","Text":"Then it\u0027s going to be multiplied by sine of our angle Theta."},{"Start":"04:34.235 ","End":"04:38.015","Text":"Then we have also our Mg,"},{"Start":"04:38.015 ","End":"04:39.560","Text":"which has some component in"},{"Start":"04:39.560 ","End":"04:43.730","Text":"the negative y direction so that\u0027s going to be equal to negative Mg"},{"Start":"04:43.730 ","End":"04:52.830","Text":"and now this time it\u0027s going to be equal to cosine of Theta because as we can see,"},{"Start":"04:52.830 ","End":"05:00.110","Text":"our Mg vector is perpendicular to our fictitious force Ma vector."},{"Start":"05:00.110 ","End":"05:07.205","Text":"As we know, cosine of an angle is perpendicular to sine of the same angle."},{"Start":"05:07.205 ","End":"05:09.890","Text":"Then of course we know that this is all equal to"},{"Start":"05:09.890 ","End":"05:13.130","Text":"0 because we know that the sum of all of the forces in"},{"Start":"05:13.130 ","End":"05:20.100","Text":"the y direction is equal to 0 and that\u0027s because we know that we have no acceleration."},{"Start":"05:20.260 ","End":"05:28.075","Text":"Now let\u0027s write out the sum of all of our forces in the x direction."},{"Start":"05:28.075 ","End":"05:34.860","Text":"Now we\u0027re going to take our component for our Mg in the x direction,"},{"Start":"05:34.860 ","End":"05:39.500","Text":"so that is going to be festival in the positive x direction."},{"Start":"05:39.500 ","End":"05:46.115","Text":"So that\u0027s Mg multiplied by sine of Theta."},{"Start":"05:46.115 ","End":"05:55.190","Text":"Then we have negative our Ma and it\u0027s again perpendicular to our Mg so now it\u0027s going to"},{"Start":"05:55.190 ","End":"06:04.324","Text":"be cosine of Theta and then minus our frictional force,"},{"Start":"06:04.324 ","End":"06:08.299","Text":"which is going in the negative x direction so that\u0027s why it\u0027s a minus."},{"Start":"06:08.299 ","End":"06:13.895","Text":"So negative fs and that is equal to 0 because again,"},{"Start":"06:13.895 ","End":"06:18.530","Text":"we\u0027re looking for an acceleration 0 so that means that the sum"},{"Start":"06:18.530 ","End":"06:24.150","Text":"of all of the forces also in this direction must be equal to 0."},{"Start":"06:25.040 ","End":"06:28.820","Text":"Now we\u0027ve finished with our free body diagram and also with"},{"Start":"06:28.820 ","End":"06:32.090","Text":"writing out our force equations."},{"Start":"06:32.090 ","End":"06:38.840","Text":"Now what we need is to write the condition that has to take place"},{"Start":"06:38.840 ","End":"06:45.785","Text":"between our frictional force and our normal force such that our mass won\u0027t slip."},{"Start":"06:45.785 ","End":"06:51.934","Text":"The condition between these 2 forces is that our frictional force"},{"Start":"06:51.934 ","End":"06:58.490","Text":"must be smaller or equal to our coefficient of static friction,"},{"Start":"06:58.490 ","End":"07:01.670","Text":"which is equal to this as we spoke about before,"},{"Start":"07:01.670 ","End":"07:05.760","Text":"multiplied by our normal force."},{"Start":"07:07.160 ","End":"07:12.530","Text":"Now what I\u0027m going to do is I\u0027m going to isolate out my fs,"},{"Start":"07:12.530 ","End":"07:15.650","Text":"my frictional force from this equation over here."},{"Start":"07:15.650 ","End":"07:22.560","Text":"I\u0027ll get that my fs is equal to Mg sine"},{"Start":"07:22.560 ","End":"07:29.435","Text":"of Theta minus Ma cosine of Theta."},{"Start":"07:29.435 ","End":"07:32.824","Text":"Next, I\u0027m going to isolate"},{"Start":"07:32.824 ","End":"07:38.075","Text":"my normal force and that I can get from this equation over here."},{"Start":"07:38.075 ","End":"07:43.800","Text":"My normal force is equal to Ma sine"},{"Start":"07:43.800 ","End":"07:49.950","Text":"of Theta plus Mg cosine of Theta."},{"Start":"07:49.950 ","End":"07:52.185","Text":"Of course my Mu s,"},{"Start":"07:52.185 ","End":"07:54.585","Text":"I know the value of."},{"Start":"07:54.585 ","End":"08:00.905","Text":"Now I can plug in these values into this equation over here."},{"Start":"08:00.905 ","End":"08:06.090","Text":"I\u0027ll have that fs, which is Mg sine of"},{"Start":"08:06.090 ","End":"08:15.015","Text":"Theta minus Ma cosine of Theta is smaller or equal to Mu."},{"Start":"08:15.015 ","End":"08:20.655","Text":"We said that our Mu s is also can be called Mu multiplied by N,"},{"Start":"08:20.655 ","End":"08:25.440","Text":"which is Ma sine of"},{"Start":"08:25.440 ","End":"08:32.655","Text":"Theta plus Mg cosine of Theta."},{"Start":"08:32.655 ","End":"08:38.375","Text":"Now what we can do is we can divide both sides by the mass M,"},{"Start":"08:38.375 ","End":"08:43.670","Text":"and we can see that our condition for the acceleration is independent of"},{"Start":"08:43.670 ","End":"08:49.685","Text":"M. Now what we can do is we can get our g\u0027s,"},{"Start":"08:49.685 ","End":"08:56.155","Text":"all the terms with a on one side and all the terms with a on the other side."},{"Start":"08:56.155 ","End":"09:00.830","Text":"We\u0027re just going to rearrange this and we\u0027ll get that g multiplied"},{"Start":"09:00.830 ","End":"09:06.020","Text":"by sine of Theta minus Mu cosine of"},{"Start":"09:06.020 ","End":"09:12.215","Text":"Theta is smaller or equal to a multiplied by"},{"Start":"09:12.215 ","End":"09:19.600","Text":"Mu sine of Theta plus cosine Theta."},{"Start":"09:19.600 ","End":"09:27.205","Text":"Now what we want to do is we want to plug in our values."},{"Start":"09:27.205 ","End":"09:29.570","Text":"We know that our Mu=0.2."},{"Start":"09:33.060 ","End":"09:40.540","Text":"We also know that our sine of Theta is equal to 0.6."},{"Start":"09:40.540 ","End":"09:46.150","Text":"If we say that our sine of Theta is equal to 0.6,"},{"Start":"09:46.150 ","End":"09:49.810","Text":"we can work out what our cosine of Theta is equal to."},{"Start":"09:49.810 ","End":"09:59.455","Text":"As we know, cosine^2 Theta plus sine^2 Theta is equal to 1."},{"Start":"09:59.455 ","End":"10:01.555","Text":"That means therefore,"},{"Start":"10:01.555 ","End":"10:10.160","Text":"that cosine of Theta is equal to the square root of 1 minus sine^2 Theta,"},{"Start":"10:10.350 ","End":"10:15.340","Text":"which we know that our sine of Theta is equal to 0.6."},{"Start":"10:15.340 ","End":"10:21.737","Text":"We know that that\u0027s the square root of 1-0.6^2,"},{"Start":"10:21.737 ","End":"10:26.420","Text":"and that is equal to 0.8."},{"Start":"10:27.360 ","End":"10:32.830","Text":"Now let\u0027s scroll down a little bit to give us some more space."},{"Start":"10:32.830 ","End":"10:37.090","Text":"Now we can plug in these values into this equation over here."},{"Start":"10:37.090 ","End":"10:41.955","Text":"We\u0027ll get that g multiplied by sine of Theta,"},{"Start":"10:41.955 ","End":"10:47.730","Text":"which is equal to 0.6 minus our Mu,"},{"Start":"10:47.730 ","End":"10:55.555","Text":"which is 0.2, multiplied by cosine Theta, which is 0.8,"},{"Start":"10:55.555 ","End":"11:00.880","Text":"must be smaller or equal to a multiplied by Mu,"},{"Start":"11:00.880 ","End":"11:03.959","Text":"which is 0.2 sine of Theta,"},{"Start":"11:03.959 ","End":"11:10.240","Text":"so that\u0027s 0.6 plus cosine of Theta, which is 0.8."},{"Start":"11:10.240 ","End":"11:17.275","Text":"Then we can divide both sides by what\u0027s in this bracket."},{"Start":"11:17.275 ","End":"11:24.860","Text":"We\u0027ll get that a must be bigger or equal to 0.48g."},{"Start":"11:28.680 ","End":"11:34.195","Text":"Question 1 was to find the condition on the acceleration,"},{"Start":"11:34.195 ","End":"11:38.575","Text":"a, such that the mass M won\u0027t slip downwards."},{"Start":"11:38.575 ","End":"11:42.235","Text":"This is the answer to question number 1."},{"Start":"11:42.235 ","End":"11:46.000","Text":"The next video we\u0027ll answer question number 2."},{"Start":"11:46.000 ","End":"11:49.015","Text":"Hello. In the previous video,"},{"Start":"11:49.015 ","End":"11:51.385","Text":"we answered question number 1."},{"Start":"11:51.385 ","End":"11:54.625","Text":"Now we\u0027re going to answer question number 2."},{"Start":"11:54.625 ","End":"11:58.465","Text":"We\u0027re now told that the acceleration of the car,"},{"Start":"11:58.465 ","End":"12:01.495","Text":"a, is equal to 0.2."},{"Start":"12:01.495 ","End":"12:04.870","Text":"So a is equal to 0.2."},{"Start":"12:04.870 ","End":"12:11.590","Text":"We can see that our acceleration of the car is smaller than this critical acceleration,"},{"Start":"12:11.590 ","End":"12:18.850","Text":"such that our mass placed on top of the car wouldn\u0027t slide downwards."},{"Start":"12:18.850 ","End":"12:22.360","Text":"We can see that in this question,"},{"Start":"12:22.360 ","End":"12:25.285","Text":"our mass is in fact going to slide downwards."},{"Start":"12:25.285 ","End":"12:29.140","Text":"We\u0027re being told to calculate the acceleration of the mass"},{"Start":"12:29.140 ","End":"12:34.340","Text":"itself in the car\u0027s frame of reference."},{"Start":"12:34.350 ","End":"12:36.562","Text":"Sorry, my mistake,"},{"Start":"12:36.562 ","End":"12:40.940","Text":"the acceleration is 0.2g. Corrected it."},{"Start":"12:42.510 ","End":"12:46.045","Text":"What does this mean for us?"},{"Start":"12:46.045 ","End":"12:51.025","Text":"Now we know that our mass is going to be accelerating."},{"Start":"12:51.025 ","End":"12:56.125","Text":"We can cross out that our a tag is equal to 0."},{"Start":"12:56.125 ","End":"13:03.370","Text":"Now our a tag is going to have some x-component because it\u0027s slipping downwards."},{"Start":"13:03.370 ","End":"13:07.795","Text":"We said that that is our positive x direction."},{"Start":"13:07.795 ","End":"13:10.435","Text":"If our mass is slipping downwards,"},{"Start":"13:10.435 ","End":"13:13.420","Text":"that means that our coefficient of friction,"},{"Start":"13:13.420 ","End":"13:14.830","Text":"or our frictional force,"},{"Start":"13:14.830 ","End":"13:21.985","Text":"is now going to be a kinetic frictional force rather than a static frictional force."},{"Start":"13:21.985 ","End":"13:28.510","Text":"That also means that the sum of all of our forces is not going to be equal to 0,"},{"Start":"13:28.510 ","End":"13:37.045","Text":"but it\u0027s going to be equal to the mass multiplied by the x-component of its acceleration."},{"Start":"13:37.045 ","End":"13:41.980","Text":"Now notice there\u0027s a tag here because this is relative acceleration."},{"Start":"13:41.980 ","End":"13:48.520","Text":"This is the acceleration relative to the viewer in the car."},{"Start":"13:50.010 ","End":"13:58.120","Text":"In this question, what we\u0027re trying to work out is what a_x tag is equal to."},{"Start":"13:58.120 ","End":"13:59.440","Text":"Question number 2,"},{"Start":"13:59.440 ","End":"14:02.360","Text":"we\u0027re trying to work this out."},{"Start":"14:03.660 ","End":"14:06.925","Text":"As we said, our mass is moving,"},{"Start":"14:06.925 ","End":"14:14.080","Text":"which means that we\u0027re going to be using our kinetic friction."},{"Start":"14:14.080 ","End":"14:17.830","Text":"Our f_k is equal to"},{"Start":"14:17.830 ","End":"14:25.960","Text":"our coefficient or kinetic coefficient of friction multiplied by the normal force."},{"Start":"14:26.160 ","End":"14:31.750","Text":"Our Mu_k is still the same, it\u0027s 0.2."},{"Start":"14:31.750 ","End":"14:35.250","Text":"Now all we have to do is plug this in,"},{"Start":"14:35.250 ","End":"14:38.530","Text":"isolate out over here our f_k,"},{"Start":"14:38.530 ","End":"14:43.555","Text":"and solve the equations like we did in the last question."},{"Start":"14:43.555 ","End":"14:49.150","Text":"What I\u0027ve done over here is just written out the algebra really quickly,"},{"Start":"14:49.150 ","End":"14:52.045","Text":"and I solved it just like we did for question number 1."},{"Start":"14:52.045 ","End":"14:54.858","Text":"I isolated out my f_k from here,"},{"Start":"14:54.858 ","End":"14:58.370","Text":"and substituted it to this side of the equation."},{"Start":"14:58.370 ","End":"15:01.885","Text":"My Mu_k is equal to 0.2."},{"Start":"15:01.885 ","End":"15:05.035","Text":"Then I multiplied all of that by my normal,"},{"Start":"15:05.035 ","End":"15:09.024","Text":"which I just isolated out from this equation,"},{"Start":"15:09.024 ","End":"15:10.540","Text":"plug that in over here."},{"Start":"15:10.540 ","End":"15:13.720","Text":"Similar situation to what went on over here."},{"Start":"15:13.720 ","End":"15:17.830","Text":"Then I simply canceled out like terms,"},{"Start":"15:17.830 ","End":"15:21.280","Text":"and went according to this,"},{"Start":"15:21.280 ","End":"15:25.165","Text":"but according to this equation over here for kinetic friction,"},{"Start":"15:25.165 ","End":"15:29.469","Text":"rather than this equation over here for static friction."},{"Start":"15:29.469 ","End":"15:33.370","Text":"Then I plugged in all of my variables,"},{"Start":"15:33.370 ","End":"15:36.220","Text":"the same variables that we have here,"},{"Start":"15:36.220 ","End":"15:38.260","Text":"where sine of Theta is 0.6,"},{"Start":"15:38.260 ","End":"15:40.180","Text":"and cosine of Theta is 0.8,"},{"Start":"15:40.180 ","End":"15:42.235","Text":"and my Mu is 0.2,"},{"Start":"15:42.235 ","End":"15:50.365","Text":"and I got that my a_x tag is equal to 0.256g."},{"Start":"15:50.365 ","End":"15:53.455","Text":"This is the answer to Question 2."},{"Start":"15:53.455 ","End":"15:54.535","Text":"In the next video,"},{"Start":"15:54.535 ","End":"15:57.550","Text":"we\u0027re going to be answering Question 3."},{"Start":"15:57.550 ","End":"16:00.115","Text":"Hello. In the previous video,"},{"Start":"16:00.115 ","End":"16:01.960","Text":"we answered question number 2,"},{"Start":"16:01.960 ","End":"16:04.555","Text":"where this was our final answer."},{"Start":"16:04.555 ","End":"16:07.930","Text":"Now we\u0027re going to answer question number 3."},{"Start":"16:07.930 ","End":"16:11.530","Text":"Now we\u0027re being told to calculate the acceleration of the mass,"},{"Start":"16:11.530 ","End":"16:16.640","Text":"but this time in the lab\u0027s frame of reference."},{"Start":"16:16.710 ","End":"16:22.855","Text":"We\u0027ll see that now I am able to answer this question."},{"Start":"16:22.855 ","End":"16:24.670","Text":"However, earlier if I would have"},{"Start":"16:24.670 ","End":"16:33.100","Text":"automatically or straightaway tried"},{"Start":"16:33.100 ","End":"16:37.240","Text":"to work out the acceleration of this mass in the labs frame of reference,"},{"Start":"16:37.240 ","End":"16:41.215","Text":"it would\u0027ve been a very difficult and complicated calculation"},{"Start":"16:41.215 ","End":"16:44.905","Text":"because our mass is sliding down in this way but"},{"Start":"16:44.905 ","End":"16:54.595","Text":"our car is also accelerating in this direction at 0.2g."},{"Start":"16:54.595 ","End":"17:00.490","Text":"Which means that our mass\u0027s motion is in the x direction relative to the car,"},{"Start":"17:00.490 ","End":"17:04.060","Text":"but it\u0027s also moving along in this direction."},{"Start":"17:04.060 ","End":"17:07.555","Text":"Now let\u0027s see how to solve this question."},{"Start":"17:07.555 ","End":"17:15.310","Text":"Now, the way we\u0027re going to do this is we\u0027re simply going to use a transformation"},{"Start":"17:15.310 ","End":"17:24.055","Text":"of our a_x tag in order to put it within the terms of the reference frame for the lab."},{"Start":"17:24.055 ","End":"17:31.060","Text":"As we can see, our axis over here are going in the diagonal direction."},{"Start":"17:31.060 ","End":"17:37.070","Text":"Now our lab\u0027s frame of reference uses a different set of axis system."},{"Start":"17:37.350 ","End":"17:42.655","Text":"Our transformation is going to be a little bit confusing."},{"Start":"17:42.655 ","End":"17:45.970","Text":"In our lab\u0027s frame of reference in the blue dotted line,"},{"Start":"17:45.970 ","End":"17:48.828","Text":"we have our x going this way,"},{"Start":"17:48.828 ","End":"17:54.290","Text":"and our y going in this direction."},{"Start":"17:55.440 ","End":"17:58.000","Text":"Just to simplify this,"},{"Start":"17:58.000 ","End":"18:00.220","Text":"let\u0027s call the blue X, X_N,"},{"Start":"18:00.220 ","End":"18:02.935","Text":"so X Mu,"},{"Start":"18:02.935 ","End":"18:06.980","Text":"and the blue Y, Y_N for Y Mu."},{"Start":"18:07.920 ","End":"18:17.300","Text":"Now I\u0027m trying to find what my acceleration is according to my blue axis."},{"Start":"18:17.300 ","End":"18:27.760","Text":"We know that my a_x tag was along the green line and only in the x direction."},{"Start":"18:27.760 ","End":"18:32.950","Text":"This red vector is my a_x tag."},{"Start":"18:32.950 ","End":"18:34.795","Text":"It\u0027s not very clear."},{"Start":"18:34.795 ","End":"18:41.230","Text":"A_x tag is this red arrow over here going along the green X-dotted line."},{"Start":"18:41.230 ","End":"18:43.090","Text":"Now, as we can see,"},{"Start":"18:43.090 ","End":"18:49.325","Text":"it\u0027s going to be relative to our new reference frame."},{"Start":"18:49.325 ","End":"18:51.395","Text":"That\u0027s the blue-dotted line."},{"Start":"18:51.395 ","End":"18:55.345","Text":"We\u0027re going to have an X Mu component,"},{"Start":"18:55.345 ","End":"18:59.815","Text":"and a Y Mu component. Can we see that?"},{"Start":"18:59.815 ","End":"19:07.930","Text":"Now, we can see that this angle over here is equal to, of course, Theta."},{"Start":"19:07.930 ","End":"19:11.795","Text":"Now all we have to do is we have to break up"},{"Start":"19:11.795 ","End":"19:18.530","Text":"our a_x tag into the different X_N and Y_N components."},{"Start":"19:18.530 ","End":"19:25.520","Text":"I can say that my acceleration in the X_N direction,"},{"Start":"19:25.520 ","End":"19:30.490","Text":"so this is going to be equal to"},{"Start":"19:30.490 ","End":"19:37.490","Text":"our a_x tag multiplied by cosine of the angle,"},{"Start":"19:37.490 ","End":"19:46.190","Text":"which is going to be equal to 0.256g multiplied by cosine of our angle Theta,"},{"Start":"19:46.190 ","End":"19:51.665","Text":"which we worked out in previous videos was equal to 0.8."},{"Start":"19:51.665 ","End":"20:00.430","Text":"This is going to be around about equal to 0.2g to 1 decimal place."},{"Start":"20:00.430 ","End":"20:07.026","Text":"Now, similarly, our a_yN,"},{"Start":"20:07.026 ","End":"20:12.800","Text":"so our component of our a_x in our Y_N direction is"},{"Start":"20:12.800 ","End":"20:18.845","Text":"going to be equal to a_x tag multiplied by sine of the angle,"},{"Start":"20:18.845 ","End":"20:26.720","Text":"which is equal to 0.256g multiplied by 0.6."},{"Start":"20:26.720 ","End":"20:36.365","Text":"This is around about equal to 0.15g to 2 decimal places."},{"Start":"20:36.365 ","End":"20:40.710","Text":"Now, an important point is that this a_xN and"},{"Start":"20:40.710 ","End":"20:46.945","Text":"a_yN are still meant to be denoted with tags on top of them."},{"Start":"20:46.945 ","End":"20:48.295","Text":"Now why is that?"},{"Start":"20:48.295 ","End":"20:54.190","Text":"Because we still haven\u0027t done the transformation into our labs reference frame."},{"Start":"20:54.190 ","End":"20:58.615","Text":"We\u0027re still in this observer in the car\u0027s reference frame."},{"Start":"20:58.615 ","End":"21:03.445","Text":"All we\u0027ve done is we\u0027ve written out our acceleration"},{"Start":"21:03.445 ","End":"21:09.460","Text":"in the components in a Mu axis setup."},{"Start":"21:09.460 ","End":"21:13.090","Text":"That\u0027s what we\u0027ve done so far and now we\u0027re going to"},{"Start":"21:13.090 ","End":"21:17.480","Text":"transform this into our lab\u0027s frame of reference."},{"Start":"21:18.120 ","End":"21:22.090","Text":"Now we\u0027re going to do the actual transformation,"},{"Start":"21:22.090 ","End":"21:31.615","Text":"so the equation for the acceleration is that our acceleration of our mass relative to"},{"Start":"21:31.615 ","End":"21:39.715","Text":"the car is equal to the acceleration of the mass relative to"},{"Start":"21:39.715 ","End":"21:43.300","Text":"the lab\u0027s frame of reference and this is in fact what we\u0027re trying to"},{"Start":"21:43.300 ","End":"21:48.475","Text":"find minus our car\u0027s acceleration,"},{"Start":"21:48.475 ","End":"21:52.550","Text":"which is given to us in the question."},{"Start":"21:53.700 ","End":"21:59.890","Text":"Let\u0027s work out our acceleration in the x direction."},{"Start":"21:59.890 ","End":"22:03.340","Text":"That is going to be equal to the acceleration of"},{"Start":"22:03.340 ","End":"22:08.620","Text":"the mass relative to the car in the x direction so that was"},{"Start":"22:08.620 ","End":"22:17.620","Text":"equal to 0.2g plus our acceleration of the car,"},{"Start":"22:17.620 ","End":"22:22.300","Text":"which is also equal to 0.2g."},{"Start":"22:22.300 ","End":"22:30.800","Text":"Therefore, our acceleration of the mass in the x direction is equal to 0.4g."},{"Start":"22:31.080 ","End":"22:34.884","Text":"Now similarly in the y direction,"},{"Start":"22:34.884 ","End":"22:39.805","Text":"so our a_y is going to be equal to"},{"Start":"22:39.805 ","End":"22:47.950","Text":"our acceleration of our mass relative to the car\u0027s frame of reference,"},{"Start":"22:47.950 ","End":"22:49.300","Text":"which we worked out here,"},{"Start":"22:49.300 ","End":"22:54.025","Text":"was equal to 0.15g plus,"},{"Start":"22:54.025 ","End":"22:56.200","Text":"so I\u0027m moving this acceleration of"},{"Start":"22:56.200 ","End":"22:59.140","Text":"the car over to this side of the equation like we did over"},{"Start":"22:59.140 ","End":"23:05.995","Text":"here but my acceleration of the car in the y direction is equal to 0, so plus 0."},{"Start":"23:05.995 ","End":"23:12.130","Text":"We can see that my acceleration of the mass in the y direction relative to"},{"Start":"23:12.130 ","End":"23:19.130","Text":"my lab\u0027s frame of reference is simply going to be equal to 0.15g."},{"Start":"23:20.130 ","End":"23:24.880","Text":"This is our answer to question number 3."},{"Start":"23:24.880 ","End":"23:26.320","Text":"As we can see,"},{"Start":"23:26.320 ","End":"23:29.919","Text":"this is a lot simpler than going straight"},{"Start":"23:29.919 ","End":"23:36.080","Text":"for this question without doing the first 2 steps."},{"Start":"23:36.930 ","End":"23:41.260","Text":"This question is actually a good example of why"},{"Start":"23:41.260 ","End":"23:44.590","Text":"fictitious forces are super useful in"},{"Start":"23:44.590 ","End":"23:48.324","Text":"solving questions because if we didn\u0027t do our first stages,"},{"Start":"23:48.324 ","End":"23:51.580","Text":"which to use the idea of fictitious forces,"},{"Start":"23:51.580 ","End":"23:56.785","Text":"then we wouldn\u0027t have been able to have solved this question 3 with such ease."},{"Start":"23:56.785 ","End":"23:58.960","Text":"That\u0027s the end of question number 3."},{"Start":"23:58.960 ","End":"23:59.995","Text":"In the next video,"},{"Start":"23:59.995 ","End":"24:02.875","Text":"we\u0027re going to solve question number 4."},{"Start":"24:02.875 ","End":"24:07.555","Text":"Hello, now we\u0027re going to answer question number 4."},{"Start":"24:07.555 ","End":"24:12.400","Text":"We\u0027re now told that the car is traveling to the left,"},{"Start":"24:12.400 ","End":"24:18.910","Text":"so this is the direction of travel and we\u0027re being asked what must be"},{"Start":"24:18.910 ","End":"24:22.015","Text":"the car\u0027s acceleration in the left direction"},{"Start":"24:22.015 ","End":"24:26.005","Text":"such that the mass M will detach from the car."},{"Start":"24:26.005 ","End":"24:28.660","Text":"We\u0027re again trying to find the acceleration,"},{"Start":"24:28.660 ","End":"24:31.960","Text":"but now that the mass will simply detach."},{"Start":"24:31.960 ","End":"24:34.580","Text":"Let\u0027s see how we do this."},{"Start":"24:35.550 ","End":"24:42.970","Text":"The condition that has to be met is that our normal force"},{"Start":"24:42.970 ","End":"24:50.530","Text":"has to be equal to 0 if our mass is going to be detached from the car."},{"Start":"24:50.530 ","End":"24:52.600","Text":"That makes a lot of sense."},{"Start":"24:52.600 ","End":"24:56.410","Text":"If the mass isn\u0027t exerting a force on the car,"},{"Start":"24:56.410 ","End":"25:01.750","Text":"then there\u0027s going to be no equal and opposite normal force exerted back."},{"Start":"25:01.750 ","End":"25:05.450","Text":"This is the condition which has to be met."},{"Start":"25:07.350 ","End":"25:10.870","Text":"We\u0027re going to have to write our equations for"},{"Start":"25:10.870 ","End":"25:14.470","Text":"force but the first thing that we\u0027re going to have to do is we\u0027re going to"},{"Start":"25:14.470 ","End":"25:16.840","Text":"have to flip the side of"},{"Start":"25:16.840 ","End":"25:23.095","Text":"this fictitious force and now our arrow is pointing in this direction,"},{"Start":"25:23.095 ","End":"25:26.740","Text":"Ma and that\u0027s because our fictitious force is acting"},{"Start":"25:26.740 ","End":"25:30.865","Text":"in the opposite direction to our acceleration."},{"Start":"25:30.865 ","End":"25:34.760","Text":"Now we can write out our force equations."},{"Start":"25:36.320 ","End":"25:41.985","Text":"Let\u0027s write the sum of all of the forces in the y direction."},{"Start":"25:41.985 ","End":"25:45.090","Text":"Now notice in the original y direction,"},{"Start":"25:45.090 ","End":"25:48.785","Text":"so this green dotted line."},{"Start":"25:48.785 ","End":"25:55.390","Text":"We have our normal force in the positive y direction and then we have"},{"Start":"25:55.390 ","End":"26:03.685","Text":"plus our Ma sine of Theta because now it\u0027s flipped over."},{"Start":"26:03.685 ","End":"26:14.319","Text":"Then we have minus our Mg cosine of Theta and this is going to be equal to 0."},{"Start":"26:14.319 ","End":"26:17.470","Text":"Now, why is this equal to 0?"},{"Start":"26:17.470 ","End":"26:21.085","Text":"If our mass is going to detach from our car,"},{"Start":"26:21.085 ","End":"26:24.640","Text":"so we can assume that there\u0027s going to be 0 acceleration."},{"Start":"26:24.640 ","End":"26:32.900","Text":"We don\u0027t need acceleration of the mass M itself in order for it to detach from the car."},{"Start":"26:34.050 ","End":"26:38.335","Text":"Now, we don\u0027t have to work out the sum of the forces in"},{"Start":"26:38.335 ","End":"26:44.920","Text":"our x direction because we can see that our condition for a mass M detaching"},{"Start":"26:44.920 ","End":"26:53.290","Text":"from the car is that our normal force is equal to 0 and only in our force equation in"},{"Start":"26:53.290 ","End":"26:56.845","Text":"the y direction do we have a normal force appear"},{"Start":"26:56.845 ","End":"27:02.950","Text":"and then that means that we have one equation with one unknown,"},{"Start":"27:02.950 ","End":"27:08.140","Text":"which is our a, which is what we\u0027re trying to calculate so we can solve this."},{"Start":"27:08.140 ","End":"27:11.455","Text":"Now what we can do is we can isolate out our N,"},{"Start":"27:11.455 ","End":"27:14.776","Text":"so we can see that N is going to be equal to"},{"Start":"27:14.776 ","End":"27:24.760","Text":"Mg cosine of Theta minus Ma sine of Theta."},{"Start":"27:24.760 ","End":"27:32.709","Text":"I\u0027ve just isolated out my N and it has to fulfill this condition that it is equal to 0."},{"Start":"27:32.709 ","End":"27:40.090","Text":"Then I can say that my Mg cosine of Theta is equal to Ma sine of Theta."},{"Start":"27:40.090 ","End":"27:46.075","Text":"I can move this over to this side and then I can divide both sides by M. Then I\u0027ll get"},{"Start":"27:46.075 ","End":"27:54.590","Text":"that g cosine of Theta is equal to a sine of Theta."},{"Start":"27:54.870 ","End":"27:58.330","Text":"Now all I have to do is I have to isolate out"},{"Start":"27:58.330 ","End":"28:01.135","Text":"my a because that\u0027s what I\u0027m trying to find."},{"Start":"28:01.135 ","End":"28:11.410","Text":"My a is going to be equal to g multiplied by cosine of Theta divided by sine of Theta,"},{"Start":"28:11.410 ","End":"28:14.245","Text":"which as we said, was equal to,"},{"Start":"28:14.245 ","End":"28:21.250","Text":"so cosine of Theta is equal to 0.8 and sine of Theta is equal to 0.6,"},{"Start":"28:21.250 ","End":"28:28.060","Text":"as seen in the question and from previous answers that we\u0027ve gotten."},{"Start":"28:28.060 ","End":"28:31.525","Text":"Once we plug this into the calculator,"},{"Start":"28:31.525 ","End":"28:38.420","Text":"we\u0027ll get that our a is approximately equal to 1.33g."},{"Start":"28:38.640 ","End":"28:41.935","Text":"This is the answer to question 4,"},{"Start":"28:41.935 ","End":"28:46.810","Text":"so our acceleration in the leftwards direction has to be equal to"},{"Start":"28:46.810 ","End":"28:52.690","Text":"1.33g in order for a mass M to disconnect from the car."},{"Start":"28:52.690 ","End":"28:55.400","Text":"That\u0027s the end of this question."}],"ID":10775}],"Thumbnail":null,"ID":5351},{"Name":"Fictitious Forces in a Rotating System – Centrifugal and Coriolis","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explaining the Forces","Duration":"10m 5s","ChapterTopicVideoID":9007,"CourseChapterTopicPlaylistID":5352,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9007.jpeg","UploadDate":"2017-03-21T09:48:30.2270000","DurationForVideoObject":"PT10M5S","Description":null,"MetaTitle":"Explaining the Forces: Video + Workbook | Proprep","MetaDescription":"D Alemberts Principle-Fictitious Forces - Fictitious Forces in a Rotating System – Centrifugal and Coriolis. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/d-alemberts-principle_fictitious-forces/fictitious-forces-in-a-rotating-system-%e2%80%93-centrifugal-and-coriolis/vid9288","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.190","Text":"Hello. In this video,"},{"Start":"00:02.190 ","End":"00:05.835","Text":"I\u0027d like to talk about fictitious forces in a rotating system,"},{"Start":"00:05.835 ","End":"00:10.900","Text":"specifically discussing the centrifugal force and the Coriolis force."},{"Start":"00:11.930 ","End":"00:15.240","Text":"In our first discussion of fictitious forces,"},{"Start":"00:15.240 ","End":"00:16.709","Text":"we took 2 systems,"},{"Start":"00:16.709 ","End":"00:18.405","Text":"1 which was at rest,"},{"Start":"00:18.405 ","End":"00:21.300","Text":"and the other which relative to our first system,"},{"Start":"00:21.300 ","End":"00:25.575","Text":"was moving with an acceleration of a on a straight line,"},{"Start":"00:25.575 ","End":"00:30.700","Text":"in this case, across the x-axis relative to our first system."},{"Start":"00:31.310 ","End":"00:33.735","Text":"Now in this example,"},{"Start":"00:33.735 ","End":"00:38.670","Text":"we again have 2 systems where 1 is at rest and the other isn\u0027t relative motion."},{"Start":"00:38.670 ","End":"00:41.040","Text":"Our first system, the system in black,"},{"Start":"00:41.040 ","End":"00:43.325","Text":"x, y, is our system at rest."},{"Start":"00:43.325 ","End":"00:44.900","Text":"That\u0027s where our laboratory is,"},{"Start":"00:44.900 ","End":"00:46.414","Text":"where we take measurements."},{"Start":"00:46.414 ","End":"00:52.400","Text":"The red system is rotating around the black system and they share the same origin point."},{"Start":"00:52.400 ","End":"00:55.415","Text":"Standing on the same point where the laboratory is,"},{"Start":"00:55.415 ","End":"00:57.500","Text":"we can imagine there\u0027s a ballerina."},{"Start":"00:57.500 ","End":"01:03.565","Text":"This ballerina is rotating with an angular velocity in a counterclockwise direction."},{"Start":"01:03.565 ","End":"01:09.275","Text":"Let\u0027s imagine again that the ballerina and the laboratory are at the exact same point,"},{"Start":"01:09.275 ","End":"01:11.015","Text":"they share an origin."},{"Start":"01:11.015 ","End":"01:14.195","Text":"The angle between the x-axis of the ballerina,"},{"Start":"01:14.195 ","End":"01:15.665","Text":"that is x tag,"},{"Start":"01:15.665 ","End":"01:17.825","Text":"and the x-axis of the laboratory,"},{"Start":"01:17.825 ","End":"01:20.455","Text":"that is x, is the angle Theta."},{"Start":"01:20.455 ","End":"01:23.290","Text":"Well that means that the integral of Theta,"},{"Start":"01:23.290 ","End":"01:27.045","Text":"Theta dot equals Omega."},{"Start":"01:27.045 ","End":"01:32.420","Text":"This Omega or Theta dot represents the change in the angle over time."},{"Start":"01:32.420 ","End":"01:35.840","Text":"We can also call this the angular velocity."},{"Start":"01:35.840 ","End":"01:37.835","Text":"In a case like this,"},{"Start":"01:37.835 ","End":"01:40.430","Text":"where we have 1 system that\u0027s stationary and"},{"Start":"01:40.430 ","End":"01:43.220","Text":"another rotating around it and they share an origin,"},{"Start":"01:43.220 ","End":"01:47.980","Text":"we consider the rotating system to be in a non-inertial system."},{"Start":"01:47.980 ","End":"01:53.075","Text":"That means that we have to account for fictitious forces in this rotating system."},{"Start":"01:53.075 ","End":"01:55.430","Text":"If we were to measure an object,"},{"Start":"01:55.430 ","End":"01:57.305","Text":"say this triangle in blue,"},{"Start":"01:57.305 ","End":"02:00.875","Text":"from the perspective of both the ballerina and the laboratory,"},{"Start":"02:00.875 ","End":"02:03.065","Text":"we would get 2 different measurements."},{"Start":"02:03.065 ","End":"02:05.120","Text":"To account for these different measurements,"},{"Start":"02:05.120 ","End":"02:08.000","Text":"we would need to add fictitious forces to"},{"Start":"02:08.000 ","End":"02:12.005","Text":"our system to make Newton\u0027s second law applicable."},{"Start":"02:12.005 ","End":"02:16.760","Text":"Keep in mind this is only the case if one of the observers is rotating."},{"Start":"02:16.760 ","End":"02:19.640","Text":"If this triangle were to rotate around the laboratory,"},{"Start":"02:19.640 ","End":"02:23.090","Text":"it would not mean we need to add fictitious forces and"},{"Start":"02:23.090 ","End":"02:26.825","Text":"we would not need to account for the centrifugal force or the Coriolis force."},{"Start":"02:26.825 ","End":"02:29.030","Text":"Only when one of the observers,"},{"Start":"02:29.030 ","End":"02:33.500","Text":"in this case the ballerina is rotating do we have to add the fictitious forces of"},{"Start":"02:33.500 ","End":"02:38.840","Text":"the centrifugal force and the Coriolis force to make Newton\u0027s second law work."},{"Start":"02:38.840 ","End":"02:43.930","Text":"Let\u0027s get down to brass tacks and talk about the actual forces we\u0027re measuring."},{"Start":"02:43.930 ","End":"02:47.960","Text":"First let\u0027s assume this blue triangle is moving with a velocity"},{"Start":"02:47.960 ","End":"02:52.880","Text":"v in a direction more or less parallel to the x-axis."},{"Start":"02:52.880 ","End":"02:56.900","Text":"If that\u0027s measured from the origin by the laboratory,"},{"Start":"02:56.900 ","End":"03:00.035","Text":"by the stationary system,"},{"Start":"03:00.035 ","End":"03:02.645","Text":"we\u0027ll get a measurement of v. However,"},{"Start":"03:02.645 ","End":"03:05.869","Text":"if we\u0027re measuring from the perspective of the spinning ballerina,"},{"Start":"03:05.869 ","End":"03:07.730","Text":"we\u0027re going to get a different measurement,"},{"Start":"03:07.730 ","End":"03:11.885","Text":"v tag and v tag will be a different velocity altogether,"},{"Start":"03:11.885 ","End":"03:15.590","Text":"meaning it has both a different magnitude and a different direction."},{"Start":"03:15.590 ","End":"03:19.505","Text":"To be honest, we can forget about the stationary system for now."},{"Start":"03:19.505 ","End":"03:21.935","Text":"We know how to measure in a stationary system,"},{"Start":"03:21.935 ","End":"03:25.250","Text":"what I care about now is the ballerina, the relative system."},{"Start":"03:25.250 ","End":"03:29.225","Text":"Just to reiterate, v is what\u0027s measured from the laboratory,"},{"Start":"03:29.225 ","End":"03:32.475","Text":"and v tag is what\u0027s measured by the ballerina."},{"Start":"03:32.475 ","End":"03:35.900","Text":"In order to do a list of forces for v tag,"},{"Start":"03:35.900 ","End":"03:39.950","Text":"what we first want to do is account for the sum of forces."},{"Start":"03:39.950 ","End":"03:41.735","Text":"These are the real forces."},{"Start":"03:41.735 ","End":"03:44.570","Text":"Those are the forces that we normally talk about such as"},{"Start":"03:44.570 ","End":"03:47.600","Text":"gravity or friction or something like that."},{"Start":"03:47.600 ","End":"03:48.950","Text":"We\u0027re going to add to that"},{"Start":"03:48.950 ","End":"03:53.135","Text":"2 different fictitious forces in the case of a rotating system."},{"Start":"03:53.135 ","End":"03:55.445","Text":"First, the centrifugal force."},{"Start":"03:55.445 ","End":"04:02.840","Text":"We\u0027re going to first symbolize that with this f. Secondly,"},{"Start":"04:02.840 ","End":"04:06.630","Text":"we\u0027re going to add the Coriolis force."},{"Start":"04:08.840 ","End":"04:13.120","Text":"Now of course, each of these are vectors."},{"Start":"04:13.400 ","End":"04:20.630","Text":"In the end they have to equal ma tag vector."},{"Start":"04:20.630 ","End":"04:27.235","Text":"This a tag vector is the acceleration relative to the rotating observer,"},{"Start":"04:27.235 ","End":"04:29.020","Text":"that is the ballerina."},{"Start":"04:29.020 ","End":"04:33.965","Text":"Now it\u0027s time to incorporate our centrifugal and Coriolis forces."},{"Start":"04:33.965 ","End":"04:36.956","Text":"Let\u0027s start with the centrifugal force,"},{"Start":"04:36.956 ","End":"04:39.550","Text":"1 note to make is that the centrifugal force and"},{"Start":"04:39.550 ","End":"04:42.460","Text":"the Coriolis force have no relationship between each other."},{"Start":"04:42.460 ","End":"04:47.970","Text":"They are 2 separate forces with separate equations and no relationship 1 to the other."},{"Start":"04:47.970 ","End":"04:50.745","Text":"Starting with the centrifugal force,"},{"Start":"04:50.745 ","End":"04:56.375","Text":"the f vector equals m Omega^2 rr-hat."},{"Start":"04:56.375 ","End":"05:00.545","Text":"The first thing to note is that this Omega, the angular velocity,"},{"Start":"05:00.545 ","End":"05:03.095","Text":"is the angular velocity of our ballerina,"},{"Start":"05:03.095 ","End":"05:06.305","Text":"of our observer relative to the laboratory."},{"Start":"05:06.305 ","End":"05:12.980","Text":"Just to reiterate, Omega is the angular velocity of the observer,"},{"Start":"05:12.980 ","End":"05:17.095","Text":"the ballerina, relative to the inertial system, the laboratory."},{"Start":"05:17.095 ","End":"05:20.255","Text":"What this means is even if the triangle,"},{"Start":"05:20.255 ","End":"05:24.110","Text":"our object being observed has some Omega,"},{"Start":"05:24.110 ","End":"05:27.660","Text":"that is a radial velocity,"},{"Start":"05:27.660 ","End":"05:30.035","Text":"we\u0027re still not interested in that."},{"Start":"05:30.035 ","End":"05:34.430","Text":"What we\u0027re measuring is the radial velocity of the ballerina, of the observer."},{"Start":"05:34.430 ","End":"05:38.380","Text":"We\u0027re measuring the change in the angle Theta, this angle Theta."},{"Start":"05:38.380 ","End":"05:40.240","Text":"Now the next thing is r,"},{"Start":"05:40.240 ","End":"05:44.645","Text":"which is the distance from the origin to the object being measured."},{"Start":"05:44.645 ","End":"05:47.255","Text":"It\u0027s coming from the origin to the object,"},{"Start":"05:47.255 ","End":"05:50.510","Text":"and that\u0027s r, and it\u0027s in the direction of r-hat,"},{"Start":"05:50.510 ","End":"05:56.375","Text":"which of course is the continuation or the same direction as this line"},{"Start":"05:56.375 ","End":"06:04.145","Text":"r. What we have is Omega^2 r in the direction of r-hat,"},{"Start":"06:04.145 ","End":"06:06.505","Text":"and that\u0027s what this line represents."},{"Start":"06:06.505 ","End":"06:08.810","Text":"This essentially solves for me what we\u0027re"},{"Start":"06:08.810 ","End":"06:11.525","Text":"talking about when we talk about the centrifugal force."},{"Start":"06:11.525 ","End":"06:13.925","Text":"Usually we\u0027re going to use this formula."},{"Start":"06:13.925 ","End":"06:17.210","Text":"However, the fuller version is the lower formula here."},{"Start":"06:17.210 ","End":"06:21.605","Text":"Now, what really is going on here is these are equal of course."},{"Start":"06:21.605 ","End":"06:25.415","Text":"As soon as Omega is on the xy-plane,"},{"Start":"06:25.415 ","End":"06:26.585","Text":"as we normally have it,"},{"Start":"06:26.585 ","End":"06:30.440","Text":"and r is on the z-plane or on the z-axis,"},{"Start":"06:30.440 ","End":"06:33.825","Text":"we end up getting these 2 equaling each other."},{"Start":"06:33.825 ","End":"06:37.130","Text":"In fact, you can even see that the magnitudes are the same."},{"Start":"06:37.130 ","End":"06:42.545","Text":"You have Omega times Omega or Omega^2 times r. The magnitude is the same,"},{"Start":"06:42.545 ","End":"06:46.490","Text":"as long as the angles work out the same which they do if we\u0027re working with Omega on"},{"Start":"06:46.490 ","End":"06:51.310","Text":"the xy-plane and r on the z-axis, then we\u0027re set."},{"Start":"06:51.310 ","End":"06:53.525","Text":"Moving on to the Coriolis force,"},{"Start":"06:53.525 ","End":"06:55.055","Text":"it may seem confusing,"},{"Start":"06:55.055 ","End":"06:58.025","Text":"but once you understand what each variable stands for,"},{"Start":"06:58.025 ","End":"07:01.205","Text":"it\u0027s rather simple and we can incorporate it easily."},{"Start":"07:01.205 ","End":"07:06.420","Text":"We have negative 2m Omega cross v tag."},{"Start":"07:06.420 ","End":"07:07.485","Text":"So what\u0027s each of these?"},{"Start":"07:07.485 ","End":"07:10.725","Text":"Negative 2m is negative 2 times the mass."},{"Start":"07:10.725 ","End":"07:12.510","Text":"Omega, as always,"},{"Start":"07:12.510 ","End":"07:16.715","Text":"is the angular velocity of the observer relative to the inertial system,"},{"Start":"07:16.715 ","End":"07:17.930","Text":"that is the laboratory."},{"Start":"07:17.930 ","End":"07:21.115","Text":"We can actually get rid of this relative here."},{"Start":"07:21.115 ","End":"07:24.650","Text":"v tag, because it has the tag,"},{"Start":"07:24.650 ","End":"07:27.215","Text":"we know that we\u0027re talking about a relative velocity."},{"Start":"07:27.215 ","End":"07:30.110","Text":"It is the velocity of the object being measured, in our case,"},{"Start":"07:30.110 ","End":"07:35.695","Text":"the triangle, relative to our relative observer, the ballerina."},{"Start":"07:35.695 ","End":"07:38.705","Text":"Let\u0027s look at the example of our problem here."},{"Start":"07:38.705 ","End":"07:42.170","Text":"We already have v tag symbolized here."},{"Start":"07:42.170 ","End":"07:45.775","Text":"What we need to do is find Omega."},{"Start":"07:45.775 ","End":"07:48.680","Text":"According to the right-hand rule,"},{"Start":"07:48.680 ","End":"07:52.040","Text":"we can see that Omega should be coming towards us."},{"Start":"07:52.040 ","End":"07:54.535","Text":"This is the Omega vector."},{"Start":"07:54.535 ","End":"08:01.175","Text":"Now I need to perform a scalar multiplication between Omega and v tag."},{"Start":"08:01.175 ","End":"08:04.565","Text":"I know based on the right hand rule that because"},{"Start":"08:04.565 ","End":"08:07.505","Text":"Omega is going in a counterclockwise direction,"},{"Start":"08:07.505 ","End":"08:10.055","Text":"that what I need is something that is"},{"Start":"08:10.055 ","End":"08:14.365","Text":"perpendicular to v tag and it\u0027s going to come towards us."},{"Start":"08:14.365 ","End":"08:18.740","Text":"This is the result of Omega cross v tag."},{"Start":"08:18.740 ","End":"08:21.470","Text":"However, over here we have negative 2,"},{"Start":"08:21.470 ","End":"08:24.800","Text":"meaning that the actual result we\u0027ll use in our problem,"},{"Start":"08:24.800 ","End":"08:27.035","Text":"will be going the opposite direction."},{"Start":"08:27.035 ","End":"08:32.145","Text":"In fact, this line here is going to be the result of our problem."},{"Start":"08:32.145 ","End":"08:35.280","Text":"This is F_c."},{"Start":"08:35.280 ","End":"08:39.215","Text":"This is the end of the explanation mathematically."},{"Start":"08:39.215 ","End":"08:41.315","Text":"If this is long and complicated,"},{"Start":"08:41.315 ","End":"08:45.170","Text":"I want to do a quick summary just to make sure everything\u0027s clear."},{"Start":"08:45.170 ","End":"08:47.795","Text":"When we\u0027re dealing with a rotating system,"},{"Start":"08:47.795 ","End":"08:52.595","Text":"that is where we have 1 observer that\u0027s rotating around an inertial system."},{"Start":"08:52.595 ","End":"08:56.000","Text":"Then we need to immediately add 2 fictitious forces."},{"Start":"08:56.000 ","End":"08:59.630","Text":"First, the centrifugal force and second the Coriolis force."},{"Start":"08:59.630 ","End":"09:03.485","Text":"We do our sum of forces and it has to equal ma tag,"},{"Start":"09:03.485 ","End":"09:05.025","Text":"of course, with all vectors,"},{"Start":"09:05.025 ","End":"09:07.185","Text":"and to find the centrifugal force,"},{"Start":"09:07.185 ","End":"09:10.550","Text":"we take Omega^2 r outwards from"},{"Start":"09:10.550 ","End":"09:15.575","Text":"the center and multiply that by m. For the Coriolis force,"},{"Start":"09:15.575 ","End":"09:21.560","Text":"we need to multiply out Omega and v tag and figure out which direction it takes us."},{"Start":"09:21.560 ","End":"09:25.130","Text":"Now you have a mathematical and a theoretical explanation."},{"Start":"09:25.130 ","End":"09:29.690","Text":"The last thing I want to do is take a step back and talk about why this is important."},{"Start":"09:29.690 ","End":"09:32.075","Text":"Frankly, this is really important because"},{"Start":"09:32.075 ","End":"09:37.090","Text":"our most basic measurements are taken from a rotating system."},{"Start":"09:37.090 ","End":"09:40.160","Text":"If we\u0027re standing on Earth and we want to measure"},{"Start":"09:40.160 ","End":"09:43.610","Text":"a star or a satellite or anything outside of the Earth,"},{"Start":"09:43.610 ","End":"09:46.850","Text":"we\u0027re functionally standing on a rotating system."},{"Start":"09:46.850 ","End":"09:49.820","Text":"Therefore, we need to account for the Coriolis and"},{"Start":"09:49.820 ","End":"09:53.755","Text":"centrifugal forces to measure these objects accurately."},{"Start":"09:53.755 ","End":"09:59.105","Text":"Now I want to look at a case of basic circular motion and analyze"},{"Start":"09:59.105 ","End":"10:01.100","Text":"this problem so that it\u0027ll give us maybe"},{"Start":"10:01.100 ","End":"10:05.910","Text":"a better understanding of fictitious forces in a rotating system."}],"ID":9288},{"Watched":false,"Name":"Example 1 part a","Duration":"2m 22s","ChapterTopicVideoID":24724,"CourseChapterTopicPlaylistID":5352,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:01.040 ","End":"00:05.055","Text":"Here we have our object and we have our circle,"},{"Start":"00:05.055 ","End":"00:08.400","Text":"and we have a radius of R. Let\u0027s assume for"},{"Start":"00:08.400 ","End":"00:11.820","Text":"the sake of simplicity that we have someone swinging of stone"},{"Start":"00:11.820 ","End":"00:16.170","Text":"around in a circular motion using a string with the length of"},{"Start":"00:16.170 ","End":"00:24.000","Text":"R. Now what I want to do is look at this from the perspective of the inertial observer."},{"Start":"00:24.000 ","End":"00:26.250","Text":"Now, I\u0027ve added here the x and"},{"Start":"00:26.250 ","End":"00:30.555","Text":"the y-axis and our laboratory from which we take measurements."},{"Start":"00:30.555 ","End":"00:33.180","Text":"Let\u0027s say that our object is rotating with"},{"Start":"00:33.180 ","End":"00:40.560","Text":"an angular velocity of Omega_1 relative to the lab, of course."},{"Start":"00:40.560 ","End":"00:44.270","Text":"Now, I want to make a list of my forces and in this case,"},{"Start":"00:44.270 ","End":"00:47.210","Text":"we only have T, that is tension,"},{"Start":"00:47.210 ","End":"00:50.390","Text":"acting towards the center of the circle."},{"Start":"00:50.390 ","End":"00:53.960","Text":"Now, when I do my sum of forces,"},{"Start":"00:53.960 ","End":"00:56.025","Text":"that is my force equation,"},{"Start":"00:56.025 ","End":"01:05.360","Text":"I have sum of forces equals Tr hat or really negative Tr hat."},{"Start":"01:05.360 ","End":"01:06.605","Text":"Remember, from before,"},{"Start":"01:06.605 ","End":"01:10.595","Text":"we define r hat is going outwards in this direction."},{"Start":"01:10.595 ","End":"01:13.755","Text":"But in our case T is going inwards."},{"Start":"01:13.755 ","End":"01:15.375","Text":"Instead of being positive T,"},{"Start":"01:15.375 ","End":"01:22.170","Text":"it will be negative T. This has to equal a_r, r hat."},{"Start":"01:22.170 ","End":"01:26.825","Text":"That is the rotational acceleration or the angular acceleration."},{"Start":"01:26.825 ","End":"01:33.425","Text":"That equals negative Omega^2 times R and we can use big R here."},{"Start":"01:33.425 ","End":"01:37.035","Text":"We\u0027re talking about this length r hat."},{"Start":"01:37.035 ","End":"01:42.625","Text":"Again, it\u0027s negative because while we define r hat is going outwards,"},{"Start":"01:42.625 ","End":"01:48.515","Text":"the angular acceleration goes inwards towards the center of the circle."},{"Start":"01:48.515 ","End":"01:52.685","Text":"Just to reiterate normally r hat that is,"},{"Start":"01:52.685 ","End":"01:55.175","Text":"is going to be towards the inside of the circle."},{"Start":"01:55.175 ","End":"01:58.980","Text":"In this case, we\u0027ve chosen to define r hat is going outwards."},{"Start":"01:58.980 ","End":"02:04.775","Text":"We\u0027d been consistent with our negative directions for Omega over here and T over here."},{"Start":"02:04.775 ","End":"02:08.530","Text":"Now remember, the Omega we\u0027re talking about is Omega_1."},{"Start":"02:08.530 ","End":"02:14.925","Text":"Ultimately, we can solve for T and our tension equals Omega_1"},{"Start":"02:14.925 ","End":"02:22.770","Text":"squared times R. This is the value of the tension in the rope."}],"ID":25637},{"Watched":false,"Name":"Example 1 part b","Duration":"3m 19s","ChapterTopicVideoID":24725,"CourseChapterTopicPlaylistID":5352,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.955","Text":"Now, let\u0027s do the same breakdown from the perspective of a non-inertial observer."},{"Start":"00:05.955 ","End":"00:09.150","Text":"Let\u0027s assume I have the same ballerina that\u0027s"},{"Start":"00:09.150 ","End":"00:12.735","Text":"rotating around the origin in the middle of my circle,"},{"Start":"00:12.735 ","End":"00:16.020","Text":"and it just so happens that the angular velocity with which"},{"Start":"00:16.020 ","End":"00:19.710","Text":"the ballerina rotates is also Omega 1,"},{"Start":"00:19.710 ","End":"00:21.690","Text":"meaning it\u0027s identical to"},{"Start":"00:21.690 ","End":"00:27.180","Text":"the angular velocity with which our rock rotates around the center of the circle."},{"Start":"00:27.180 ","End":"00:32.175","Text":"The next step is to do a force equation for my ballerina."},{"Start":"00:32.175 ","End":"00:36.070","Text":"First I want to add the real forces just like I did over here on the left,"},{"Start":"00:36.070 ","End":"00:38.400","Text":"and the real forces are the same as they are"},{"Start":"00:38.400 ","End":"00:41.295","Text":"above from the perspective of the inertial observer."},{"Start":"00:41.295 ","End":"00:46.155","Text":"That is negative Tr hat."},{"Start":"00:46.155 ","End":"00:49.730","Text":"Now I need to add to that my fictitious forces."},{"Start":"00:49.730 ","End":"00:53.030","Text":"First, the centrifugal force and we\u0027re going to do,"},{"Start":"00:53.030 ","End":"00:58.090","Text":"remember, it is m Omega squared rr hat."},{"Start":"00:58.370 ","End":"01:04.895","Text":"We\u0027re going to add m Omega squared rr hat."},{"Start":"01:04.895 ","End":"01:07.250","Text":"Now, it just so happens that in this case,"},{"Start":"01:07.250 ","End":"01:11.630","Text":"the Omega of the ballerina is the same as the Omega of the stone."},{"Start":"01:11.630 ","End":"01:14.875","Text":"However, we\u0027re always going to use the Omega of the ballerina."},{"Start":"01:14.875 ","End":"01:17.620","Text":"In this case, it\u0027s still Omega 1."},{"Start":"01:17.620 ","End":"01:19.430","Text":"However, if these 2 are different,"},{"Start":"01:19.430 ","End":"01:21.815","Text":"we would be selecting the Omega of the ballerina,"},{"Start":"01:21.815 ","End":"01:24.500","Text":"the non-inertial observer relative to"},{"Start":"01:24.500 ","End":"01:29.405","Text":"the inertial observer rather than the Omega of the stone."},{"Start":"01:29.405 ","End":"01:33.245","Text":"Lastly, I need to add my Coriolis force."},{"Start":"01:33.245 ","End":"01:37.580","Text":"However, if you think about it from the perspective of the ballerina,"},{"Start":"01:37.580 ","End":"01:39.965","Text":"V tag equals 0."},{"Start":"01:39.965 ","End":"01:41.405","Text":"Now why is that?"},{"Start":"01:41.405 ","End":"01:44.900","Text":"Well, as the ballerinas spins, so does the stone."},{"Start":"01:44.900 ","End":"01:49.760","Text":"Because they\u0027re moving in concert in sync when with the other V tag,"},{"Start":"01:49.760 ","End":"01:51.500","Text":"according to the ballerina, that is,"},{"Start":"01:51.500 ","End":"01:54.935","Text":"is 0, the ballerina does not see the stone moving at all."},{"Start":"01:54.935 ","End":"01:59.635","Text":"It\u0027s spins with the ballerina as she spins around the origin."},{"Start":"01:59.635 ","End":"02:04.865","Text":"Because V tag equals 0 and we\u0027re doing a cross multiplication,"},{"Start":"02:04.865 ","End":"02:07.350","Text":"then the whole Coriolis force equals 0."},{"Start":"02:07.350 ","End":"02:11.230","Text":"We don\u0027t have to add any Coriolis force at all."},{"Start":"02:11.230 ","End":"02:14.435","Text":"Because of V tag equals 0,"},{"Start":"02:14.435 ","End":"02:18.005","Text":"we don\u0027t have to account for the Coriolis force."},{"Start":"02:18.005 ","End":"02:22.010","Text":"Now this may explain why earlier when we talked about circular motion"},{"Start":"02:22.010 ","End":"02:25.940","Text":"and we talked about fictitious forces and the centrifugal force."},{"Start":"02:25.940 ","End":"02:28.265","Text":"We hadn\u0027t brought up the Coriolis force."},{"Start":"02:28.265 ","End":"02:32.120","Text":"That\u0027s because as soon as you have the observer spinning,"},{"Start":"02:32.120 ","End":"02:36.760","Text":"rotating at the same angular velocity as the object it\u0027s measuring,"},{"Start":"02:36.760 ","End":"02:40.210","Text":"there is no need to account for the Coriolis force."},{"Start":"02:40.210 ","End":"02:46.450","Text":"Now our next step is we need to set this equal to ma tag."},{"Start":"02:46.450 ","End":"02:52.295","Text":"If we see this equals ma tag vector,"},{"Start":"02:52.295 ","End":"02:54.965","Text":"but if V tag equals 0,"},{"Start":"02:54.965 ","End":"02:58.610","Text":"therefore, a tag has to also equal 0."},{"Start":"02:58.610 ","End":"03:01.745","Text":"That means that if a tag equals 0,"},{"Start":"03:01.745 ","End":"03:04.715","Text":"then ma tag also equals 0."},{"Start":"03:04.715 ","End":"03:08.630","Text":"In the end we get that Tr hat,"},{"Start":"03:08.630 ","End":"03:11.690","Text":"although we can simplify that out,"},{"Start":"03:11.690 ","End":"03:19.720","Text":"equals m Omega squared r, just like above."}],"ID":25638},{"Watched":false,"Name":"Example 1 part c","Duration":"8m 1s","ChapterTopicVideoID":24726,"CourseChapterTopicPlaylistID":5352,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Now, let\u0027s look at 1 more observer,"},{"Start":"00:03.030 ","End":"00:06.000","Text":"and this will probably be the most interesting observer."},{"Start":"00:06.000 ","End":"00:10.780","Text":"Let\u0027s for a second, take our ballerina and put her on the side."},{"Start":"00:10.880 ","End":"00:13.170","Text":"We\u0027re going to have another observer,"},{"Start":"00:13.170 ","End":"00:15.045","Text":"this observer is going to be a clown."},{"Start":"00:15.045 ","End":"00:16.380","Text":"I\u0027ll draw clown hat,"},{"Start":"00:16.380 ","End":"00:18.180","Text":"because of my limited drawing skills,"},{"Start":"00:18.180 ","End":"00:21.960","Text":"and the clown is also rotating on the same point."},{"Start":"00:21.960 ","End":"00:27.180","Text":"But the clown has a different angular velocity, that is."},{"Start":"00:27.180 ","End":"00:32.040","Text":"His angular velocity is Omega subzero."},{"Start":"00:32.040 ","End":"00:37.840","Text":"Let\u0027s assume for the sake of the problem that the clown\u0027s angular velocity,"},{"Start":"00:37.840 ","End":"00:41.975","Text":"Omega subzero is a little bit slower than Omega sub 1."},{"Start":"00:41.975 ","End":"00:44.835","Text":"Let\u0027s make some room down here and write out what we know."},{"Start":"00:44.835 ","End":"00:47.480","Text":"In order to find the velocity of the stone,"},{"Start":"00:47.480 ","End":"00:49.100","Text":"that is the object being measured,"},{"Start":"00:49.100 ","End":"00:51.650","Text":"relative to our non-inertial observer,"},{"Start":"00:51.650 ","End":"00:53.915","Text":"spinning with a unique angular velocity,"},{"Start":"00:53.915 ","End":"00:56.975","Text":"that is our clown, we\u0027re going to use an equation,"},{"Start":"00:56.975 ","End":"00:59.745","Text":"a formula of angular motion,"},{"Start":"00:59.745 ","End":"01:02.745","Text":"and this allows us to convert velocities."},{"Start":"01:02.745 ","End":"01:04.954","Text":"We know that Omega tag,"},{"Start":"01:04.954 ","End":"01:09.340","Text":"the relative velocity equals Omega_1,"},{"Start":"01:09.340 ","End":"01:11.105","Text":"which is the velocity of the object,"},{"Start":"01:11.105 ","End":"01:14.450","Text":"the stone minus Omega_0,"},{"Start":"01:14.450 ","End":"01:18.895","Text":"the angular velocity of the clown as it rotates around the origin."},{"Start":"01:18.895 ","End":"01:22.520","Text":"For our purposes, this new variable, Omega tag,"},{"Start":"01:22.520 ","End":"01:25.070","Text":"is the angular velocity of the stone as it"},{"Start":"01:25.070 ","End":"01:28.940","Text":"rotates around the circle from the perspective of the clown."},{"Start":"01:28.940 ","End":"01:32.195","Text":"Now if I want to know the linear velocity"},{"Start":"01:32.195 ","End":"01:35.180","Text":"of the same stone from the perspective of the clown,"},{"Start":"01:35.180 ","End":"01:37.235","Text":"I need to find V tag."},{"Start":"01:37.235 ","End":"01:39.500","Text":"That because we know that R,"},{"Start":"01:39.500 ","End":"01:41.300","Text":"the radius is staying the same,"},{"Start":"01:41.300 ","End":"01:46.595","Text":"that is going to equal Omega tag times R, the radius."},{"Start":"01:46.595 ","End":"01:49.145","Text":"Now I can move on to my force equation"},{"Start":"01:49.145 ","End":"01:51.380","Text":"and we\u0027ll use the same force equation we used above,"},{"Start":"01:51.380 ","End":"01:53.585","Text":"the same force equation from over here."},{"Start":"01:53.585 ","End":"01:56.045","Text":"First, we find the sum of real forces."},{"Start":"01:56.045 ","End":"02:01.345","Text":"This hasn\u0027t changed, it\u0027s still negative Tr hat."},{"Start":"02:01.345 ","End":"02:04.175","Text":"Then we add to that the centrifugal force,"},{"Start":"02:04.175 ","End":"02:08.645","Text":"which in our case is m Omega 0^2."},{"Start":"02:08.645 ","End":"02:10.700","Text":"Remember, we take Omega_0,"},{"Start":"02:10.700 ","End":"02:16.295","Text":"because that is the angular velocity of the observer of the clown."},{"Start":"02:16.295 ","End":"02:19.130","Text":"We\u0027re going to multiply that by large R,"},{"Start":"02:19.130 ","End":"02:22.060","Text":"the radius r hat."},{"Start":"02:22.370 ","End":"02:25.215","Text":"Now, to add the Coriolis force,"},{"Start":"02:25.215 ","End":"02:26.355","Text":"I need to add, remember,"},{"Start":"02:26.355 ","End":"02:30.675","Text":"negative 2m Omega cross V tag."},{"Start":"02:30.675 ","End":"02:34.070","Text":"First of all, we know V tag is going in this direction,"},{"Start":"02:34.070 ","End":"02:38.645","Text":"because Omega subzero is slightly slower than Omega sub 1."},{"Start":"02:38.645 ","End":"02:43.000","Text":"Still, V tag will be going in this direction."},{"Start":"02:43.000 ","End":"02:46.460","Text":"From there, we can deduce just like we did above,"},{"Start":"02:46.460 ","End":"02:50.435","Text":"that if Omega is on the z-axis,"},{"Start":"02:50.435 ","End":"02:57.590","Text":"then Omega cross V will be coming towards the center of the graph."},{"Start":"02:57.590 ","End":"03:00.440","Text":"However, because we have negative 2m,"},{"Start":"03:00.440 ","End":"03:04.570","Text":"in fact, the result here will again be pointing outwards like so."},{"Start":"03:04.570 ","End":"03:07.565","Text":"Again, in this case our FC,"},{"Start":"03:07.565 ","End":"03:10.475","Text":"our Coriolis force will be going outwards."},{"Start":"03:10.475 ","End":"03:13.680","Text":"We can add 2m"},{"Start":"03:14.750 ","End":"03:22.295","Text":"Omega subzero V tag r hat."},{"Start":"03:22.295 ","End":"03:30.235","Text":"This in the end has to equal ma tag."},{"Start":"03:30.235 ","End":"03:32.345","Text":"Before we get too far ahead of ourselves,"},{"Start":"03:32.345 ","End":"03:36.545","Text":"I should remind you that the reason we can just do this multiplication so simply,"},{"Start":"03:36.545 ","End":"03:39.150","Text":"is we\u0027ve accounted for the direction above here."},{"Start":"03:39.150 ","End":"03:42.785","Text":"This is no longer a cross multiplication of vector multiplication,"},{"Start":"03:42.785 ","End":"03:44.345","Text":"this can be a simple scalar."},{"Start":"03:44.345 ","End":"03:49.265","Text":"We\u0027ve accounted for r hat and we can now do a simple scalar multiplication."},{"Start":"03:49.265 ","End":"03:52.565","Text":"Now, back to my ma tag,"},{"Start":"03:52.565 ","End":"03:54.485","Text":"I can rewrite this a different way."},{"Start":"03:54.485 ","End":"03:56.960","Text":"The acceleration that is observed,"},{"Start":"03:56.960 ","End":"04:02.510","Text":"can also be written as Omega squared large R. But remember,"},{"Start":"04:02.510 ","End":"04:07.240","Text":"this is the relative angular velocity and therefore it\u0027s Omega tag."},{"Start":"04:07.240 ","End":"04:10.055","Text":"Now, remember, we have to account for direction here."},{"Start":"04:10.055 ","End":"04:14.090","Text":"We\u0027re adding r hats and if r hat is facing outwards,"},{"Start":"04:14.090 ","End":"04:16.040","Text":"then this has to be negative,"},{"Start":"04:16.040 ","End":"04:20.645","Text":"because the acceleration is going inwards towards the center of the circle."},{"Start":"04:20.645 ","End":"04:22.970","Text":"My next step is to simplify,"},{"Start":"04:22.970 ","End":"04:25.910","Text":"and I\u0027m going to do this by first getting rid of my r hats,"},{"Start":"04:25.910 ","End":"04:28.250","Text":"because everything is in terms of r hat."},{"Start":"04:28.250 ","End":"04:35.525","Text":"I\u0027m going to take every V tag and turn it into Omega tag R and every Omega tag,"},{"Start":"04:35.525 ","End":"04:41.930","Text":"including the V tags that become Omega tags and turn them into Omega_1 minus Omega_0."},{"Start":"04:41.930 ","End":"04:47.745","Text":"All of this can be simplified to equal negative T plus"},{"Start":"04:47.745 ","End":"04:57.240","Text":"m Omega sub 0 squared R plus 2m Omega sub 0."},{"Start":"04:57.240 ","End":"04:58.500","Text":"Instead of V tag,"},{"Start":"04:58.500 ","End":"05:01.065","Text":"we replaced it with Omega tag,"},{"Start":"05:01.065 ","End":"05:05.490","Text":"which then equals Omega_1 minus"},{"Start":"05:05.490 ","End":"05:12.720","Text":"Omega_0 times R. All of this equals m times,"},{"Start":"05:12.720 ","End":"05:14.910","Text":"instead of Omega tags squared,"},{"Start":"05:14.910 ","End":"05:21.830","Text":"(Omega_1 minus Omega_0)^2 times R. Now,"},{"Start":"05:21.830 ","End":"05:24.080","Text":"we can simplify this even further by opening"},{"Start":"05:24.080 ","End":"05:27.485","Text":"the parentheses and this should be minus negative."},{"Start":"05:27.485 ","End":"05:30.114","Text":"What we\u0027ll do is write it out as follows,"},{"Start":"05:30.114 ","End":"05:38.035","Text":"Negative T plus m Omega sub 0 squared R plus."},{"Start":"05:38.035 ","End":"05:39.860","Text":"Now it gets a little more complicated."},{"Start":"05:39.860 ","End":"05:41.870","Text":"We have to open up our parentheses,"},{"Start":"05:41.870 ","End":"05:49.520","Text":"2m Omega sub 0 times Omega_1,"},{"Start":"05:49.520 ","End":"05:53.480","Text":"excuse me, R minus"},{"Start":"05:53.480 ","End":"05:57.330","Text":"2m Omega sub zero"},{"Start":"05:57.330 ","End":"06:03.620","Text":"squared R. This is just opening this parentheses and multiplying by the like terms."},{"Start":"06:03.620 ","End":"06:18.900","Text":"This equals negative m Omega sub 1 squared R plus 2m"},{"Start":"06:19.700 ","End":"06:24.735","Text":"Omega_1 times Omega 0 times R"},{"Start":"06:24.735 ","End":"06:33.763","Text":"minus m Omega sub"},{"Start":"06:33.763 ","End":"06:40.070","Text":"zero squared R. There\u0027s obviously a lot going on here,"},{"Start":"06:40.070 ","End":"06:41.825","Text":"but when we eliminate like terms,"},{"Start":"06:41.825 ","End":"06:43.220","Text":"it gets a lot simpler."},{"Start":"06:43.220 ","End":"06:46.625","Text":"First of all, 2m Omega 0,"},{"Start":"06:46.625 ","End":"06:56.330","Text":"Omega 1R is the same as 2m Omega 1 Omega 0 R. This and this can drop out."},{"Start":"06:56.330 ","End":"07:00.980","Text":"Next, we have m Omega_0^2 times R minus"},{"Start":"07:00.980 ","End":"07:05.720","Text":"2m Omega_0^2 R. We can assume that this is just negative"},{"Start":"07:05.720 ","End":"07:15.724","Text":"1m Omega 0 squared R. That is the same as negative m Omega_0^2 R. This and this dropout,"},{"Start":"07:15.724 ","End":"07:17.315","Text":"this is already dropped out."},{"Start":"07:17.315 ","End":"07:27.319","Text":"What we\u0027re left with is that T equals m Omega_1^2 R,"},{"Start":"07:27.319 ","End":"07:31.460","Text":"which is the exact same thing that we have here and here,"},{"Start":"07:31.460 ","End":"07:35.405","Text":"once we add in our mass that we forgot earlier, of course."},{"Start":"07:35.405 ","End":"07:40.565","Text":"This is just another proof that the Coriolis force and the centrifugal force,"},{"Start":"07:40.565 ","End":"07:42.230","Text":"help all of this work,"},{"Start":"07:42.230 ","End":"07:45.200","Text":"so that we can then get the same answers"},{"Start":"07:45.200 ","End":"07:49.495","Text":"regardless of whether we have a rotating observer."},{"Start":"07:49.495 ","End":"07:51.010","Text":"We\u0027ll put these in red,"},{"Start":"07:51.010 ","End":"07:52.610","Text":"because we need to remember this formula,"},{"Start":"07:52.610 ","End":"07:54.200","Text":"because the centrifugal force and"},{"Start":"07:54.200 ","End":"07:58.460","Text":"the Coriolis force are 2 fictitious forces in a rotating system,"},{"Start":"07:58.460 ","End":"08:01.650","Text":"make Newton\u0027s second law work."}],"ID":25639}],"Thumbnail":null,"ID":5352},{"Name":"Exercises with Coriolis and Centrifugal Forces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise - A Ship Firing a Shell","Duration":"10m 28s","ChapterTopicVideoID":9018,"CourseChapterTopicPlaylistID":5353,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.489","Text":"Hello. In this exercise,"},{"Start":"00:02.489 ","End":"00:06.225","Text":"we\u0027re given ship A which is sailing at latitude Lambda"},{"Start":"00:06.225 ","End":"00:11.145","Text":"and fires a shell at a velocity of v towards ship B,"},{"Start":"00:11.145 ","End":"00:14.970","Text":"which is sailing at a distance of d South of ship A."},{"Start":"00:14.970 ","End":"00:17.820","Text":"What we\u0027re asked to do is find the deviation of"},{"Start":"00:17.820 ","End":"00:21.930","Text":"the position of the shell caused by the Coriolis force."},{"Start":"00:21.930 ","End":"00:27.045","Text":"We\u0027re also given that the velocity of the Earth\u0027s rotation is Omega,"},{"Start":"00:27.045 ","End":"00:31.020","Text":"and that we can disregard the effect of the force on the East to"},{"Start":"00:31.020 ","End":"00:35.835","Text":"West velocity of the shell and on the orthogonal velocity relative to Earth."},{"Start":"00:35.835 ","End":"00:39.290","Text":"We\u0027re also told we can assume that the shell is moving in a straight line"},{"Start":"00:39.290 ","End":"00:44.009","Text":"and that we don\u0027t have to think of it in terms of ballistic trajectory."},{"Start":"00:44.450 ","End":"00:47.000","Text":"Let\u0027s begin. First of all,"},{"Start":"00:47.000 ","End":"00:54.080","Text":"we have a shell that\u0027s traveling North to South along the latitude line Lambda."},{"Start":"00:54.080 ","End":"00:57.020","Text":"It\u0027s going a distance of d,"},{"Start":"00:57.020 ","End":"00:59.225","Text":"and we\u0027ll disregard the effect of d,"},{"Start":"00:59.225 ","End":"01:00.750","Text":"the distance d,"},{"Start":"01:00.750 ","End":"01:02.935","Text":"on the angle Lambda."},{"Start":"01:02.935 ","End":"01:06.810","Text":"Now what I\u0027m going to do is assign my axis."},{"Start":"01:06.810 ","End":"01:10.560","Text":"I\u0027ll have y going away from the center of Earth,"},{"Start":"01:10.560 ","End":"01:13.760","Text":"that is orthogonal to the surface of the Earth,"},{"Start":"01:13.760 ","End":"01:17.584","Text":"and I\u0027ll have x going in the same direction,"},{"Start":"01:17.584 ","End":"01:21.480","Text":"North to South as our shell."},{"Start":"01:21.520 ","End":"01:26.040","Text":"Now let\u0027s start accounting for the Coriolis force."},{"Start":"01:26.950 ","End":"01:29.510","Text":"For the sake of review,"},{"Start":"01:29.510 ","End":"01:36.350","Text":"the formula for the Coriolis force is F equals negative 2m for mass,"},{"Start":"01:36.350 ","End":"01:38.960","Text":"Omega, cross multiplied by v,"},{"Start":"01:38.960 ","End":"01:41.575","Text":"or perhaps more accurately v tag."},{"Start":"01:41.575 ","End":"01:46.520","Text":"In this case, m is the mass of the object being measured, which is our shell."},{"Start":"01:46.520 ","End":"01:49.955","Text":"Omega is the angular velocity of the observer."},{"Start":"01:49.955 ","End":"01:53.360","Text":"That would be a person standing on Earth watching this happen,"},{"Start":"01:53.360 ","End":"01:57.910","Text":"and we can use the same Omega as the rotational velocity of Earth."},{"Start":"01:57.910 ","End":"02:01.586","Text":"V or v tag is the velocity of the object,"},{"Start":"02:01.586 ","End":"02:05.400","Text":"that is the shell, in the rotating system."},{"Start":"02:06.070 ","End":"02:10.145","Text":"Now, the mass m is not given to us in this problem,"},{"Start":"02:10.145 ","End":"02:11.825","Text":"but in the end it will drop out,"},{"Start":"02:11.825 ","End":"02:14.255","Text":"so we shouldn\u0027t be too concerned."},{"Start":"02:14.255 ","End":"02:19.235","Text":"Now what I want to do is execute this vector multiplication."},{"Start":"02:19.235 ","End":"02:22.430","Text":"There are 2 ways to do this."},{"Start":"02:22.430 ","End":"02:24.965","Text":"I can do this using the right-hand rule,"},{"Start":"02:24.965 ","End":"02:26.990","Text":"multiplying Omega by v,"},{"Start":"02:26.990 ","End":"02:30.440","Text":"the magnitude and finding the angle between the 2"},{"Start":"02:30.440 ","End":"02:34.805","Text":"or I can alternatively use a classic vector multiplication,"},{"Start":"02:34.805 ","End":"02:40.085","Text":"where I break down each vector into its components and multiply them by each other."},{"Start":"02:40.085 ","End":"02:43.840","Text":"In this example, I\u0027m going to multiply the 2 components,"},{"Start":"02:43.840 ","End":"02:46.640","Text":"but feel free to try the right-hand rule on your"},{"Start":"02:46.640 ","End":"02:50.345","Text":"own and make sure that your result equals the same result we find here."},{"Start":"02:50.345 ","End":"02:52.380","Text":"They should be equal."},{"Start":"02:53.030 ","End":"02:58.430","Text":"Using this method, we need to break down each of these vectors into its components."},{"Start":"02:58.430 ","End":"03:02.270","Text":"Let\u0027s start with v. V vector equals"},{"Start":"03:02.270 ","End":"03:06.470","Text":"the magnitude of the velocity v in the direction of the x-axis."},{"Start":"03:06.470 ","End":"03:08.690","Text":"Because if you recall, the object,"},{"Start":"03:08.690 ","End":"03:12.920","Text":"our shell is going in the exact same direction as the x-axis with"},{"Start":"03:12.920 ","End":"03:17.555","Text":"a magnitude of v. Going back to the information given in the question,"},{"Start":"03:17.555 ","End":"03:19.550","Text":"I want to assume that the shell is going in"},{"Start":"03:19.550 ","End":"03:22.685","Text":"a directly straight line and not with ballistic trajectory,"},{"Start":"03:22.685 ","End":"03:24.879","Text":"that would be an arc to trajectory."},{"Start":"03:24.879 ","End":"03:27.670","Text":"Also according to the information I\u0027m given,"},{"Start":"03:27.670 ","End":"03:31.450","Text":"I\u0027m going to disregard the force in the direction of the y-axis,"},{"Start":"03:31.450 ","End":"03:35.465","Text":"and I\u0027m going to disregard the forces in the direction of the z-axis,"},{"Start":"03:35.465 ","End":"03:38.420","Text":"that is the East to West axis as shown here."},{"Start":"03:38.420 ","End":"03:40.160","Text":"In terms of the Coriolis force,"},{"Start":"03:40.160 ","End":"03:42.395","Text":"it would have an effect going towards the East,"},{"Start":"03:42.395 ","End":"03:45.155","Text":"going away from us through the paper."},{"Start":"03:45.155 ","End":"03:48.590","Text":"Now I want to write out Omega as a vector."},{"Start":"03:48.590 ","End":"03:51.350","Text":"I know that if the rotation is to the right,"},{"Start":"03:51.350 ","End":"03:53.000","Text":"then using the right-hand rule,"},{"Start":"03:53.000 ","End":"03:56.705","Text":"Omega will be acting upwards on a vertical axis."},{"Start":"03:56.705 ","End":"04:00.650","Text":"I know this because if my fingers on my right hand are rotating"},{"Start":"04:00.650 ","End":"04:04.430","Text":"the same way that the force is,"},{"Start":"04:04.430 ","End":"04:07.860","Text":"then my thumb should be facing upwards."},{"Start":"04:08.210 ","End":"04:12.470","Text":"If our Omega vector points upwards like this and"},{"Start":"04:12.470 ","End":"04:16.460","Text":"our y vector is going from the center of the Earth outwards like so,"},{"Start":"04:16.460 ","End":"04:21.910","Text":"and the x vector is orthogonal to the y vector along the same line as the v vector,"},{"Start":"04:21.910 ","End":"04:25.935","Text":"our z vector will come at us like this,"},{"Start":"04:25.935 ","End":"04:29.775","Text":"and we\u0027ll explain why later in the problem,"},{"Start":"04:29.775 ","End":"04:34.315","Text":"and our Omega vector will point upwards like so."},{"Start":"04:34.315 ","End":"04:38.165","Text":"I\u0027m going to erase the z so we have a little more room here."},{"Start":"04:38.165 ","End":"04:45.550","Text":"Now I want to break this down into its components just like I did with v. First of all,"},{"Start":"04:45.550 ","End":"04:49.550","Text":"we know that if we were to continue this line here,"},{"Start":"04:49.550 ","End":"04:55.645","Text":"that the angle between the y-axis and the horizontal axis would be Lambda."},{"Start":"04:55.645 ","End":"04:58.040","Text":"If this angle is Lambda,"},{"Start":"04:58.040 ","End":"05:04.765","Text":"then that means that the angle between Omega and y is 90 degrees minus Lambda."},{"Start":"05:04.765 ","End":"05:11.550","Text":"That also means that the angle between Omega and x is Lambda."},{"Start":"05:12.880 ","End":"05:17.615","Text":"Broken down into components Omega equals,"},{"Start":"05:17.615 ","End":"05:21.800","Text":"Omega cosine Lambda in the direction of x,"},{"Start":"05:21.800 ","End":"05:25.790","Text":"plus Omega sine Lambda in the direction of y."},{"Start":"05:25.790 ","End":"05:28.970","Text":"We start with negative Omega in the direction of x,"},{"Start":"05:28.970 ","End":"05:32.795","Text":"because if you look here you\u0027ll see that Omega goes against the direction of x,"},{"Start":"05:32.795 ","End":"05:34.400","Text":"it goes in a negative direction."},{"Start":"05:34.400 ","End":"05:36.020","Text":"Whereas in terms of y,"},{"Start":"05:36.020 ","End":"05:37.795","Text":"it\u0027s going in a positive direction."},{"Start":"05:37.795 ","End":"05:42.695","Text":"Now that I have Omega and v in terms of their components,"},{"Start":"05:42.695 ","End":"05:46.462","Text":"I can do my cross multiplication, my vector multiplication."},{"Start":"05:46.462 ","End":"05:53.625","Text":"What I\u0027ll do is I\u0027m going to take Omega and cross multiply it by vx-hat."},{"Start":"05:53.625 ","End":"05:57.680","Text":"The way it works is exactly as if I was doing a normal multiplication."},{"Start":"05:57.680 ","End":"06:00.650","Text":"I take my x component multiplied by this,"},{"Start":"06:00.650 ","End":"06:02.930","Text":"and then my y component multiplied by this."},{"Start":"06:02.930 ","End":"06:05.525","Text":"When I multiply x-hat by x-hat,"},{"Start":"06:05.525 ","End":"06:08.855","Text":"it equals 0, and when I multiply y by x-hat,"},{"Start":"06:08.855 ","End":"06:10.685","Text":"I get negative z-hat."},{"Start":"06:10.685 ","End":"06:14.045","Text":"Remember, I\u0027ll put this in the corner as a note,"},{"Start":"06:14.045 ","End":"06:18.350","Text":"x-hat times y-hat equals z-hat,"},{"Start":"06:18.350 ","End":"06:24.765","Text":"and y-hat times x-hat equals negative z-hat."},{"Start":"06:24.765 ","End":"06:29.652","Text":"My solution is as follows,"},{"Start":"06:29.652 ","End":"06:32.745","Text":"negative 2m, that comes from above,"},{"Start":"06:32.745 ","End":"06:35.915","Text":"times Omega sine Lambda v,"},{"Start":"06:35.915 ","End":"06:41.055","Text":"which is the cross multiplication of my y component of Omega and my v vector,"},{"Start":"06:41.055 ","End":"06:43.395","Text":"in the direction of negative z."},{"Start":"06:43.395 ","End":"06:47.030","Text":"If you notice, things are only going in the direction of negative z,"},{"Start":"06:47.030 ","End":"06:51.685","Text":"meaning that f is only a force along the z-axis."},{"Start":"06:51.685 ","End":"06:58.371","Text":"My next step is to set this equal to the acceleration along the z-axis."},{"Start":"06:58.371 ","End":"07:03.390","Text":"You see I have 2m Omega sine Lambda v,"},{"Start":"07:03.390 ","End":"07:05.170","Text":"and I can forget that z-hat,"},{"Start":"07:05.170 ","End":"07:08.855","Text":"because we\u0027re only talking in terms of z equals ma_z."},{"Start":"07:08.855 ","End":"07:11.340","Text":"We can say that F_z=ma_z."},{"Start":"07:13.480 ","End":"07:15.634","Text":"If I look at these 2,"},{"Start":"07:15.634 ","End":"07:17.450","Text":"notice I have m on both sides."},{"Start":"07:17.450 ","End":"07:20.030","Text":"As I said before, the mass will fall out."},{"Start":"07:20.030 ","End":"07:23.390","Text":"What we\u0027re left with is 2 constants,"},{"Start":"07:23.390 ","End":"07:27.855","Text":"Omega and v. Lambda obviously is an angle measurement,"},{"Start":"07:27.855 ","End":"07:29.750","Text":"and that equals the acceleration."},{"Start":"07:29.750 ","End":"07:31.985","Text":"We know that the acceleration is constant,"},{"Start":"07:31.985 ","End":"07:34.580","Text":"and if we have constant acceleration,"},{"Start":"07:34.580 ","End":"07:39.210","Text":"then we can use a formula to find the position as a function of time."},{"Start":"07:39.350 ","End":"07:43.910","Text":"This formula will find us the position on the z-axis in terms"},{"Start":"07:43.910 ","End":"07:47.859","Text":"of time as long as we know that az is constant."},{"Start":"07:47.859 ","End":"07:49.535","Text":"Z(t) equals z_0,"},{"Start":"07:49.535 ","End":"07:54.965","Text":"plus v_0z t plus 1/2 a_zt^2."},{"Start":"07:54.965 ","End":"07:57.280","Text":"Now we know z_0=0,"},{"Start":"07:57.280 ","End":"08:00.435","Text":"our initial z position is 0."},{"Start":"08:00.435 ","End":"08:04.490","Text":"In addition our velocity along the z-axis is initially 0."},{"Start":"08:04.490 ","End":"08:07.800","Text":"We have no movement along the z-axis."},{"Start":"08:07.800 ","End":"08:12.720","Text":"What we\u0027re left with is 1/2a_zt^2."},{"Start":"08:12.720 ","End":"08:19.930","Text":"If you look one line above you\u0027ll notice that 2 Omega v sine Lambda equals a_ z."},{"Start":"08:19.930 ","End":"08:22.745","Text":"If we assume these 2 components are the same,"},{"Start":"08:22.745 ","End":"08:27.538","Text":"then if we multiply this by 1/2 and by t^2,"},{"Start":"08:27.538 ","End":"08:28.850","Text":"the 2 are equal."},{"Start":"08:28.850 ","End":"08:33.500","Text":"If I find out the time t that it takes my shell to go from A to B,"},{"Start":"08:33.500 ","End":"08:38.670","Text":"then I can find the deviation along the z-axis over the course of its flight."},{"Start":"08:38.670 ","End":"08:42.490","Text":"Now, what I need to do is figure out my axis real quick."},{"Start":"08:42.490 ","End":"08:45.365","Text":"If x is here and y is here,"},{"Start":"08:45.365 ","End":"08:48.530","Text":"I know that z is somehow perpendicular to x."},{"Start":"08:48.530 ","End":"08:53.435","Text":"If my fingers are curving in the direction from x to y like so,"},{"Start":"08:53.435 ","End":"08:56.000","Text":"that means that in accordance with the right hand"},{"Start":"08:56.000 ","End":"08:59.120","Text":"rule my thumb has to be coming back at me."},{"Start":"08:59.120 ","End":"09:01.895","Text":"Imagine that this hand is rotated"},{"Start":"09:01.895 ","End":"09:04.775","Text":"a little bit so that the thumb is coming back at your face,"},{"Start":"09:04.775 ","End":"09:07.985","Text":"out of the paper or out of your screen and back towards you."},{"Start":"09:07.985 ","End":"09:10.730","Text":"Perhaps it\u0027s easier to imagine if all your fingers are lined up"},{"Start":"09:10.730 ","End":"09:14.410","Text":"like this and the thumb is coming right back at us."},{"Start":"09:14.410 ","End":"09:19.520","Text":"The reason I did that is it\u0027s easier to solve t in terms of x."},{"Start":"09:19.520 ","End":"09:23.405","Text":"Using what we know about v along the x-axis,"},{"Start":"09:23.405 ","End":"09:29.885","Text":"we can solve for t. We know that the movement along the x-axis equals v times t,"},{"Start":"09:29.885 ","End":"09:32.480","Text":"that is the velocity times time."},{"Start":"09:32.480 ","End":"09:35.920","Text":"That\u0027s going to equal d, which is the distance traveled."},{"Start":"09:35.920 ","End":"09:37.899","Text":"Therefore, we know that t=d divided by"},{"Start":"09:37.899 ","End":"09:42.703","Text":"v. Imagine dividing v from each of these 2 sides,"},{"Start":"09:42.703 ","End":"09:44.995","Text":"and you end up with t=dv."},{"Start":"09:44.995 ","End":"09:51.885","Text":"What we can now do is replace t with d over v and solve a little easier."},{"Start":"09:51.885 ","End":"09:54.855","Text":"When we plug in d over v,"},{"Start":"09:54.855 ","End":"10:02.200","Text":"what we get is z equals Omega v sine Lambda times (d over v)^2."},{"Start":"10:02.200 ","End":"10:03.570","Text":"Because this is squared,"},{"Start":"10:03.570 ","End":"10:10.740","Text":"one of your v\u0027s will drop out and you end up with Omega d^2 over v sine Lambda."},{"Start":"10:10.740 ","End":"10:13.145","Text":"This is pretty much your final answer."},{"Start":"10:13.145 ","End":"10:17.192","Text":"It\u0027ll tell you how much our projectile, our shell,"},{"Start":"10:17.192 ","End":"10:22.520","Text":"will deviate along the z-axis over the course of d distance."},{"Start":"10:22.520 ","End":"10:24.635","Text":"If I was a sailor on ship A,"},{"Start":"10:24.635 ","End":"10:26.780","Text":"I would need to account for this deviation."},{"Start":"10:26.780 ","End":"10:29.610","Text":"With that, we end the exercise."}],"ID":9292},{"Watched":false,"Name":"Exercise - River","Duration":"10m 22s","ChapterTopicVideoID":9019,"CourseChapterTopicPlaylistID":5353,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.740 ","End":"00:05.415","Text":"Let\u0027s jump into an exercise dealing with the Coriolis force."},{"Start":"00:05.415 ","End":"00:07.620","Text":"In this exercise, we\u0027re given Earth,"},{"Start":"00:07.620 ","End":"00:13.120","Text":"our home planet, and it has a radius of RE."},{"Start":"00:13.370 ","End":"00:19.740","Text":"On the surface of Earth is a river and it\u0027s flowing in this direction with"},{"Start":"00:19.740 ","End":"00:25.185","Text":"a velocity of v. It has an angle with relation to the core of the Earth,"},{"Start":"00:25.185 ","End":"00:28.360","Text":"the center of the earth of Theta."},{"Start":"00:29.240 ","End":"00:33.840","Text":"Now, of course, the earth is rotating around its center in"},{"Start":"00:33.840 ","End":"00:38.250","Text":"this direction and the rotational velocity is given,"},{"Start":"00:38.250 ","End":"00:43.870","Text":"omega that is, as 2 Pi over 24 hours."},{"Start":"00:43.870 ","End":"00:47.990","Text":"The goal of this exercise is to find the difference in"},{"Start":"00:47.990 ","End":"00:53.615","Text":"height between the two banks of the river."},{"Start":"00:53.615 ","End":"00:57.440","Text":"If you imagine that we\u0027re sitting in the river with our feet in"},{"Start":"00:57.440 ","End":"01:02.675","Text":"the water and the water is flowing at us and it comes at this angle."},{"Start":"01:02.675 ","End":"01:08.315","Text":"Now, these two lines here represent the banks of the river."},{"Start":"01:08.315 ","End":"01:11.315","Text":"If this still isn\u0027t clear in terms of the perspective,"},{"Start":"01:11.315 ","End":"01:13.820","Text":"we\u0027re looking based on the width of the river here."},{"Start":"01:13.820 ","End":"01:17.630","Text":"Imagine that you\u0027re standing in the river and the water is going"},{"Start":"01:17.630 ","End":"01:22.185","Text":"through the screen or through your paper towards the back of the paper."},{"Start":"01:22.185 ","End":"01:27.845","Text":"Our goal is to find out the angle of the water between the two banks."},{"Start":"01:27.845 ","End":"01:32.040","Text":"This angle and we\u0027re going to call this angle Phi."},{"Start":"01:33.670 ","End":"01:37.250","Text":"Now that we understand the essence of our problem,"},{"Start":"01:37.250 ","End":"01:39.815","Text":"the next step is to figure out how to solve it."},{"Start":"01:39.815 ","End":"01:41.780","Text":"In order to find the angle Phi,"},{"Start":"01:41.780 ","End":"01:45.590","Text":"we need to figure out which direction the water should be facing."},{"Start":"01:45.590 ","End":"01:47.105","Text":"One way we can do that,"},{"Start":"01:47.105 ","End":"01:49.910","Text":"there are a couple of ways to do that is to think"},{"Start":"01:49.910 ","End":"01:53.630","Text":"which direction the sum of forces is facing."},{"Start":"01:53.630 ","End":"01:54.950","Text":"If the sum of forces,"},{"Start":"01:54.950 ","End":"01:57.635","Text":"in this case, it\u0027s facing something like this."},{"Start":"01:57.635 ","End":"02:00.410","Text":"We know that the water has to be at a right angle,"},{"Start":"02:00.410 ","End":"02:03.035","Text":"perpendicular, that is to the sum of forces."},{"Start":"02:03.035 ","End":"02:05.045","Text":"You can think about this intuitively."},{"Start":"02:05.045 ","End":"02:07.250","Text":"Imagine up in the corner here,"},{"Start":"02:07.250 ","End":"02:11.540","Text":"that we have water and the force of gravity"},{"Start":"02:11.540 ","End":"02:15.995","Text":"is pointing downwards while the water is of course at a 90-degree angle with that."},{"Start":"02:15.995 ","End":"02:18.920","Text":"But if the force of gravity was somehow in"},{"Start":"02:18.920 ","End":"02:23.850","Text":"this direction then the water would have to flow like this."},{"Start":"02:24.220 ","End":"02:26.360","Text":"In order to figure this out,"},{"Start":"02:26.360 ","End":"02:28.895","Text":"we have to first assign our axis."},{"Start":"02:28.895 ","End":"02:31.610","Text":"Let\u0027s start by saying that the y-axis is"},{"Start":"02:31.610 ","End":"02:34.160","Text":"the continuation of the radius going outwards from"},{"Start":"02:34.160 ","End":"02:39.650","Text":"the center of Earth and the x-axis is going in the same direction as the water."},{"Start":"02:39.650 ","End":"02:42.530","Text":"There\u0027s a y-axis and there\u0027s a x-axis and"},{"Start":"02:42.530 ","End":"02:47.810","Text":"the z-axis is coming straight at us through the paper and towards our faces."},{"Start":"02:47.810 ","End":"02:50.690","Text":"Transposing this to our drawing on the right,"},{"Start":"02:50.690 ","End":"02:54.020","Text":"that means that our y-axis has to be vertical,"},{"Start":"02:54.020 ","End":"02:56.800","Text":"coming from the center of the Earth going upwards."},{"Start":"02:56.800 ","End":"03:00.350","Text":"Our x-axis will be the direction that the water is flowing."},{"Start":"03:00.350 ","End":"03:04.750","Text":"Imagine that the water is flowing through the screen away from us."},{"Start":"03:04.750 ","End":"03:10.670","Text":"The z-axis is in that case going to be going towards the right."},{"Start":"03:10.670 ","End":"03:14.960","Text":"Whereas before the z-axis was coming at us from this perspective on"},{"Start":"03:14.960 ","End":"03:19.400","Text":"the left and the x-axis was going downwards at an angle."},{"Start":"03:19.400 ","End":"03:21.680","Text":"In terms of our picture on the right,"},{"Start":"03:21.680 ","End":"03:23.720","Text":"the y-axis is going upwards,"},{"Start":"03:23.720 ","End":"03:25.190","Text":"the z-axis to the right,"},{"Start":"03:25.190 ","End":"03:28.210","Text":"and the x-axis away from us."},{"Start":"03:28.210 ","End":"03:30.995","Text":"Now let\u0027s start drawing out our forces."},{"Start":"03:30.995 ","End":"03:33.365","Text":"The first force is obviously gravity."},{"Start":"03:33.365 ","End":"03:37.295","Text":"Gravity is going towards the center of the Earth along the y-axis."},{"Start":"03:37.295 ","End":"03:40.100","Text":"This is mg. Next,"},{"Start":"03:40.100 ","End":"03:42.665","Text":"because we\u0027re dealing with a rotating system,"},{"Start":"03:42.665 ","End":"03:46.310","Text":"the Earth, we need to now add in our fictitious forces."},{"Start":"03:46.310 ","End":"03:48.625","Text":"We have to add in the centrifugal force."},{"Start":"03:48.625 ","End":"03:53.060","Text":"We know the centrifugal force is m Omega^2r."},{"Start":"03:53.060 ","End":"03:56.450","Text":"That it\u0027s going to be acting in this general direction."},{"Start":"03:56.450 ","End":"03:59.810","Text":"m Omega^2r."},{"Start":"03:59.810 ","End":"04:02.090","Text":"The reason we know that it will be acting in"},{"Start":"04:02.090 ","End":"04:05.450","Text":"this direction is that the centrifugal force will act along"},{"Start":"04:05.450 ","End":"04:08.900","Text":"the latitudinal line with the radius r slightly smaller"},{"Start":"04:08.900 ","End":"04:12.530","Text":"than the radius Re that goes like this."},{"Start":"04:12.530 ","End":"04:16.670","Text":"This is the rotation along this point latitudinally of the Earth."},{"Start":"04:16.670 ","End":"04:21.770","Text":"That gives us the centrifugal force in this direction, m Omega^2r."},{"Start":"04:21.770 ","End":"04:25.190","Text":"Now our last force is the Coriolis force."},{"Start":"04:25.190 ","End":"04:28.620","Text":"F_c, the Coriolis force=mv,"},{"Start":"04:30.770 ","End":"04:36.390","Text":"vector cross multiplied by Omega vector."},{"Start":"04:36.390 ","End":"04:37.900","Text":"Now we know that the v,"},{"Start":"04:37.900 ","End":"04:41.780","Text":"vector goes in the direction of the x-axis and"},{"Start":"04:41.780 ","End":"04:46.270","Text":"the relevant aspect of the Omega vector goes in the direction of the y-axis."},{"Start":"04:46.270 ","End":"04:52.210","Text":"If we do a cross multiplication of vector multiplication we\u0027ll end up with the F_c,"},{"Start":"04:52.210 ","End":"04:56.855","Text":"the Coriolis force coming in the direction of z out towards us."},{"Start":"04:56.855 ","End":"04:59.440","Text":"Now that we know the direction of F_c,"},{"Start":"04:59.440 ","End":"05:02.630","Text":"the Coriolis force, we can now deal with the magnitude."},{"Start":"05:02.630 ","End":"05:10.220","Text":"The magnitude of the Coriolis force is going to equal to m times the absolute value of"},{"Start":"05:10.220 ","End":"05:13.520","Text":"v times also the magnitude or"},{"Start":"05:13.520 ","End":"05:18.700","Text":"absolute value of Omega times the Sine of the angle between the two of them."},{"Start":"05:18.700 ","End":"05:20.690","Text":"We\u0027ll call that angle Alpha."},{"Start":"05:20.690 ","End":"05:26.465","Text":"Now we need to think of this angle Alpha in terms that are easier for us to understand."},{"Start":"05:26.465 ","End":"05:29.855","Text":"What is the angle between Omega and v?"},{"Start":"05:29.855 ","End":"05:33.800","Text":"Well, we know that v is in direction of the x-axis."},{"Start":"05:33.800 ","End":"05:38.780","Text":"This is the direction of v. We know that Omega is going along"},{"Start":"05:38.780 ","End":"05:44.840","Text":"the central pole here because Omega is always along the axis of rotation."},{"Start":"05:44.840 ","End":"05:46.475","Text":"This is Omega here."},{"Start":"05:46.475 ","End":"05:51.035","Text":"We can think of the angle between the two as this angle here."},{"Start":"05:51.035 ","End":"05:59.530","Text":"Now, we know that this angle using some basic trigonometry is 90 degrees minus Theta."},{"Start":"05:59.530 ","End":"06:01.140","Text":"We have here Theta,"},{"Start":"06:01.140 ","End":"06:03.575","Text":"we have 90 minus Theta here."},{"Start":"06:03.575 ","End":"06:09.415","Text":"Once again, here we have Theta and here we have 90 minus Theta."},{"Start":"06:09.415 ","End":"06:13.955","Text":"Knowing that this angle Alpha is 90 minus Theta helps us out a lot"},{"Start":"06:13.955 ","End":"06:18.785","Text":"because the sine of 90 minus Theta equals the cosine of Theta."},{"Start":"06:18.785 ","End":"06:21.095","Text":"We can think of the same thing."},{"Start":"06:21.095 ","End":"06:25.820","Text":"The absolute value or magnitude of F_c that is as"},{"Start":"06:25.820 ","End":"06:35.350","Text":"2m v times Omega times the cosine of Theta."},{"Start":"06:36.260 ","End":"06:41.865","Text":"Now we know that this F_c is in direction of z."},{"Start":"06:41.865 ","End":"06:45.500","Text":"Now we can list out our forces using a force equation."},{"Start":"06:45.500 ","End":"06:49.714","Text":"Let\u0027s start with the sum of forces along the y axis."},{"Start":"06:49.714 ","End":"06:52.805","Text":"Along the y-axis, we have negative mg."},{"Start":"06:52.805 ","End":"06:54.350","Text":"This is the force of gravity."},{"Start":"06:54.350 ","End":"06:57.950","Text":"We\u0027re going to add to that m Omega^2r."},{"Start":"06:57.950 ","End":"07:02.870","Text":"But we need to first transpose that from this axis up to the y-axis."},{"Start":"07:02.870 ","End":"07:05.570","Text":"Now if we look below the centrifugal force,"},{"Start":"07:05.570 ","End":"07:11.270","Text":"that is m Omega^2r is going in the same direction as our axis here."},{"Start":"07:11.270 ","End":"07:17.275","Text":"We know that the angle between the two is 90 degrees minus Theta."},{"Start":"07:17.275 ","End":"07:19.610","Text":"Doing some basic geometry,"},{"Start":"07:19.610 ","End":"07:27.145","Text":"we find that we add m Omega^2r times the sine of Theta."},{"Start":"07:27.145 ","End":"07:29.900","Text":"We can simplify things even a little further by taking"},{"Start":"07:29.900 ","End":"07:33.835","Text":"r from our solution and realizing that it\u0027s this radius here."},{"Start":"07:33.835 ","End":"07:37.450","Text":"That this radius equals Re,"},{"Start":"07:37.450 ","End":"07:38.915","Text":"the radius of the Earth,"},{"Start":"07:38.915 ","End":"07:41.970","Text":"times the sine of Theta."},{"Start":"07:42.370 ","End":"07:46.895","Text":"These are the forces in the direction of the y-axis."},{"Start":"07:46.895 ","End":"07:49.640","Text":"Now, the x-axis isn\u0027t really relevant to"},{"Start":"07:49.640 ","End":"07:52.220","Text":"us here because of the direction of x relative to us,"},{"Start":"07:52.220 ","End":"07:53.420","Text":"it\u0027s going with the river."},{"Start":"07:53.420 ","End":"07:57.110","Text":"However, we do want to know the sum of the forces along"},{"Start":"07:57.110 ","End":"08:00.990","Text":"the z-axis and that is just the Coriolis force."},{"Start":"08:00.990 ","End":"08:06.074","Text":"What we have is 2m times v"},{"Start":"08:06.074 ","End":"08:13.240","Text":"Omega times cosine of Theta in the direction of z."},{"Start":"08:13.310 ","End":"08:17.885","Text":"Now let\u0027s bring this up to our diagram in the upper right-hand side."},{"Start":"08:17.885 ","End":"08:23.015","Text":"If the forces in the direction of the z-axis are going to a right like this."},{"Start":"08:23.015 ","End":"08:27.530","Text":"This is the sum of the forces z and the sum of"},{"Start":"08:27.530 ","End":"08:32.645","Text":"the forces in the direction of the y-axis ultimately point downwards like so."},{"Start":"08:32.645 ","End":"08:34.955","Text":"Sum of the forces y,"},{"Start":"08:34.955 ","End":"08:39.020","Text":"then the sum of total forces has to be something like this."},{"Start":"08:39.020 ","End":"08:42.454","Text":"The sum of total forces."},{"Start":"08:42.454 ","End":"08:45.530","Text":"Now because the sum of total forces has to be at"},{"Start":"08:45.530 ","End":"08:49.835","Text":"a 90-degree angle with the angle of the surface of the water."},{"Start":"08:49.835 ","End":"08:52.055","Text":"We know that this angle is Phi,"},{"Start":"08:52.055 ","End":"08:55.640","Text":"then essentially what we\u0027re doing is because y goes downwards instead of upwards,"},{"Start":"08:55.640 ","End":"08:57.995","Text":"we know that this is also Phi."},{"Start":"08:57.995 ","End":"09:00.230","Text":"In order to solve for Phi,"},{"Start":"09:00.230 ","End":"09:02.120","Text":"what we\u0027re going to do is down here,"},{"Start":"09:02.120 ","End":"09:07.160","Text":"take the tangent of Phi and that equals the sum of"},{"Start":"09:07.160 ","End":"09:13.565","Text":"the forces along the z-axis over the sum of forces along the y-axis."},{"Start":"09:13.565 ","End":"09:20.510","Text":"That equals 2mv Omega cosine Theta."},{"Start":"09:20.510 ","End":"09:25.670","Text":"That\u0027s the sum of the forces along the z-axis divided by"},{"Start":"09:25.670 ","End":"09:36.175","Text":"negative mg plus m Omega^2 and instead of r we\u0027re going to put in Re."},{"Start":"09:36.175 ","End":"09:42.235","Text":"Now because we have two sines of Theta, sine squared Theta."},{"Start":"09:42.235 ","End":"09:46.055","Text":"Next we need to see if the solution really makes sense."},{"Start":"09:46.055 ","End":"09:49.160","Text":"Well, so what we\u0027re going to do is imagine a scenario where"},{"Start":"09:49.160 ","End":"09:53.125","Text":"our river is parallel to the central pole of the earth."},{"Start":"09:53.125 ","End":"09:55.115","Text":"v is in this direction."},{"Start":"09:55.115 ","End":"09:57.050","Text":"Now what should happen is there should be"},{"Start":"09:57.050 ","End":"10:01.105","Text":"no difference of the height between the two banks of the river."},{"Start":"10:01.105 ","End":"10:03.455","Text":"That means this has to equal 0."},{"Start":"10:03.455 ","End":"10:05.135","Text":"Now at this point,"},{"Start":"10:05.135 ","End":"10:08.135","Text":"the angle Theta equals Pi over 2."},{"Start":"10:08.135 ","End":"10:10.645","Text":"If Theta equals Pi over 2,"},{"Start":"10:10.645 ","End":"10:14.060","Text":"then the sum of forces along the z-axis zeros out,"},{"Start":"10:14.060 ","End":"10:16.340","Text":"which makes sense there should be no Coriolis effect."},{"Start":"10:16.340 ","End":"10:19.100","Text":"That means that no matter what our bottom equals,"},{"Start":"10:19.100 ","End":"10:22.260","Text":"everything zeros out and we should be okay."}],"ID":9293}],"Thumbnail":null,"ID":5353}]

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