Drag- Explanation and Example with a Skydiver
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Buoyancy
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Stoke's Law
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[{"Name":"Drag- Explanation and Example with a Skydiver","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation","Duration":"2m 5s","ChapterTopicVideoID":9021,"CourseChapterTopicPlaylistID":5334,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.110 ","End":"00:02.670","Text":"Hello. In this lecture,"},{"Start":"00:02.670 ","End":"00:04.620","Text":"we\u0027re going to talk about the drag force."},{"Start":"00:04.620 ","End":"00:08.175","Text":"I\u0027ll give you an explanation and then we\u0027ll walk through an example together."},{"Start":"00:08.175 ","End":"00:11.730","Text":"Drag force can be described with the following formula here."},{"Start":"00:11.730 ","End":"00:13.740","Text":"That is negative k, first,"},{"Start":"00:13.740 ","End":"00:17.055","Text":"remember it has to be negative and k is a given constant."},{"Start":"00:17.055 ","End":"00:19.245","Text":"We\u0027ll talk more about that later."},{"Start":"00:19.245 ","End":"00:23.360","Text":"That will be multiplied by V, the velocity vector."},{"Start":"00:23.360 ","End":"00:25.595","Text":"Usually, when we use drag force,"},{"Start":"00:25.595 ","End":"00:28.730","Text":"we\u0027re talking about some friction in the case of the parachute,"},{"Start":"00:28.730 ","End":"00:30.275","Text":"or that\u0027s air resistance."},{"Start":"00:30.275 ","End":"00:32.795","Text":"In other cases, we\u0027ll oftentimes talk about"},{"Start":"00:32.795 ","End":"00:36.155","Text":"resistance within water or some other liquid medium."},{"Start":"00:36.155 ","End":"00:38.210","Text":"When describing drag force,"},{"Start":"00:38.210 ","End":"00:43.205","Text":"keep in mind that the constant k is reliant on a lot of different factors."},{"Start":"00:43.205 ","End":"00:47.060","Text":"In our example, something important for the parachute might be"},{"Start":"00:47.060 ","End":"00:51.780","Text":"the shape or the size or the structure of the parachute itself."},{"Start":"00:59.300 ","End":"01:03.095","Text":"We need to incorporate this negative sign because"},{"Start":"01:03.095 ","End":"01:06.700","Text":"our drag force is acting in the opposite direction, the velocity."},{"Start":"01:06.700 ","End":"01:08.885","Text":"If our parachute is falling,"},{"Start":"01:08.885 ","End":"01:12.470","Text":"that means that our drag force is acting in an upwards direction."},{"Start":"01:12.470 ","End":"01:13.985","Text":"This is the force of the drag."},{"Start":"01:13.985 ","End":"01:19.010","Text":"We also need to make sure not to confuse between drag force and other forms of friction,"},{"Start":"01:19.010 ","End":"01:21.050","Text":"traditional friction, if you will."},{"Start":"01:21.050 ","End":"01:23.720","Text":"If you recall, traditional friction is a constant."},{"Start":"01:23.720 ","End":"01:27.065","Text":"It always equals Mu k or Fk."},{"Start":"01:27.065 ","End":"01:29.570","Text":"But when we talk about drag force,"},{"Start":"01:29.570 ","End":"01:33.785","Text":"the drag force changes based on the velocity of the given object."},{"Start":"01:33.785 ","End":"01:35.675","Text":"As the velocity increases,"},{"Start":"01:35.675 ","End":"01:37.145","Text":"so does the drag force."},{"Start":"01:37.145 ","End":"01:38.960","Text":"Whereas with traditional friction,"},{"Start":"01:38.960 ","End":"01:42.775","Text":"the friction remains the same regardless of the velocity of the object."},{"Start":"01:42.775 ","End":"01:49.765","Text":"Remember that f_k our traditional friction equals always Mu kN,"},{"Start":"01:49.765 ","End":"01:54.970","Text":"whereas our drag force equals this negative kV."},{"Start":"01:54.970 ","End":"02:00.410","Text":"In other words, Fk are normal version of friction is constant,"},{"Start":"02:00.410 ","End":"02:01.610","Text":"whereas this F,"},{"Start":"02:01.610 ","End":"02:05.790","Text":"drag force is not constant. It can change."}],"ID":9294},{"Watched":false,"Name":"Example Part A","Duration":"3m 33s","ChapterTopicVideoID":9022,"CourseChapterTopicPlaylistID":5334,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.100","Text":"Now that we understand the drag force,"},{"Start":"00:03.100 ","End":"00:07.000","Text":"let\u0027s analyze the skydiver\u0027s motion and utilize drag force."},{"Start":"00:07.000 ","End":"00:11.050","Text":"If we\u0027re going to analyze the skydiver\u0027s motion to find the equation of motion."},{"Start":"00:11.050 ","End":"00:15.565","Text":"First, we need to do a free body diagram and we\u0027re going to count for gravity,"},{"Start":"00:15.565 ","End":"00:18.355","Text":"that is mg, and that\u0027ll be going downwards."},{"Start":"00:18.355 ","End":"00:19.780","Text":"Let\u0027s assume that M,"},{"Start":"00:19.780 ","End":"00:23.290","Text":"the mass of the parachute or of the skydiver, is given to us."},{"Start":"00:23.290 ","End":"00:26.030","Text":"Now we need to account for the drag force."},{"Start":"00:26.030 ","End":"00:29.140","Text":"Oftentimes students will confuse themselves"},{"Start":"00:29.140 ","End":"00:32.260","Text":"by trying to draw the drag force in a negative direction from"},{"Start":"00:32.260 ","End":"00:34.960","Text":"the start are trying to figure out its direction before"},{"Start":"00:34.960 ","End":"00:38.695","Text":"figuring out the real forces that are acting upon our object."},{"Start":"00:38.695 ","End":"00:42.450","Text":"What I would suggest is first choose a positive direction"},{"Start":"00:42.450 ","End":"00:46.310","Text":"and start by having your forces go in that positive direction."},{"Start":"00:46.310 ","End":"00:48.380","Text":"Then when you use your equations,"},{"Start":"00:48.380 ","End":"00:51.620","Text":"you\u0027ll find that the forces write themselves, so to speak."},{"Start":"00:51.620 ","End":"00:57.780","Text":"For example, let\u0027s say that the y our vertical axis is going downwards."},{"Start":"00:57.780 ","End":"01:00.185","Text":"Downwards is our positive direction."},{"Start":"01:00.185 ","End":"01:05.345","Text":"In that case, we\u0027re going to just throw mg in that positive direction as well and F_D,"},{"Start":"01:05.345 ","End":"01:07.430","Text":"which will be our drag force,"},{"Start":"01:07.430 ","End":"01:09.950","Text":"will also go in a positive direction."},{"Start":"01:09.950 ","End":"01:11.570","Text":"Now let\u0027s check if this works."},{"Start":"01:11.570 ","End":"01:13.820","Text":"If mg is positive and in fact,"},{"Start":"01:13.820 ","End":"01:17.915","Text":"it is in real life the skydiver will be falling downwards,"},{"Start":"01:17.915 ","End":"01:23.015","Text":"then negative Kv is going to throw us in a negative direction and ultimately"},{"Start":"01:23.015 ","End":"01:28.100","Text":"F_D will force upwards and that is in fact correct as you can see here,"},{"Start":"01:28.100 ","End":"01:30.680","Text":"the equation is right at itself."},{"Start":"01:30.680 ","End":"01:33.350","Text":"If this is still confusing for you,"},{"Start":"01:33.350 ","End":"01:37.565","Text":"the simplest thing to do is not worry about direction at all and diagrams,"},{"Start":"01:37.565 ","End":"01:42.655","Text":"but just keep your negative sign in the equations and things should work out correctly."},{"Start":"01:42.655 ","End":"01:48.740","Text":"Moving on, let\u0027s look at part A here and try to find the equation of motion."},{"Start":"01:48.740 ","End":"01:53.540","Text":"To remind you, when we\u0027re looking in part A for the equation of motion,"},{"Start":"01:53.540 ","End":"01:56.525","Text":"we\u0027re not looking for something in terms of t,"},{"Start":"01:56.525 ","End":"02:00.045","Text":"y, and time."},{"Start":"02:00.045 ","End":"02:04.775","Text":"Rather we\u0027re looking for something that is in terms of y, v,"},{"Start":"02:04.775 ","End":"02:07.025","Text":"that is velocity and acceleration,"},{"Start":"02:07.025 ","End":"02:10.580","Text":"or put in other words y and its derivatives."},{"Start":"02:10.580 ","End":"02:15.275","Text":"This doesn\u0027t necessarily mean that we need to use y, v, and a,"},{"Start":"02:15.275 ","End":"02:19.250","Text":"but we should have something in terms of y and v or in terms of y and a,"},{"Start":"02:19.250 ","End":"02:21.380","Text":"or in terms of v and a."},{"Start":"02:21.380 ","End":"02:23.795","Text":"When we have all those things together,"},{"Start":"02:23.795 ","End":"02:28.760","Text":"we can then solve and the solution will be something y in terms of t. This"},{"Start":"02:28.760 ","End":"02:34.690","Text":"is the solution of the equation of motion."},{"Start":"02:34.730 ","End":"02:39.150","Text":"Now let\u0027s move on and find the equation of motion."},{"Start":"02:39.640 ","End":"02:45.200","Text":"In general, the way I\u0027m going to find the equation of motion is using the sum of forces."},{"Start":"02:45.200 ","End":"02:48.935","Text":"I want to list out my forces and from there I find my solution."},{"Start":"02:48.935 ","End":"02:53.240","Text":"In this case, we\u0027re going to do some of F_y because we\u0027re talking about a vertical fall."},{"Start":"02:53.240 ","End":"02:56.705","Text":"We\u0027re not accounting for the x-axis or the z-axis."},{"Start":"02:56.705 ","End":"03:00.380","Text":"The sum of F_y=mg,"},{"Start":"03:00.380 ","End":"03:02.340","Text":"which is the gravity of course,"},{"Start":"03:02.340 ","End":"03:05.780","Text":"and negative Kv, and of course,"},{"Start":"03:05.780 ","End":"03:08.555","Text":"this v is in the direction of y,"},{"Start":"03:08.555 ","End":"03:11.750","Text":"and that will equal may."},{"Start":"03:11.750 ","End":"03:16.090","Text":"Again, we\u0027re using Newton\u0027s second law to find our answer here."},{"Start":"03:16.090 ","End":"03:18.965","Text":"Now as I write this y a little more clearly,"},{"Start":"03:18.965 ","End":"03:21.710","Text":"you can see that I have an equation in terms of"},{"Start":"03:21.710 ","End":"03:25.625","Text":"v and in terms of a and while I don\u0027t have y itself here,"},{"Start":"03:25.625 ","End":"03:28.835","Text":"I do have an equation in terms of v and a is this qualifies"},{"Start":"03:28.835 ","End":"03:33.540","Text":"as an equation of motion and I\u0027ve solved Part A."}],"ID":9295},{"Watched":false,"Name":"Example Part B","Duration":"1m 11s","ChapterTopicVideoID":9023,"CourseChapterTopicPlaylistID":5334,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.155","Text":"In Part B, I\u0027m asked to find my terminal velocity."},{"Start":"00:04.155 ","End":"00:06.045","Text":"Now I can use a couple of tricks here."},{"Start":"00:06.045 ","End":"00:10.545","Text":"First of all, I know that my terminal velocity is a constant velocity."},{"Start":"00:10.545 ","End":"00:12.659","Text":"If the velocity is constant,"},{"Start":"00:12.659 ","End":"00:18.945","Text":"that means that the derivative of the velocity, that is a=0."},{"Start":"00:18.945 ","End":"00:21.045","Text":"If I know that a=0,"},{"Start":"00:21.045 ","End":"00:23.580","Text":"I can plug this into my equation above,"},{"Start":"00:23.580 ","End":"00:25.695","Text":"that is my equation of motion,"},{"Start":"00:25.695 ","End":"00:27.435","Text":"and find the results."},{"Start":"00:27.435 ","End":"00:32.550","Text":"When I have a=0 up here may goes to 0,"},{"Start":"00:32.550 ","End":"00:40.170","Text":"a nd what I have is mg minus kvy=0."},{"Start":"00:40.170 ","End":"00:42.495","Text":"If I want to solve for vy,"},{"Start":"00:42.495 ","End":"00:45.379","Text":"what I have now is vy final,"},{"Start":"00:45.379 ","End":"00:53.280","Text":"that is my terminal velocity equals mg divided by k, and there we\u0027ve solved it."},{"Start":"00:53.280 ","End":"00:57.350","Text":"Our terminal velocity equals mg over k. Again,"},{"Start":"00:57.350 ","End":"01:00.874","Text":"the trick to remember here is that when looking for terminal velocity,"},{"Start":"01:00.874 ","End":"01:06.445","Text":"set a=0 and plug that in to your equation of motion."},{"Start":"01:06.445 ","End":"01:09.300","Text":"Now that I\u0027ve solved Part B,"},{"Start":"01:09.300 ","End":"01:11.770","Text":"Let\u0027s move on to Part C."}],"ID":9296},{"Watched":false,"Name":"Example Part C","Duration":"10m 52s","ChapterTopicVideoID":9024,"CourseChapterTopicPlaylistID":5334,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"In part C, I\u0027m asked to find"},{"Start":"00:03.480 ","End":"00:07.650","Text":"the velocity as a function of time if the skydivers starts at rest."},{"Start":"00:07.650 ","End":"00:11.890","Text":"Really, what I need to do is solve the differential equation,"},{"Start":"00:11.890 ","End":"00:16.660","Text":"and we\u0027ll find in a second that this is a differential equation of motion."},{"Start":"00:17.510 ","End":"00:20.535","Text":"The question is, how do we do this?"},{"Start":"00:20.535 ","End":"00:21.720","Text":"We\u0027ll use a little trick here,"},{"Start":"00:21.720 ","End":"00:27.420","Text":"which is to remember that ay=dvy, dt."},{"Start":"00:27.420 ","End":"00:32.115","Text":"We can now convert this into a differential equation and solve."},{"Start":"00:32.115 ","End":"00:35.070","Text":"If we plug this into our equation,"},{"Start":"00:35.070 ","End":"00:38.190","Text":"we get mg minus"},{"Start":"00:38.190 ","End":"00:44.410","Text":"kvy=m dvy,"},{"Start":"00:46.250 ","End":"00:48.840","Text":"dt."},{"Start":"00:48.840 ","End":"00:51.830","Text":"The way we approach this resulting equation is using"},{"Start":"00:51.830 ","End":"00:55.520","Text":"the method of separation of variables."},{"Start":"00:55.520 ","End":"01:01.295","Text":"I want to separate my variables to get an equation with the following general form."},{"Start":"01:01.295 ","End":"01:03.835","Text":"F(t) times"},{"Start":"01:03.835 ","End":"01:13.700","Text":"dt=f(v)dv. Really,"},{"Start":"01:13.700 ","End":"01:16.580","Text":"what I\u0027m doing is separating all of my t variables to"},{"Start":"01:16.580 ","End":"01:20.440","Text":"the left side and my v variables to the right side,"},{"Start":"01:20.440 ","End":"01:24.080","Text":"and the way I\u0027ll do that is by multiplying first by dt."},{"Start":"01:24.080 ","End":"01:27.560","Text":"Step 1, multiply by dt."},{"Start":"01:27.560 ","End":"01:29.270","Text":"What I end up with is"},{"Start":"01:29.270 ","End":"01:36.980","Text":"mg minus kvy"},{"Start":"01:36.980 ","End":"01:45.140","Text":"multiplied by dt=mdvy."},{"Start":"01:45.140 ","End":"01:49.310","Text":"The next step is to divide by the entire coefficient preceding dt."},{"Start":"01:49.310 ","End":"01:54.350","Text":"The idea here is to clear this side of anything based in v"},{"Start":"01:54.350 ","End":"02:00.245","Text":"as opposed to t. I\u0027ll divide in our case by mg minus kvy."},{"Start":"02:00.245 ","End":"02:04.475","Text":"In a related note, if I have any t variables on my right side,"},{"Start":"02:04.475 ","End":"02:05.855","Text":"which I don\u0027t in this case,"},{"Start":"02:05.855 ","End":"02:10.010","Text":"I would also have to divide by them to move them to the t side."},{"Start":"02:10.010 ","End":"02:19.160","Text":"In our case, step 2 is to divide by mg minus kvy."},{"Start":"02:19.160 ","End":"02:26.850","Text":"What I end up with is dt=m divided"},{"Start":"02:26.850 ","End":"02:35.680","Text":"by mg minus kvy multiplied by dvy."},{"Start":"02:36.860 ","End":"02:41.630","Text":"Now I\u0027ve gotten my function to the form I want where all of"},{"Start":"02:41.630 ","End":"02:46.520","Text":"my t variables are on the left and all of my v variables are on the right."},{"Start":"02:46.520 ","End":"02:50.090","Text":"What I have is a function of t times dt."},{"Start":"02:50.090 ","End":"02:52.220","Text":"What we can say is the function here is 1."},{"Start":"02:52.220 ","End":"02:54.200","Text":"It\u0027s totally alright that it\u0027s just a"},{"Start":"02:54.200 ","End":"02:57.395","Text":"constant and on the right side we have this function,"},{"Start":"02:57.395 ","End":"03:03.005","Text":"m over mg minus kvy times dvy."},{"Start":"03:03.005 ","End":"03:06.490","Text":"What I can now do is take an integral of each side."},{"Start":"03:06.490 ","End":"03:09.650","Text":"I\u0027m going to take an integral on the right in terms of"},{"Start":"03:09.650 ","End":"03:12.800","Text":"t and on the left I\u0027ll take an integral in terms of"},{"Start":"03:12.800 ","End":"03:19.925","Text":"v. What I need to do is I can do either a definite or an indefinite integral."},{"Start":"03:19.925 ","End":"03:22.370","Text":"With a definite integral, I add limits,"},{"Start":"03:22.370 ","End":"03:23.975","Text":"with an indefinite integral,"},{"Start":"03:23.975 ","End":"03:27.305","Text":"I need to multiply it by some constant."},{"Start":"03:27.305 ","End":"03:30.680","Text":"In this case, we\u0027ll do a definite integral."},{"Start":"03:30.680 ","End":"03:32.450","Text":"If you want to try the other method,"},{"Start":"03:32.450 ","End":"03:34.490","Text":"of course, you\u0027re welcome to."},{"Start":"03:34.490 ","End":"03:38.255","Text":"In order to make these 2 integrals compatible,"},{"Start":"03:38.255 ","End":"03:40.415","Text":"I need them to share their limits."},{"Start":"03:40.415 ","End":"03:43.910","Text":"On the left side we\u0027ll do from 0-t,"},{"Start":"03:43.910 ","End":"03:46.730","Text":"time 0 to time t, and then on the right,"},{"Start":"03:46.730 ","End":"03:54.210","Text":"what I need to do is at v(t)=0,"},{"Start":"03:54.210 ","End":"03:58.260","Text":"and that would actually equal 0 and on top"},{"Start":"03:58.260 ","End":"04:02.910","Text":"we\u0027ll have v of the same t that we have on the left."},{"Start":"04:02.990 ","End":"04:08.770","Text":"As I solve this, keep in mind that v(t) is where I\u0027m going to find my answer,"},{"Start":"04:08.770 ","End":"04:10.660","Text":"so I\u0027ll try to isolate that."},{"Start":"04:10.660 ","End":"04:14.455","Text":"Now if I solve, once I plug in my limits on the left side,"},{"Start":"04:14.455 ","End":"04:15.940","Text":"I\u0027m left with t,"},{"Start":"04:15.940 ","End":"04:18.150","Text":"and t equals,"},{"Start":"04:18.150 ","End":"04:23.200","Text":"on the right side, I can first take my m and put it on"},{"Start":"04:23.200 ","End":"04:29.890","Text":"the outside and multiply it by the natural log ln of the entire denominator."},{"Start":"04:29.890 ","End":"04:36.110","Text":"That is mg minus kvy,"},{"Start":"04:36.520 ","End":"04:40.435","Text":"and I need to divide this by the inner derivative,"},{"Start":"04:40.435 ","End":"04:46.150","Text":"which is negative k. Now I do need to account for my boundaries here,"},{"Start":"04:46.150 ","End":"04:51.230","Text":"my limits, so I have to go from 0-v(t)."},{"Start":"04:51.230 ","End":"04:54.140","Text":"Now I can start incorporating this below,"},{"Start":"04:54.140 ","End":"05:00.215","Text":"so I get t equals and I can account for 1 over negative k here,"},{"Start":"05:00.215 ","End":"05:07.360","Text":"m divided by negative k times ln of,"},{"Start":"05:07.360 ","End":"05:12.065","Text":"and what I can do here is account for both of my limits already with the v,"},{"Start":"05:12.065 ","End":"05:15.559","Text":"so I get mg minus"},{"Start":"05:15.559 ","End":"05:21.455","Text":"kv(t) over"},{"Start":"05:21.455 ","End":"05:25.890","Text":"mg. Now"},{"Start":"05:25.890 ","End":"05:28.990","Text":"I can start rearranging my terms."},{"Start":"05:29.270 ","End":"05:34.570","Text":"If I multiply each side by negative k and divide by m,"},{"Start":"05:34.570 ","End":"05:38.755","Text":"what I\u0027m left with is negative k over m on the left side."},{"Start":"05:38.755 ","End":"05:41.560","Text":"The next step is to put e on both sides and"},{"Start":"05:41.560 ","End":"05:44.260","Text":"take e to the power of each of these 2 things."},{"Start":"05:44.260 ","End":"05:50.560","Text":"On the left, I\u0027m left with e to the power of negative k over mt and on the right,"},{"Start":"05:50.560 ","End":"05:52.595","Text":"my ln falls out."},{"Start":"05:52.595 ","End":"05:58.395","Text":"Once I rearrange my terms with mg minus kv and mg,"},{"Start":"05:58.395 ","End":"06:05.550","Text":"what I\u0027m left with is that vt=mg"},{"Start":"06:05.550 ","End":"06:13.485","Text":"over k times 1 minus e to the power of negative k,"},{"Start":"06:13.485 ","End":"06:20.120","Text":"m times t. If this was a little confusing,"},{"Start":"06:20.120 ","End":"06:22.220","Text":"we can verify that it\u0027s true by checking"},{"Start":"06:22.220 ","End":"06:25.360","Text":"our results against the values of t that we know."},{"Start":"06:25.360 ","End":"06:29.850","Text":"When t=0, e^0 is 1."},{"Start":"06:29.850 ","End":"06:31.958","Text":"So we get 1 minus 1,"},{"Start":"06:31.958 ","End":"06:35.143","Text":"mg over k times 0=0,"},{"Start":"06:35.143 ","End":"06:36.830","Text":"and v(t) should equal 0,"},{"Start":"06:36.830 ","End":"06:40.475","Text":"which is in fact true as we know that we\u0027re starting at rest,"},{"Start":"06:40.475 ","End":"06:42.460","Text":"and when t equals infinity,"},{"Start":"06:42.460 ","End":"06:47.000","Text":"e to the power of negative infinity equals 0,"},{"Start":"06:47.000 ","End":"06:50.070","Text":"1 minus 0=1,"},{"Start":"06:50.070 ","End":"06:55.370","Text":"and mg over k times 1 is mg over k and we know that our terminal velocity,"},{"Start":"06:55.370 ","End":"06:58.010","Text":"which would be v at t equals infinity,"},{"Start":"06:58.010 ","End":"07:01.985","Text":"equals mg over k, so this should work."},{"Start":"07:01.985 ","End":"07:04.880","Text":"Now that we\u0027ve found our answers to A,"},{"Start":"07:04.880 ","End":"07:07.090","Text":"B, and C, let\u0027s do a little summary."},{"Start":"07:07.090 ","End":"07:12.485","Text":"First, remember, we talked about the drag force and said that it equals negative kv,"},{"Start":"07:12.485 ","End":"07:14.795","Text":"k being a given constant,"},{"Start":"07:14.795 ","End":"07:17.240","Text":"and v being the velocity vector."},{"Start":"07:17.240 ","End":"07:20.315","Text":"We said that when we want to find the equation of motion,"},{"Start":"07:20.315 ","End":"07:24.905","Text":"the first thing we need to do is set our force in a positive direction."},{"Start":"07:24.905 ","End":"07:27.350","Text":"That means writing negative kv in"},{"Start":"07:27.350 ","End":"07:30.320","Text":"our equations and letting the negative do its work for us."},{"Start":"07:30.320 ","End":"07:33.260","Text":"It doesn\u0027t matter which direction we choose to be positive,"},{"Start":"07:33.260 ","End":"07:35.755","Text":"we have to just start in a positive direction,"},{"Start":"07:35.755 ","End":"07:38.190","Text":"it\u0027ll make everything less confusing."},{"Start":"07:38.190 ","End":"07:40.760","Text":"Then when we want to find the equation of motion,"},{"Start":"07:40.760 ","End":"07:44.825","Text":"remember it\u0027s going to be in terms of y and v and a,"},{"Start":"07:44.825 ","End":"07:47.390","Text":"or really any coordinate and its derivatives,"},{"Start":"07:47.390 ","End":"07:49.115","Text":"it could be x, it could be y,"},{"Start":"07:49.115 ","End":"07:51.380","Text":"it could be z, it could be other coordinates,"},{"Start":"07:51.380 ","End":"07:54.935","Text":"but in terms of any given coordinate and its derivatives."},{"Start":"07:54.935 ","End":"07:58.325","Text":"Once we do our equation of motion,"},{"Start":"07:58.325 ","End":"08:01.790","Text":"remember that we\u0027re generally going to use Newton\u0027s second law,"},{"Start":"08:01.790 ","End":"08:04.505","Text":"setting something equal to ma."},{"Start":"08:04.505 ","End":"08:08.945","Text":"By setting up a sum of forces equation and using Newton\u0027s second law,"},{"Start":"08:08.945 ","End":"08:11.000","Text":"we can get our answer to part A."},{"Start":"08:11.000 ","End":"08:12.770","Text":"Now in part B,"},{"Start":"08:12.770 ","End":"08:15.718","Text":"if you recall, we\u0027re looking for our terminal velocity,"},{"Start":"08:15.718 ","End":"08:20.645","Text":"and the first thing we established is that when our velocity is terminal, v is constant."},{"Start":"08:20.645 ","End":"08:24.170","Text":"We did is set a equal to 0 and we"},{"Start":"08:24.170 ","End":"08:28.100","Text":"plug that into our force equation above into Newton\u0027s second law."},{"Start":"08:28.100 ","End":"08:33.399","Text":"Using that, we very quickly found our final or terminal velocity."},{"Start":"08:33.399 ","End":"08:35.560","Text":"Finally, in part C,"},{"Start":"08:35.560 ","End":"08:37.940","Text":"which is really the central part of this question,"},{"Start":"08:37.940 ","End":"08:41.080","Text":"we wanted to find the velocity as a function of time."},{"Start":"08:41.080 ","End":"08:45.485","Text":"We said the basic way to do this was to solve the equation of motion,"},{"Start":"08:45.485 ","End":"08:47.750","Text":"and the easy way to go about this is to"},{"Start":"08:47.750 ","End":"08:50.390","Text":"realize that it is in fact a differential equation."},{"Start":"08:50.390 ","End":"08:52.685","Text":"We set ay equal to dvy,"},{"Start":"08:52.685 ","End":"08:54.770","Text":"dt in our case."},{"Start":"08:54.770 ","End":"08:58.010","Text":"By using the method of separation of variables,"},{"Start":"08:58.010 ","End":"09:00.710","Text":"we look for an equation in this format with"},{"Start":"09:00.710 ","End":"09:05.575","Text":"our t variables on the left side and our v variables on the right side."},{"Start":"09:05.575 ","End":"09:10.280","Text":"Then, using some clever multiplication and division once we had that result,"},{"Start":"09:10.280 ","End":"09:12.125","Text":"we took an integral on each side,"},{"Start":"09:12.125 ","End":"09:15.590","Text":"on the left side using t from the limits of"},{"Start":"09:15.590 ","End":"09:19.700","Text":"0-t and on the right side using the according limits for v,"},{"Start":"09:19.700 ","End":"09:23.465","Text":"v at t=0 and v at t=t."},{"Start":"09:23.465 ","End":"09:27.050","Text":"Now we could have also done an indefinite integral,"},{"Start":"09:27.050 ","End":"09:29.680","Text":"but in our case, we chose to use limits."},{"Start":"09:29.680 ","End":"09:33.320","Text":"Once we had that, all we needed to do was simplify"},{"Start":"09:33.320 ","End":"09:36.860","Text":"and isolate and end up with something in terms of vt,"},{"Start":"09:36.860 ","End":"09:38.885","Text":"which we did again by using e,"},{"Start":"09:38.885 ","End":"09:43.040","Text":"by multiplying a little bit and we ended up with our solution."},{"Start":"09:43.040 ","End":"09:45.515","Text":"If you recall, in our problem,"},{"Start":"09:45.515 ","End":"09:47.915","Text":"when time was 0 at the initial time,"},{"Start":"09:47.915 ","End":"09:51.680","Text":"v also equaled 0 because our skydiver starts at rest."},{"Start":"09:51.680 ","End":"09:53.840","Text":"When we wanted to test things,"},{"Start":"09:53.840 ","End":"10:00.900","Text":"we tested at t=0 and at t equals infinity and found that when t=0 and v=0,"},{"Start":"10:00.900 ","End":"10:03.225","Text":"then in fact we get 0."},{"Start":"10:03.225 ","End":"10:05.915","Text":"We also found that when t equals infinity,"},{"Start":"10:05.915 ","End":"10:08.810","Text":"we get the same result as our terminal velocity,"},{"Start":"10:08.810 ","End":"10:15.875","Text":"which was mg over k. That ends the explanation and example of the drag force."},{"Start":"10:15.875 ","End":"10:21.860","Text":"One last note I wanted to make is we will encounter some cases where f, the drag force,"},{"Start":"10:21.860 ","End":"10:24.305","Text":"does not equal negative kv,"},{"Start":"10:24.305 ","End":"10:26.420","Text":"a constant times the velocity vector,"},{"Start":"10:26.420 ","End":"10:30.440","Text":"rather it will equal negative kv squared."},{"Start":"10:30.440 ","End":"10:33.800","Text":"We\u0027ll deal with this sometimes when talking about air resistance,"},{"Start":"10:33.800 ","End":"10:37.850","Text":"other times when working with liquid media but just so you know,"},{"Start":"10:37.850 ","End":"10:40.280","Text":"sometimes we\u0027ll be working with negative kv^2"},{"Start":"10:40.280 ","End":"10:43.220","Text":"and you should be aware that it works slightly differently,"},{"Start":"10:43.220 ","End":"10:45.760","Text":"but the principles remain the same."},{"Start":"10:45.760 ","End":"10:49.520","Text":"With that, I hope you now understand the whole concept,"},{"Start":"10:49.520 ","End":"10:52.350","Text":"and let\u0027s move on to the next subject."}],"ID":9297}],"Thumbnail":null,"ID":5334},{"Name":"Buoyancy","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Force of Buoyancy","Duration":"6m 59s","ChapterTopicVideoID":9025,"CourseChapterTopicPlaylistID":5335,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:02.685","Text":"Hello. In this lecture,"},{"Start":"00:02.685 ","End":"00:05.625","Text":"we\u0027re going to talk about Buoyancy and Stokes Force."},{"Start":"00:05.625 ","End":"00:11.025","Text":"These are 2 forces that act upon an object that is moving in a liquid medium."},{"Start":"00:11.025 ","End":"00:13.109","Text":"For the sake of explanation,"},{"Start":"00:13.109 ","End":"00:20.020","Text":"let\u0027s look at an example where we have a container of water and a ball inside of it."},{"Start":"00:20.660 ","End":"00:24.660","Text":"If this isn\u0027t clear, we have a container here that\u0027s filled with"},{"Start":"00:24.660 ","End":"00:29.230","Text":"water and inside of the water is a ball."},{"Start":"00:29.240 ","End":"00:35.715","Text":"Now, Stokes Force and the Force of Buoyancy will both be acting upon this ball."},{"Start":"00:35.715 ","End":"00:39.710","Text":"What I want to do is first talk about the force of buoyancy."},{"Start":"00:39.710 ","End":"00:44.465","Text":"The first thing to know about buoyancy is it\u0027s always acting upwards."},{"Start":"00:44.465 ","End":"00:46.445","Text":"We\u0027re going to symbolize this with Fb,"},{"Start":"00:46.445 ","End":"00:48.995","Text":"of course b for buoyancy."},{"Start":"00:48.995 ","End":"00:51.409","Text":"The formula for buoyancy,"},{"Start":"00:51.409 ","End":"00:53.060","Text":"as you can see on the left here,"},{"Start":"00:53.060 ","End":"00:57.025","Text":"is Fb equals Rho Vg."},{"Start":"00:57.025 ","End":"00:59.690","Text":"Rho is the density of the liquid,"},{"Start":"00:59.690 ","End":"01:02.170","Text":"the medium in which our object sits,"},{"Start":"01:02.170 ","End":"01:05.370","Text":"V is the volume of the object, and g,"},{"Start":"01:05.370 ","End":"01:09.690","Text":"of course, is the constant of gravitational acceleration."},{"Start":"01:09.890 ","End":"01:13.760","Text":"Again, it\u0027s important to emphasize that Fb,"},{"Start":"01:13.760 ","End":"01:15.110","Text":"the force of buoyancy,"},{"Start":"01:15.110 ","End":"01:17.795","Text":"is always going to be pointing upwards."},{"Start":"01:17.795 ","End":"01:20.330","Text":"It\u0027s like mg, but backwards."},{"Start":"01:20.330 ","End":"01:24.754","Text":"Remember mg, is the force of gravity which is always pointing downwards."},{"Start":"01:24.754 ","End":"01:29.300","Text":"The force of buoyancy will be the opposite of that, always going upwards."},{"Start":"01:29.300 ","End":"01:32.570","Text":"Let\u0027s look at what\u0027s happening here a little more closely"},{"Start":"01:32.570 ","End":"01:36.290","Text":"to understand where the force of buoyancy comes from."},{"Start":"01:36.290 ","End":"01:40.925","Text":"As you can see, we\u0027re now looking at the same container of water,"},{"Start":"01:40.925 ","End":"01:44.450","Text":"but we\u0027ve taken the ball and we\u0027ve taken it outside of the container."},{"Start":"01:44.450 ","End":"01:47.675","Text":"We\u0027re looking at a point before we drop the ball into the water,"},{"Start":"01:47.675 ","End":"01:50.545","Text":"and our water level is here."},{"Start":"01:50.545 ","End":"01:54.275","Text":"Now if we drop the ball into the water,"},{"Start":"01:54.275 ","End":"01:58.070","Text":"the water level will rise to this point."},{"Start":"01:58.070 ","End":"02:01.265","Text":"If you look at the region between these 2 lines,"},{"Start":"02:01.265 ","End":"02:02.585","Text":"these 2 water levels,"},{"Start":"02:02.585 ","End":"02:05.150","Text":"which represents the rise in water level,"},{"Start":"02:05.150 ","End":"02:09.780","Text":"you\u0027ll notice that it equals the volume of the ball."},{"Start":"02:10.100 ","End":"02:13.435","Text":"We know this because of Archimedes\u0027 Law,"},{"Start":"02:13.435 ","End":"02:16.250","Text":"which states that the volume of water that rises"},{"Start":"02:16.250 ","End":"02:19.805","Text":"equals the volume of the object that has entered the medium."},{"Start":"02:19.805 ","End":"02:23.600","Text":"What we\u0027re left with is an untenable situation where the water that\u0027s"},{"Start":"02:23.600 ","End":"02:27.485","Text":"been displaced wants to return and go downwards."},{"Start":"02:27.485 ","End":"02:30.845","Text":"In fact mg, the force of gravity is"},{"Start":"02:30.845 ","End":"02:35.210","Text":"acting upon this water trying to bring it down. What is mg?"},{"Start":"02:35.210 ","End":"02:38.375","Text":"Mg is mass times the gravity constant,"},{"Start":"02:38.375 ","End":"02:40.730","Text":"but we can also rewrite m, mass,"},{"Start":"02:40.730 ","End":"02:43.340","Text":"as Rho l,"},{"Start":"02:43.340 ","End":"02:45.965","Text":"that is the density of the liquid,"},{"Start":"02:45.965 ","End":"02:49.580","Text":"times Vl, the volume of the liquid."},{"Start":"02:49.580 ","End":"02:51.905","Text":"Of course, we multiply that by g,"},{"Start":"02:51.905 ","End":"02:57.170","Text":"the gravitational constant, and we get mg. Now we also know that Vl,"},{"Start":"02:57.170 ","End":"02:59.150","Text":"the volume of the liquid displaced,"},{"Start":"02:59.150 ","End":"03:00.770","Text":"equals V ball,"},{"Start":"03:00.770 ","End":"03:05.765","Text":"the volume of the ball or object that entered the water using Archimedes\u0027 Law."},{"Start":"03:05.765 ","End":"03:08.405","Text":"We can rewrite this again as"},{"Start":"03:08.405 ","End":"03:16.870","Text":"Rho L times V bar times g. Now,"},{"Start":"03:16.870 ","End":"03:19.040","Text":"it just so happens that the same mg,"},{"Start":"03:19.040 ","End":"03:21.890","Text":"the force of gravity that\u0027s pushing the water downwards,"},{"Start":"03:21.890 ","End":"03:25.040","Text":"is also acting in an upwards direction on the ball."},{"Start":"03:25.040 ","End":"03:29.045","Text":"This occurs because of a complicated process with water pressure."},{"Start":"03:29.045 ","End":"03:30.710","Text":"We\u0027ll forget about that for now,"},{"Start":"03:30.710 ","End":"03:34.850","Text":"but what you need to know is that the same mg acting downwards on the water,"},{"Start":"03:34.850 ","End":"03:38.045","Text":"is acting upwards on our object inside the water,"},{"Start":"03:38.045 ","End":"03:40.040","Text":"in our case, the ball."},{"Start":"03:40.040 ","End":"03:44.300","Text":"Just to reiterate, the thing to take away here is that mg,"},{"Start":"03:44.300 ","End":"03:46.070","Text":"the force going downwards,"},{"Start":"03:46.070 ","End":"03:48.530","Text":"the force of gravity, equals Fb,"},{"Start":"03:48.530 ","End":"03:49.865","Text":"the force of buoyancy."},{"Start":"03:49.865 ","End":"03:55.980","Text":"In fact, you can see that these have the same magnitude, just like Rho L,"},{"Start":"03:55.980 ","End":"04:00.695","Text":"V ball or the volume of the object that has entered the medium,"},{"Start":"04:00.695 ","End":"04:03.475","Text":"and g, the gravitational constant."},{"Start":"04:03.475 ","End":"04:07.540","Text":"Ultimately, I can say that this equals Fb,"},{"Start":"04:07.540 ","End":"04:09.515","Text":"the force of buoyancy."},{"Start":"04:09.515 ","End":"04:12.980","Text":"Now I want to take a step back and look at another characteristic."},{"Start":"04:12.980 ","End":"04:17.545","Text":"Now remember this mg is the mg of the water acting downwards."},{"Start":"04:17.545 ","End":"04:19.165","Text":"What it gives us is Fb,"},{"Start":"04:19.165 ","End":"04:22.475","Text":"the force of buoyancy acting upwards on our object,"},{"Start":"04:22.475 ","End":"04:24.020","Text":"in our case the ball."},{"Start":"04:24.020 ","End":"04:26.915","Text":"But what if I\u0027m interested in the mg of the ball,"},{"Start":"04:26.915 ","End":"04:30.185","Text":"the downward force of gravity acting on our ball."},{"Start":"04:30.185 ","End":"04:33.740","Text":"This is mbg, the mass of the ball times gravity."},{"Start":"04:33.740 ","End":"04:36.020","Text":"If I were to calculate mbg,"},{"Start":"04:36.020 ","End":"04:39.860","Text":"I could use the same format as I used for mg;"},{"Start":"04:39.860 ","End":"04:48.700","Text":"Rho of the ball times V of the ball times gravitational acceleration."},{"Start":"04:48.700 ","End":"04:53.360","Text":"You may notice that this is very similar to our buoyancy force."},{"Start":"04:53.360 ","End":"04:57.590","Text":"The only difference is Rho L versus Rho b,"},{"Start":"04:57.590 ","End":"05:00.830","Text":"the density of the liquid and the density of the ball."},{"Start":"05:00.830 ","End":"05:03.110","Text":"If I want to set these 2 equal,"},{"Start":"05:03.110 ","End":"05:07.050","Text":"the V ball and g will both fall out,"},{"Start":"05:07.050 ","End":"05:11.240","Text":"and what I\u0027m left with is the following situation; if Rho b,"},{"Start":"05:11.240 ","End":"05:13.280","Text":"that is Rho of the ball,"},{"Start":"05:13.280 ","End":"05:16.040","Text":"is greater than Rho l,"},{"Start":"05:16.040 ","End":"05:18.200","Text":"that is the density of the liquid,"},{"Start":"05:18.200 ","End":"05:20.946","Text":"therefore, wg,"},{"Start":"05:20.946 ","End":"05:25.970","Text":"recall that the force of gravity is going to be greater than Fb,"},{"Start":"05:25.970 ","End":"05:27.755","Text":"the force of buoyancy."},{"Start":"05:27.755 ","End":"05:30.515","Text":"What happens is when my ball or my object,"},{"Start":"05:30.515 ","End":"05:34.190","Text":"whatever it may be, enters the water and may move around at first."},{"Start":"05:34.190 ","End":"05:39.425","Text":"But eventually, the force of gravity will be greater than the force of buoyancy,"},{"Start":"05:39.425 ","End":"05:42.370","Text":"and my ball, my object will sink."},{"Start":"05:42.370 ","End":"05:46.120","Text":"In this case my object will sink."},{"Start":"05:46.300 ","End":"05:49.445","Text":"Now let\u0027s look at the opposite situation;"},{"Start":"05:49.445 ","End":"05:50.960","Text":"when Rho b,"},{"Start":"05:50.960 ","End":"05:52.355","Text":"the density of the ball,"},{"Start":"05:52.355 ","End":"05:54.140","Text":"is less than Rho l,"},{"Start":"05:54.140 ","End":"05:55.865","Text":"the density of the liquid."},{"Start":"05:55.865 ","End":"06:02.730","Text":"In this case, you get Fb greater than wg."},{"Start":"06:02.730 ","End":"06:06.740","Text":"That means that buoyancy is a greater force than the force of gravity."},{"Start":"06:06.740 ","End":"06:09.380","Text":"Eventually, no matter what happens with your ball,"},{"Start":"06:09.380 ","End":"06:11.000","Text":"your object at the beginning,"},{"Start":"06:11.000 ","End":"06:14.070","Text":"your object will eventually float."},{"Start":"06:14.390 ","End":"06:17.535","Text":"Now we see both cases."},{"Start":"06:17.535 ","End":"06:21.770","Text":"I hope this explanation helps you understand buoyancy a little better."},{"Start":"06:21.770 ","End":"06:23.315","Text":"Just to summarize quickly,"},{"Start":"06:23.315 ","End":"06:25.100","Text":"buoyancy or Fb,"},{"Start":"06:25.100 ","End":"06:29.965","Text":"is a force acting upwards on our object that\u0027s in our liquid media."},{"Start":"06:29.965 ","End":"06:35.990","Text":"It is derived from mg of the water that was displaced or whatever liquid it may be,"},{"Start":"06:35.990 ","End":"06:39.020","Text":"and it\u0027s going to act against the mg,"},{"Start":"06:39.020 ","End":"06:43.150","Text":"the force of gravity on our same object that\u0027s in our liquid medium."},{"Start":"06:43.150 ","End":"06:47.435","Text":"If the force of gravity is greater than the force of buoyancy,"},{"Start":"06:47.435 ","End":"06:48.830","Text":"your object will sink."},{"Start":"06:48.830 ","End":"06:52.265","Text":"If the force of buoyancy is greater than the force of gravity,"},{"Start":"06:52.265 ","End":"06:54.600","Text":"your object will float."},{"Start":"06:55.010 ","End":"06:59.260","Text":"Now let\u0027s move on to Stokes Force."}],"ID":9298}],"Thumbnail":null,"ID":5335},{"Name":"Stoke\u0027s Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Stoke\u0027s Force","Duration":"3m 56s","ChapterTopicVideoID":9026,"CourseChapterTopicPlaylistID":5336,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9026.jpeg","UploadDate":"2017-03-22T09:44:34.2300000","DurationForVideoObject":"PT3M56S","Description":null,"MetaTitle":"Stoke\u0027s Force: Video + Workbook | Proprep","MetaDescription":"Buoyancy and Drift Forces - Stoke\u0027s Law. 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/buoyancy-and-drift-forces/stoke%27s-law/vid9299","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.595","Text":"Moving on to Stoke\u0027s Force,"},{"Start":"00:02.595 ","End":"00:05.445","Text":"what\u0027s important to know is this formula below."},{"Start":"00:05.445 ","End":"00:06.780","Text":"Now, first of all,"},{"Start":"00:06.780 ","End":"00:10.065","Text":"this formula only works if we\u0027re talking about a sphere."},{"Start":"00:10.065 ","End":"00:12.075","Text":"Our object has to be a sphere,"},{"Start":"00:12.075 ","End":"00:13.890","Text":"and it also has to be in motion."},{"Start":"00:13.890 ","End":"00:16.455","Text":"It has to have a velocity vector."},{"Start":"00:16.455 ","End":"00:19.035","Text":"If these 2 conditions are met,"},{"Start":"00:19.035 ","End":"00:22.275","Text":"we can then use Stoke\u0027s force to describe the friction"},{"Start":"00:22.275 ","End":"00:26.205","Text":"our object is experiencing inside of its liquid medium."},{"Start":"00:26.205 ","End":"00:31.380","Text":"In our example, let\u0027s assume that our sphere has a velocity in this direction."},{"Start":"00:31.380 ","End":"00:35.685","Text":"I\u0027ll designate it with v. We\u0027re talking about a velocity vector."},{"Start":"00:35.685 ","End":"00:40.505","Text":"In that case, our Stoke\u0027s force will go in the opposite direction."},{"Start":"00:40.505 ","End":"00:43.340","Text":"We\u0027ll write this F_s for now."},{"Start":"00:43.340 ","End":"00:46.865","Text":"Well, you may be confused about the directions right now."},{"Start":"00:46.865 ","End":"00:50.120","Text":"We\u0027ll get to that in a minute when we talk about the formula."},{"Start":"00:50.120 ","End":"00:53.390","Text":"Just keep in mind that we can address this as though"},{"Start":"00:53.390 ","End":"00:57.065","Text":"we were talking about friction like in any other problem."},{"Start":"00:57.065 ","End":"00:59.254","Text":"Looking at the formula,"},{"Start":"00:59.254 ","End":"01:01.670","Text":"you can see we have negative 6 Pi."},{"Start":"01:01.670 ","End":"01:04.405","Text":"These are constants and then we have Eta."},{"Start":"01:04.405 ","End":"01:10.055","Text":"Eta represents the viscosity of the liquid and viscosity means thickness or stickiness."},{"Start":"01:10.055 ","End":"01:13.415","Text":"It makes sense that the thicker or stickier our liquid is"},{"Start":"01:13.415 ","End":"01:17.560","Text":"the more resistance will have when we try to move our sphere."},{"Start":"01:17.560 ","End":"01:20.445","Text":"Next, we have R, the radius of the sphere."},{"Start":"01:20.445 ","End":"01:22.970","Text":"This makes sense as the radius grows,"},{"Start":"01:22.970 ","End":"01:26.125","Text":"that means our object is larger our sphere is larger,"},{"Start":"01:26.125 ","End":"01:28.830","Text":"and that means we\u0027re going to have to move more liquid,"},{"Start":"01:28.830 ","End":"01:30.515","Text":"and the more liquid we have to move,"},{"Start":"01:30.515 ","End":"01:32.405","Text":"the more friction will encounter."},{"Start":"01:32.405 ","End":"01:35.240","Text":"Lastly, we have v, the velocity vector,"},{"Start":"01:35.240 ","End":"01:36.950","Text":"and this also makes sense."},{"Start":"01:36.950 ","End":"01:40.220","Text":"If you think about it, the faster we try to move our object,"},{"Start":"01:40.220 ","End":"01:42.095","Text":"the more friction will encounter."},{"Start":"01:42.095 ","End":"01:47.165","Text":"What you might have noticed is that what we\u0027re dealing with is basically a constant."},{"Start":"01:47.165 ","End":"01:52.850","Text":"What I mean by that is that negative 6 Pi Eta R is a constant."},{"Start":"01:52.850 ","End":"01:55.100","Text":"These are things that are not going to change."},{"Start":"01:55.100 ","End":"01:58.370","Text":"Generally speaking, the viscosity of your liquid and the radius of"},{"Start":"01:58.370 ","End":"02:02.165","Text":"your sphere will not change as you move around in some liquid medium."},{"Start":"02:02.165 ","End":"02:05.120","Text":"The only thing that\u0027s going to change is v, your velocity."},{"Start":"02:05.120 ","End":"02:09.890","Text":"We can also think of this in terms of negative K_v,"},{"Start":"02:09.890 ","End":"02:13.670","Text":"or a negative constant times the velocity vector."},{"Start":"02:13.670 ","End":"02:17.590","Text":"This then defines Stoke\u0027s force as a drag force,"},{"Start":"02:17.590 ","End":"02:20.000","Text":"if you recall, in 1 of our previous lectures,"},{"Start":"02:20.000 ","End":"02:23.492","Text":"we talked about how to deal with drag force."},{"Start":"02:23.492 ","End":"02:25.520","Text":"Just to summarize first,"},{"Start":"02:25.520 ","End":"02:27.305","Text":"we talked about buoyancy."},{"Start":"02:27.305 ","End":"02:32.885","Text":"Buoyancy or FB, is always acting upwards on our object in liquid,"},{"Start":"02:32.885 ","End":"02:37.910","Text":"and the formula for buoyancy is rho vg."},{"Start":"02:38.560 ","End":"02:42.575","Text":"Where rho is the density of the liquid."},{"Start":"02:42.575 ","End":"02:46.730","Text":"V is the volume of the object that is entered your liquid,"},{"Start":"02:46.730 ","End":"02:50.840","Text":"and g is the gravitational acceleration constant."},{"Start":"02:50.840 ","End":"02:54.005","Text":"You should put this on your formula sheet."},{"Start":"02:54.005 ","End":"02:56.990","Text":"The second force is Stoke\u0027s force."},{"Start":"02:56.990 ","End":"03:03.575","Text":"Stoke\u0027s force is only relevant when we have a sphere that is moving in a liquid medium,"},{"Start":"03:03.575 ","End":"03:05.780","Text":"and for all intents and purposes,"},{"Start":"03:05.780 ","End":"03:07.685","Text":"this is a stand-in for friction,"},{"Start":"03:07.685 ","End":"03:12.230","Text":"meaning that it acts in the opposite direction of the velocity vector."},{"Start":"03:12.230 ","End":"03:15.935","Text":"Remember, just like with other instances of drag force,"},{"Start":"03:15.935 ","End":"03:19.070","Text":"because of the negative symbol at the beginning of our force,"},{"Start":"03:19.070 ","End":"03:21.290","Text":"it can be rather confusing the direction."},{"Start":"03:21.290 ","End":"03:23.690","Text":"My suggestion is to always initially"},{"Start":"03:23.690 ","End":"03:26.750","Text":"draw Stoke\u0027s force in the same direction as velocity,"},{"Start":"03:26.750 ","End":"03:29.375","Text":"and write in your formula negative K_V,"},{"Start":"03:29.375 ","End":"03:33.400","Text":"or in our case negative 6 Pi Eta R_V,"},{"Start":"03:33.400 ","End":"03:36.890","Text":"and then the formula should do the work for you and"},{"Start":"03:36.890 ","End":"03:39.800","Text":"eventually point Stoke\u0027s force in the correct direction,"},{"Start":"03:39.800 ","End":"03:43.465","Text":"which is against the direction of the velocity vector."},{"Start":"03:43.465 ","End":"03:46.730","Text":"These 2 forces will be relevant if we have"},{"Start":"03:46.730 ","End":"03:50.540","Text":"a problem of a sphere moving in a liquid medium."},{"Start":"03:50.540 ","End":"03:54.950","Text":"In that case, we\u0027ll use Stoke\u0027s force as our drag force."},{"Start":"03:54.950 ","End":"03:57.420","Text":"Here ends our lecture."}],"ID":9299}],"Thumbnail":null,"ID":5336},{"Name":"Exercise- a Ball is Thrown in to a Pool","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Part A","Duration":"8m 32s","ChapterTopicVideoID":9027,"CourseChapterTopicPlaylistID":5337,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.625","Text":"Hello. In this exercise,"},{"Start":"00:02.625 ","End":"00:08.460","Text":"a ball is thrown into a pool and we\u0027re given an initial velocity for the ball of V_0 and"},{"Start":"00:08.460 ","End":"00:11.550","Text":"the angle at which the ball is thrown relative to"},{"Start":"00:11.550 ","End":"00:15.225","Text":"the surface of the water or relative to a horizontal axis."},{"Start":"00:15.225 ","End":"00:17.715","Text":"That angle is given as Theta."},{"Start":"00:17.715 ","End":"00:20.490","Text":"It looks as it is in this picture here."},{"Start":"00:20.490 ","End":"00:24.075","Text":"In addition to being given V_0 and the angle Theta,"},{"Start":"00:24.075 ","End":"00:28.080","Text":"we\u0027re also given the viscosity of the water as Eta,"},{"Start":"00:28.080 ","End":"00:32.520","Text":"ball radius of R, the water density of Rho_w,"},{"Start":"00:32.520 ","End":"00:35.680","Text":"and the ball density of Rho_b."},{"Start":"00:35.680 ","End":"00:41.165","Text":"In part A, we\u0027re asked to write out the equation of motion for the ball."},{"Start":"00:41.165 ","End":"00:44.460","Text":"Let\u0027s get started with part A."},{"Start":"00:44.720 ","End":"00:46.795","Text":"In part A,"},{"Start":"00:46.795 ","End":"00:50.320","Text":"I\u0027m asked to find the equation of motion."},{"Start":"00:50.320 ","End":"00:53.270","Text":"If you recall, the way we usually find the equation of"},{"Start":"00:53.270 ","End":"00:58.075","Text":"motion is by taking the sum of forces equals ma."},{"Start":"00:58.075 ","End":"01:00.675","Text":"That\u0027s Newton\u0027s second law of motion."},{"Start":"01:00.675 ","End":"01:04.070","Text":"We\u0027re going to need that equation to involve a coordinate, say y,"},{"Start":"01:04.070 ","End":"01:06.185","Text":"it could also be x or any other coordinate,"},{"Start":"01:06.185 ","End":"01:10.280","Text":"as well as the velocity and acceleration along that coordinate."},{"Start":"01:10.280 ","End":"01:13.535","Text":"Now, it could be any 2 of these 3 elements as well."},{"Start":"01:13.535 ","End":"01:17.700","Text":"But we want some combination of y, v, and a."},{"Start":"01:17.780 ","End":"01:20.465","Text":"In order to find this equation,"},{"Start":"01:20.465 ","End":"01:25.490","Text":"the first thing we should do is figure out which forces are acting upon our ball."},{"Start":"01:25.490 ","End":"01:28.175","Text":"Let\u0027s say the ball has already been thrown in the water"},{"Start":"01:28.175 ","End":"01:31.060","Text":"and it\u0027s resting somewhere around here."},{"Start":"01:31.060 ","End":"01:34.820","Text":"The forces that will be acting on it first we know there\u0027s mg,"},{"Start":"01:34.820 ","End":"01:37.835","Text":"the force of gravity which is pulling it downwards."},{"Start":"01:37.835 ","End":"01:42.530","Text":"We also know that buoyancy will be bringing it upwards F_b."},{"Start":"01:42.530 ","End":"01:47.225","Text":"We also have to account for the friction or Stokes force."},{"Start":"01:47.225 ","End":"01:48.905","Text":"We won\u0027t write that in quite yet."},{"Start":"01:48.905 ","End":"01:51.619","Text":"If you recall, we want to first write out the equation,"},{"Start":"01:51.619 ","End":"01:53.135","Text":"but we need to remember that."},{"Start":"01:53.135 ","End":"01:55.400","Text":"Let\u0027s start with Stokes force here."},{"Start":"01:55.400 ","End":"02:05.505","Text":"Stokes force read it as F_k equals negative 6 Pi Eta times r,"},{"Start":"02:05.505 ","End":"02:09.980","Text":"the radius which is given and v vector,"},{"Start":"02:09.980 ","End":"02:12.965","Text":"the v vector, the velocity vector."},{"Start":"02:12.965 ","End":"02:16.450","Text":"As for buoyancy F_b,"},{"Start":"02:16.450 ","End":"02:19.910","Text":"we know that to be equal to Rho of the liquid."},{"Start":"02:19.910 ","End":"02:22.370","Text":"In our case, Rho w, the water density,"},{"Start":"02:22.370 ","End":"02:23.825","Text":"which is given to us,"},{"Start":"02:23.825 ","End":"02:28.395","Text":"times V, the volume of the sphere."},{"Start":"02:28.395 ","End":"02:30.080","Text":"That we\u0027ll calculate in a second,"},{"Start":"02:30.080 ","End":"02:32.990","Text":"times g the gravitational constant."},{"Start":"02:32.990 ","End":"02:37.064","Text":"As an aside, the V for volume with these little shoulders on it,"},{"Start":"02:37.064 ","End":"02:40.440","Text":"so you don\u0027t confuse it with the v for velocity."},{"Start":"02:40.880 ","End":"02:42.965","Text":"On the side here,"},{"Start":"02:42.965 ","End":"02:46.190","Text":"I\u0027ll write the equation for the volume of a sphere."},{"Start":"02:46.190 ","End":"02:55.440","Text":"The volume of a sphere equals 4 Pi r to the third over 3."},{"Start":"02:55.440 ","End":"02:58.290","Text":"Additionally for mass because we\u0027ll need that later."},{"Start":"02:58.290 ","End":"03:02.990","Text":"The mass, as long as we\u0027re talking about our ball or a sphere,"},{"Start":"03:02.990 ","End":"03:06.875","Text":"is going to be the density of the sphere, which is Rho b."},{"Start":"03:06.875 ","End":"03:08.540","Text":"Remember this is given to us,"},{"Start":"03:08.540 ","End":"03:13.390","Text":"times the volume which we will have calculated over here on the left."},{"Start":"03:13.390 ","End":"03:18.845","Text":"The last thing I want to do before I calculate my forces is choose my axis."},{"Start":"03:18.845 ","End":"03:24.309","Text":"I\u0027ll choose x going positively to the right and y going positively down,"},{"Start":"03:24.309 ","End":"03:30.515","Text":"so that gravity mg will be acting in the same direction as the y-axis."},{"Start":"03:30.515 ","End":"03:34.760","Text":"First things first, if I want to draw out F_k,"},{"Start":"03:34.760 ","End":"03:36.395","Text":"that is Stokes force,"},{"Start":"03:36.395 ","End":"03:38.615","Text":"first, I need to draw my velocity."},{"Start":"03:38.615 ","End":"03:44.900","Text":"Let\u0027s assume that our sphere is moving in this direction with a velocity V. In general,"},{"Start":"03:44.900 ","End":"03:48.260","Text":"we know that the friction will be going in the opposite direction."},{"Start":"03:48.260 ","End":"03:50.380","Text":"But remember, we can get confused with the signs."},{"Start":"03:50.380 ","End":"03:53.600","Text":"The best course of action is to start by"},{"Start":"03:53.600 ","End":"03:57.860","Text":"drawing our F_k in the same direction as the velocity,"},{"Start":"03:57.860 ","End":"04:01.030","Text":"we\u0027ll draw it in blue here, F_k."},{"Start":"04:01.030 ","End":"04:05.465","Text":"We keep our sine negatives that when we actually plug in all of our equations,"},{"Start":"04:05.465 ","End":"04:07.220","Text":"it should correct itself and eventually"},{"Start":"04:07.220 ","End":"04:11.120","Text":"point the opposite direction of the velocity vector."},{"Start":"04:11.120 ","End":"04:14.210","Text":"If this explanation is confusing to you,"},{"Start":"04:14.210 ","End":"04:15.935","Text":"don\u0027t worry so much about the drawing."},{"Start":"04:15.935 ","End":"04:17.330","Text":"This is just a visual aid."},{"Start":"04:17.330 ","End":"04:20.000","Text":"Most importantly, we need to keep our negative sign"},{"Start":"04:20.000 ","End":"04:23.090","Text":"here with our constant multiplied by the V vector."},{"Start":"04:23.090 ","End":"04:27.390","Text":"That should give us F_k correctly by the end of our problem."},{"Start":"04:27.740 ","End":"04:32.255","Text":"Let me start by writing out my sum of forces on each axis,"},{"Start":"04:32.255 ","End":"04:34.220","Text":"and let\u0027s start with the y-axis."},{"Start":"04:34.220 ","End":"04:38.210","Text":"The sum of forces along the y axis equals,"},{"Start":"04:38.210 ","End":"04:40.730","Text":"first we have mg going positively."},{"Start":"04:40.730 ","End":"04:45.005","Text":"Remember downwards is positive in this case, minus F_b."},{"Start":"04:45.005 ","End":"04:49.370","Text":"F_b is negative because it\u0027s going upwards."},{"Start":"04:49.370 ","End":"04:52.885","Text":"Lastly, we need to account for F_k."},{"Start":"04:52.885 ","End":"04:55.665","Text":"In order to incorporate F_k,"},{"Start":"04:55.665 ","End":"04:57.105","Text":"we\u0027re going to use a little trick."},{"Start":"04:57.105 ","End":"04:58.415","Text":"Because this is a vector,"},{"Start":"04:58.415 ","End":"05:01.550","Text":"we can break it down into its different elements, x and y."},{"Start":"05:01.550 ","End":"05:06.670","Text":"We could write the same vector as F_k equals negative"},{"Start":"05:06.670 ","End":"05:12.065","Text":"6 Pi Eta R. Instead of V,"},{"Start":"05:12.065 ","End":"05:20.280","Text":"we write V_x in the direction of x plus V_y in the direction of y."},{"Start":"05:20.590 ","End":"05:23.105","Text":"And if we multiply this out,"},{"Start":"05:23.105 ","End":"05:26.825","Text":"we\u0027ll find that our x elements will go in the direction of x,"},{"Start":"05:26.825 ","End":"05:29.540","Text":"our y elements will go in the direction of y."},{"Start":"05:29.540 ","End":"05:32.600","Text":"As an aside before moving on and incorporating"},{"Start":"05:32.600 ","End":"05:36.725","Text":"F_k into our equation of the sum of forces along the y axis,"},{"Start":"05:36.725 ","End":"05:39.050","Text":"I wanted to mention the reason we broke it"},{"Start":"05:39.050 ","End":"05:42.065","Text":"down this way is that it works out a lot easier."},{"Start":"05:42.065 ","End":"05:44.480","Text":"Some students make the mistake of saying, well,"},{"Start":"05:44.480 ","End":"05:47.000","Text":"the ball hits the water at an angle of Theta."},{"Start":"05:47.000 ","End":"05:49.010","Text":"Why don\u0027t we take the cosine of Theta?"},{"Start":"05:49.010 ","End":"05:50.840","Text":"In fact, in this question,"},{"Start":"05:50.840 ","End":"05:52.585","Text":"that would be the wrong approach."},{"Start":"05:52.585 ","End":"05:57.260","Text":"Why is that? This angle Theta is only the angle at which the ball hits the water."},{"Start":"05:57.260 ","End":"05:59.000","Text":"It\u0027s not a constant angle."},{"Start":"05:59.000 ","End":"06:00.635","Text":"To give it away a little bit,"},{"Start":"06:00.635 ","End":"06:04.535","Text":"the movement of the ball would go something like this along the dashed line,"},{"Start":"06:04.535 ","End":"06:07.690","Text":"eventually probably rising back up."},{"Start":"06:07.690 ","End":"06:10.160","Text":"The angle Theta is actually going to be constantly"},{"Start":"06:10.160 ","End":"06:13.340","Text":"changing and we can\u0027t use it for this purpose."},{"Start":"06:13.340 ","End":"06:16.730","Text":"Instead of using the angle Theta or trying to use another angle,"},{"Start":"06:16.730 ","End":"06:21.415","Text":"we can break down into x and y elements and make our lives a little easier."},{"Start":"06:21.415 ","End":"06:25.210","Text":"Now we can incorporate our F_ky."},{"Start":"06:25.210 ","End":"06:32.785","Text":"What we end up with is the sum of forces along the y axis equal mg"},{"Start":"06:32.785 ","End":"06:42.375","Text":"minus F_b minus 6 Pi Eta RV_y."},{"Start":"06:42.375 ","End":"06:44.535","Text":"Now pay attention here."},{"Start":"06:44.535 ","End":"06:47.165","Text":"It\u0027s important to note that it doesn\u0027t matter whether"},{"Start":"06:47.165 ","End":"06:50.150","Text":"the y\u0027s positive direction was upwards or downwards."},{"Start":"06:50.150 ","End":"06:52.780","Text":"This would still be negative 6 Pi."},{"Start":"06:52.780 ","End":"06:58.460","Text":"Whereas M_g and F_b would switch positive and negative signs based on the direction of"},{"Start":"06:58.460 ","End":"07:05.795","Text":"y are constant before our V_y and F_k would always remain the same."},{"Start":"07:05.795 ","End":"07:09.990","Text":"Now I\u0027m going to make our lives a little easier and write things out a little shorter."},{"Start":"07:09.990 ","End":"07:13.730","Text":"We\u0027re going to take the constant that comes before V and we\u0027re going to write"},{"Start":"07:13.730 ","End":"07:18.270","Text":"that out as k. This will be negative KV_y,"},{"Start":"07:18.270 ","End":"07:21.690","Text":"and we\u0027ll take M_g minus F_b."},{"Start":"07:21.690 ","End":"07:23.870","Text":"Because this is also a constant,"},{"Start":"07:23.870 ","End":"07:25.668","Text":"we\u0027ll change this to C,"},{"Start":"07:25.668 ","End":"07:28.845","Text":"so that our equation ends up being C"},{"Start":"07:28.845 ","End":"07:36.570","Text":"minus KV_y equals Ma_y."},{"Start":"07:36.570 ","End":"07:39.260","Text":"This is a rather simple differential equation"},{"Start":"07:39.260 ","End":"07:41.315","Text":"and we\u0027ll look at how to solve it in a minute."},{"Start":"07:41.315 ","End":"07:43.430","Text":"This is my y element."},{"Start":"07:43.430 ","End":"07:45.590","Text":"Now, moving on to the x element,"},{"Start":"07:45.590 ","End":"07:48.470","Text":"the sum of forces along the x axis."},{"Start":"07:48.470 ","End":"07:51.515","Text":"We don\u0027t have to worry about gravity or buoyancy."},{"Start":"07:51.515 ","End":"07:53.285","Text":"We only have the drag force,"},{"Start":"07:53.285 ","End":"07:54.700","Text":"that is Stokes\u0027 force."},{"Start":"07:54.700 ","End":"07:57.830","Text":"What we\u0027re left with is the same k from before."},{"Start":"07:57.830 ","End":"08:00.755","Text":"Remember negative K equals the constant."},{"Start":"08:00.755 ","End":"08:03.870","Text":"What we\u0027re left with is negative KV_x,"},{"Start":"08:04.180 ","End":"08:09.325","Text":"and that equals Ma_x."},{"Start":"08:09.325 ","End":"08:12.435","Text":"Now I can say I have 2 equations of motion,"},{"Start":"08:12.435 ","End":"08:16.965","Text":"1 with my y elements and 1 with my x elements."},{"Start":"08:16.965 ","End":"08:19.925","Text":"The first one you notice it has terms of v and a,"},{"Start":"08:19.925 ","End":"08:21.950","Text":"fulfilling our requirements above."},{"Start":"08:21.950 ","End":"08:24.665","Text":"And the second one is in terms of v and a,"},{"Start":"08:24.665 ","End":"08:27.220","Text":"also fulfilling our requirements above."},{"Start":"08:27.220 ","End":"08:32.760","Text":"Now we\u0027ve finished part a and we can use this answer to solve parts B and C."}],"ID":9300},{"Watched":false,"Name":"Part B","Duration":"2m 16s","ChapterTopicVideoID":9028,"CourseChapterTopicPlaylistID":5337,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.905","Text":"In Part B, I\u0027m asked to find the terminal velocity,"},{"Start":"00:04.905 ","End":"00:10.590","Text":"and we know that the terminal velocity or final velocity is a constant,"},{"Start":"00:10.590 ","End":"00:13.050","Text":"and if V is a constant,"},{"Start":"00:13.050 ","End":"00:15.165","Text":"that means that it\u0027s derivative a,"},{"Start":"00:15.165 ","End":"00:17.250","Text":"the acceleration equals 0."},{"Start":"00:17.250 ","End":"00:19.890","Text":"We can do is take a equals 0,"},{"Start":"00:19.890 ","End":"00:22.680","Text":"and plug it in above, and see what we get."},{"Start":"00:22.680 ","End":"00:28.724","Text":"In terms of y, C minus KV_y finale"},{"Start":"00:28.724 ","End":"00:34.725","Text":"equals M0, which equals 0."},{"Start":"00:34.725 ","End":"00:38.130","Text":"C minus KV_y final equals 0."},{"Start":"00:38.130 ","End":"00:42.410","Text":"If we add C to both sides and divide by K,"},{"Start":"00:42.410 ","End":"00:50.135","Text":"what we end up with is Vy final equals C over K,"},{"Start":"00:50.135 ","End":"00:52.130","Text":"and if you recall,"},{"Start":"00:52.130 ","End":"00:56.810","Text":"our C equals Mg minus Fb,"},{"Start":"00:56.810 ","End":"01:02.790","Text":"and K equals 6"},{"Start":"01:02.790 ","End":"01:08.889","Text":"Pi Eta R. Now this is my terminal velocity for my y elements."},{"Start":"01:08.889 ","End":"01:13.185","Text":"Notice that if Mg is greater than Fb,"},{"Start":"01:13.185 ","End":"01:17.428","Text":"that means the force of gravity is greater than the force of buoyancy,"},{"Start":"01:17.428 ","End":"01:19.025","Text":"then my answer will be positive,"},{"Start":"01:19.025 ","End":"01:23.060","Text":"meaning it\u0027ll go in a positive y-direction or downwards, meaning it sinks."},{"Start":"01:23.060 ","End":"01:26.615","Text":"Whereas if Fb is greater than Mg,"},{"Start":"01:26.615 ","End":"01:28.130","Text":"then it\u0027ll go in a negative direction,"},{"Start":"01:28.130 ","End":"01:29.610","Text":"and my object will float."},{"Start":"01:29.610 ","End":"01:34.430","Text":"This again makes sense because Fb has Rho_w,"},{"Start":"01:34.430 ","End":"01:35.780","Text":"the density of water,"},{"Start":"01:35.780 ","End":"01:38.300","Text":"whereas M has Rho_b,"},{"Start":"01:38.300 ","End":"01:39.725","Text":"the density of the ball."},{"Start":"01:39.725 ","End":"01:42.495","Text":"Whoever has the greater density ultimately,"},{"Start":"01:42.495 ","End":"01:46.650","Text":"will determine whether our ball floats or sinks."},{"Start":"01:46.650 ","End":"01:50.020","Text":"That solves my y terminal velocity."},{"Start":"01:50.020 ","End":"01:53.525","Text":"In terms of my x terminal velocity, it\u0027s rather simple."},{"Start":"01:53.525 ","End":"01:56.375","Text":"If you notice from my x equation,"},{"Start":"01:56.375 ","End":"01:59.840","Text":"it\u0027s negative KV_x equals Ma_X."},{"Start":"01:59.840 ","End":"02:01.595","Text":"If Ma_x equals 0,"},{"Start":"02:01.595 ","End":"02:04.565","Text":"then my terminal velocity along the x axis equals 0,"},{"Start":"02:04.565 ","End":"02:08.395","Text":"meaning that at some point my ball will stop moving forwards,"},{"Start":"02:08.395 ","End":"02:13.775","Text":"and will either float or sink based on the ratio between Mg and Fb."},{"Start":"02:13.775 ","End":"02:16.710","Text":"Let\u0027s move on to Part C."}],"ID":9301},{"Watched":false,"Name":"Part C","Duration":"9m 17s","ChapterTopicVideoID":9029,"CourseChapterTopicPlaylistID":5337,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.440 ","End":"00:04.740","Text":"In Part C, I\u0027m looking for y as a function of time,"},{"Start":"00:04.740 ","End":"00:06.615","Text":"position as a function of time."},{"Start":"00:06.615 ","End":"00:08.430","Text":"Because when I look for a maximum,"},{"Start":"00:08.430 ","End":"00:10.410","Text":"I\u0027m looking for a specific position,"},{"Start":"00:10.410 ","End":"00:13.670","Text":"my deepest position of the ball in terms of time."},{"Start":"00:13.670 ","End":"00:17.190","Text":"I need to find y as a function of t,"},{"Start":"00:17.190 ","End":"00:20.700","Text":"and then from there, find the y maximum."},{"Start":"00:20.700 ","End":"00:22.410","Text":"Figure out what that equals."},{"Start":"00:22.410 ","End":"00:29.785","Text":"The way to do that is to solve my equation of motion in terms of vy(t)."},{"Start":"00:29.785 ","End":"00:32.810","Text":"Then from there, take an integral and find y(t),"},{"Start":"00:32.810 ","End":"00:34.280","Text":"and my y maximum."},{"Start":"00:34.280 ","End":"00:37.910","Text":"Now, the way to find vy(t) is to solve"},{"Start":"00:37.910 ","End":"00:42.160","Text":"the equation of motion by taking an integral and separating our terms."},{"Start":"00:42.160 ","End":"00:43.645","Text":"First, let\u0027s write out our terms,"},{"Start":"00:43.645 ","End":"00:44.260","Text":"c"},{"Start":"00:44.260 ","End":"00:45.261","Text":"minus"},{"Start":"00:53.670 ","End":"00:54.700","Text":"kv_y"},{"Start":"00:54.700 ","End":"00:54.740","Text":"equals m^dv_y over dt."},{"Start":"00:54.740 ","End":"00:58.715","Text":"Our next step is to multiply both sides by dt."},{"Start":"00:58.715 ","End":"01:01.790","Text":"What we end up with is c"},{"Start":"01:01.790 ","End":"01:10.850","Text":"minus kv_y dt equals mdv_y."},{"Start":"01:10.850 ","End":"01:16.340","Text":"Then I\u0027ll divide both sides by the entire coefficient preceding dt."},{"Start":"01:16.340 ","End":"01:22.125","Text":"I divide both sides by c minus kv_y."},{"Start":"01:22.125 ","End":"01:25.895","Text":"This will lead to a total separation of terms."},{"Start":"01:25.895 ","End":"01:28.950","Text":"What I get from this is dt"},{"Start":"01:29.690 ","End":"01:37.290","Text":"equals m over c minus kv_y^dv_y."},{"Start":"01:38.240 ","End":"01:42.315","Text":"At this point, all of my dt terms are on the left,"},{"Start":"01:42.315 ","End":"01:45.615","Text":"and all of my dv_y terms are on the right."},{"Start":"01:45.615 ","End":"01:48.600","Text":"I can take an integral of each side."},{"Start":"01:50.150 ","End":"01:57.565","Text":"On the left, it\u0027ll be in terms of t. I\u0027ll go from the time 0 to the time t. On the right,"},{"Start":"01:57.565 ","End":"02:01.780","Text":"it will go from the time of v_y t equals 0,"},{"Start":"02:01.780 ","End":"02:04.600","Text":"because we\u0027re in terms of v over here,"},{"Start":"02:04.600 ","End":"02:09.785","Text":"and v_y t equals 0 is my velocity at the initial moment of the problem."},{"Start":"02:09.785 ","End":"02:13.585","Text":"Now, I\u0027ve chosen my initial moment to be the moment when the ball hits the water."},{"Start":"02:13.585 ","End":"02:16.660","Text":"At the initial moment, I do have the angle of Theta,"},{"Start":"02:16.660 ","End":"02:19.765","Text":"and I do have the initial velocity given at v_0."},{"Start":"02:19.765 ","End":"02:27.190","Text":"What I can say is that this equals v_0 sine Theta."},{"Start":"02:27.190 ","End":"02:31.370","Text":"On top, I have v_y(t)."},{"Start":"02:31.370 ","End":"02:33.860","Text":"This, of course, is the term I\u0027m looking for."},{"Start":"02:33.860 ","End":"02:35.315","Text":"This is what I want to isolate."},{"Start":"02:35.315 ","End":"02:37.040","Text":"So I\u0027ll highlight that."},{"Start":"02:37.040 ","End":"02:39.350","Text":"Now, let\u0027s solve this integral."},{"Start":"02:39.350 ","End":"02:40.775","Text":"We get on the left side,"},{"Start":"02:40.775 ","End":"02:43.520","Text":"t and on the right side,"},{"Start":"02:43.520 ","End":"02:44.700","Text":"we have m,"},{"Start":"02:44.700 ","End":"02:46.395","Text":"falls to the outside,"},{"Start":"02:46.395 ","End":"02:48.220","Text":"and we have Ln,"},{"Start":"02:48.220 ","End":"02:51.185","Text":"the natural log of the entire denominator."},{"Start":"02:51.185 ","End":"02:55.200","Text":"Ln of c minus kv_y."},{"Start":"02:55.520 ","End":"02:58.399","Text":"We have to incorporate our limits,"},{"Start":"02:58.399 ","End":"03:08.365","Text":"which are vy(t) on top and on the bottom, v_0 sine Theta."},{"Start":"03:08.365 ","End":"03:12.595","Text":"We also have to divide by the inner derivative."},{"Start":"03:12.595 ","End":"03:16.235","Text":"We divide by negative k. We can move"},{"Start":"03:16.235 ","End":"03:20.785","Text":"the m divided by negative k to the t side on the left."},{"Start":"03:20.785 ","End":"03:28.175","Text":"What we\u0027re left with is negative k over mt equals the natural log of c minus kv_y,"},{"Start":"03:28.175 ","End":"03:31.490","Text":"with the limits that needs to be incorporated."},{"Start":"03:31.490 ","End":"03:34.970","Text":"In fact, let\u0027s go ahead and add those limits in."},{"Start":"03:34.970 ","End":"03:42.410","Text":"We end up with is negative k over mt equals the natural log"},{"Start":"03:42.410 ","End":"03:50.490","Text":"of c minus kv_y(t) on top and on the bottom,"},{"Start":"03:50.490 ","End":"03:56.525","Text":"c minus v_0 sine Theta."},{"Start":"03:56.525 ","End":"03:59.650","Text":"I can then simplify by taking each side as"},{"Start":"03:59.650 ","End":"04:03.740","Text":"a power of e. What I end up with is the following."},{"Start":"04:03.840 ","End":"04:11.740","Text":"e to the power of negative km times t equals the argument from our natural log."},{"Start":"04:11.740 ","End":"04:14.590","Text":"At this point, I\u0027ll try to take my denominator"},{"Start":"04:14.590 ","End":"04:17.140","Text":"over to the left-hand side and clean up here,"},{"Start":"04:17.140 ","End":"04:18.775","Text":"so that I\u0027m left with only v_y."},{"Start":"04:18.775 ","End":"04:21.670","Text":"My solution is as follows,"},{"Start":"04:21.670 ","End":"04:31.335","Text":"v_y(t) equals c over k minus in parentheses,"},{"Start":"04:31.335 ","End":"04:35.570","Text":"c minus v_0 sine Theta."},{"Start":"04:35.570 ","End":"04:38.540","Text":"That\u0027s our whole denominator from the right side,"},{"Start":"04:38.540 ","End":"04:44.420","Text":"times e to the power of negative k over m times t. Well,"},{"Start":"04:44.420 ","End":"04:45.815","Text":"this may look complicated."},{"Start":"04:45.815 ","End":"04:48.185","Text":"In fact, all of the figures here are given to me."},{"Start":"04:48.185 ","End":"04:51.320","Text":"If you recall c from before is m_g minus"},{"Start":"04:51.320 ","End":"04:55.925","Text":"f_b and k is given to us as the constant from Stoke\u0027s law."},{"Start":"04:55.925 ","End":"05:00.365","Text":"Ultimately, I\u0027ve found v_y(t) and I can move on to my next steps."},{"Start":"05:00.365 ","End":"05:04.580","Text":"Now, if I was asked to find the entire velocity, not just y,"},{"Start":"05:04.580 ","End":"05:07.820","Text":"then I would also have to find my v_x in terms of t using"},{"Start":"05:07.820 ","End":"05:11.510","Text":"the same process going through the same integrals and separation of terms."},{"Start":"05:11.510 ","End":"05:14.035","Text":"I\u0027ll write out the answer here for you."},{"Start":"05:14.035 ","End":"05:16.165","Text":"If it doesn\u0027t come out the same as this,"},{"Start":"05:16.165 ","End":"05:18.535","Text":"it\u0027s possible I made a typo or a calculation error,"},{"Start":"05:18.535 ","End":"05:20.395","Text":"but I would double-check the work."},{"Start":"05:20.395 ","End":"05:23.335","Text":"V_x(t), once you do the calculation,"},{"Start":"05:23.335 ","End":"05:30.190","Text":"should equal v_0 cosine of Theta minus e to"},{"Start":"05:30.190 ","End":"05:33.610","Text":"the power of negative k over m times"},{"Start":"05:33.610 ","End":"05:38.365","Text":"t. If you don\u0027t quite feel comfortable doing integrals yet,"},{"Start":"05:38.365 ","End":"05:43.645","Text":"this is a good chance to practice one on your own and check your answers against mine."},{"Start":"05:43.645 ","End":"05:48.010","Text":"Now, let\u0027s use v_y(t) to find y in terms of"},{"Start":"05:48.010 ","End":"05:54.150","Text":"t. The way that we\u0027re going to find y(t) is take an integral of v_y(t)."},{"Start":"05:54.150 ","End":"06:03.915","Text":"Remember, y(t) equals an integral of v_y(t) dependent in dt."},{"Start":"06:03.915 ","End":"06:10.255","Text":"We\u0027ll do this as an indefinite integral and we\u0027ll add on a constant c tag later."},{"Start":"06:10.255 ","End":"06:15.460","Text":"The way that we\u0027ll find c tag is by looking at the initial velocity and going from there."},{"Start":"06:15.460 ","End":"06:18.250","Text":"Solving this integral isn\u0027t all that difficult."},{"Start":"06:18.250 ","End":"06:20.020","Text":"You have a constant at the beginning,"},{"Start":"06:20.020 ","End":"06:24.925","Text":"so you just multiply that by t and e to the power of anything remains the same."},{"Start":"06:24.925 ","End":"06:28.010","Text":"You just have to divide by the inner derivative."},{"Start":"06:28.010 ","End":"06:30.060","Text":"I\u0027ll give you the result now,"},{"Start":"06:30.060 ","End":"06:31.225","Text":"and for those who are interested,"},{"Start":"06:31.225 ","End":"06:32.410","Text":"at the end of the lecture,"},{"Start":"06:32.410 ","End":"06:34.105","Text":"I\u0027ll do the integral."},{"Start":"06:34.105 ","End":"06:41.375","Text":"Here we have y(t) equals c over k,"},{"Start":"06:41.375 ","End":"06:49.245","Text":"t plus m divided by k multiplied by"},{"Start":"06:49.245 ","End":"06:55.190","Text":"c minus v_0 sine Theta times"},{"Start":"06:55.190 ","End":"07:02.590","Text":"e to the power of negative k over m t plus c tag."},{"Start":"07:02.590 ","End":"07:10.820","Text":"The way we\u0027ll find c tag again is using y(t) equals 0. y(t) equals 0,"},{"Start":"07:10.820 ","End":"07:12.470","Text":"we know equals 0,"},{"Start":"07:12.470 ","End":"07:14.345","Text":"our initial height is 0."},{"Start":"07:14.345 ","End":"07:18.590","Text":"What this equals if we plug in t equals 0 here,"},{"Start":"07:18.590 ","End":"07:21.170","Text":"c over k times t equals 0,"},{"Start":"07:21.170 ","End":"07:30.720","Text":"what we end up with is m divided by k times c minus v_0 sine Theta."},{"Start":"07:30.720 ","End":"07:33.795","Text":"This will turn into 1 plus c tag."},{"Start":"07:33.795 ","End":"07:36.195","Text":"Notice that even when y equals 0,"},{"Start":"07:36.195 ","End":"07:39.330","Text":"c tag doesn\u0027t always equal 0."},{"Start":"07:39.330 ","End":"07:43.980","Text":"Now, I\u0027ll integrate my c tag into the equation above."},{"Start":"07:43.980 ","End":"07:47.820","Text":"All I get is negative 1 on the end here."},{"Start":"07:47.820 ","End":"07:50.620","Text":"If any of this didn\u0027t make sense now,"},{"Start":"07:50.620 ","End":"07:52.760","Text":"I\u0027ll come back to it after we do a summary and"},{"Start":"07:52.760 ","End":"07:55.795","Text":"show you how we do the nuts and bolts of this procedure."},{"Start":"07:55.795 ","End":"07:59.160","Text":"Let\u0027s just check to make sure this is right."},{"Start":"07:59.230 ","End":"08:08.600","Text":"y(t) euals0 should equal 0. So c over k times t equals 0. e^0 equals 1,1-1 equals 0,"},{"Start":"08:08.600 ","End":"08:11.750","Text":"0 times anything really doesn\u0027t matter what is 0."},{"Start":"08:11.750 ","End":"08:16.080","Text":"This works. We\u0027ve finished 2 of 3 steps."},{"Start":"08:16.080 ","End":"08:20.095","Text":"We\u0027ve found v_y(t) and y(t)."},{"Start":"08:20.095 ","End":"08:24.655","Text":"Now, we need to find the maximum depth, y max."},{"Start":"08:24.655 ","End":"08:27.610","Text":"How are we going to find our y max?"},{"Start":"08:27.610 ","End":"08:30.830","Text":"We need to find the time at which y is going to be maximum."},{"Start":"08:30.830 ","End":"08:34.340","Text":"The way we\u0027re going to do that is take a derivative"},{"Start":"08:34.340 ","End":"08:38.080","Text":"of y in terms of t and set that equal to 0."},{"Start":"08:38.080 ","End":"08:40.720","Text":"Really, what we\u0027re doing is taking y dot,"},{"Start":"08:40.720 ","End":"08:44.170","Text":"or in other words, v_y and setting that equal to 0."},{"Start":"08:44.170 ","End":"08:47.200","Text":"We use that to find the proper time to look at."},{"Start":"08:47.200 ","End":"08:51.970","Text":"Let\u0027s take our v_y(t) from above for a second."},{"Start":"08:52.670 ","End":"08:56.720","Text":"We have this and we need to now set it equal to 0."},{"Start":"08:56.720 ","End":"09:00.110","Text":"Now, at this point, all that\u0027s left for me to do is a little bit of algebra"},{"Start":"09:00.110 ","End":"09:03.685","Text":"to isolate the time t I\u0027m looking for and then plug that back in."},{"Start":"09:03.685 ","End":"09:05.195","Text":"I\u0027ll save this for the end."},{"Start":"09:05.195 ","End":"09:09.005","Text":"I\u0027ll do the algebra and also perform the integral from before."},{"Start":"09:09.005 ","End":"09:10.685","Text":"But before we move on,"},{"Start":"09:10.685 ","End":"09:13.640","Text":"I want to do a quick summary to make sure that you feel"},{"Start":"09:13.640 ","End":"09:17.940","Text":"comfortable with where we stand and the concepts that we\u0027ve talked about."}],"ID":9302},{"Watched":false,"Name":"Summary","Duration":"11m 55s","ChapterTopicVideoID":9030,"CourseChapterTopicPlaylistID":5337,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.620 ","End":"00:03.990","Text":"Let\u0027s do a summary. In our problem,"},{"Start":"00:03.990 ","End":"00:07.080","Text":"a ball is thrown into a pool with an initial velocity of"},{"Start":"00:07.080 ","End":"00:10.980","Text":"v_0 at an angle of Theta relative to the water surface."},{"Start":"00:10.980 ","End":"00:13.920","Text":"We\u0027re given all of these different figures and first we\u0027re"},{"Start":"00:13.920 ","End":"00:16.850","Text":"asked to write out the equation of motion for the ball."},{"Start":"00:16.850 ","End":"00:20.310","Text":"The way we do that is by taking the sum of forces equal to"},{"Start":"00:20.310 ","End":"00:25.010","Text":"ma in terms of a coordinate x or y, v and a."},{"Start":"00:25.010 ","End":"00:26.510","Text":"Now, to list out our forces,"},{"Start":"00:26.510 ","End":"00:27.590","Text":"we have F_k,"},{"Start":"00:27.590 ","End":"00:32.780","Text":"which is Stokes force or the force of friction with the water or liquid medium,"},{"Start":"00:32.780 ","End":"00:35.480","Text":"and that equals negative 6Pi Eta R,"},{"Start":"00:35.480 ","End":"00:36.649","Text":"which is a constant,"},{"Start":"00:36.649 ","End":"00:38.825","Text":"times v, the velocity vector."},{"Start":"00:38.825 ","End":"00:40.670","Text":"Now remember in the Stokes force,"},{"Start":"00:40.670 ","End":"00:45.620","Text":"we broke down our v vector into x elements and y elements so that we could split it in"},{"Start":"00:45.620 ","End":"00:51.950","Text":"half and take the x elements over to our x equation and the y elements to our y equation."},{"Start":"00:51.950 ","End":"00:55.115","Text":"We also have the force of buoyancy,"},{"Start":"00:55.115 ","End":"00:58.160","Text":"F_b, which is Rho w, the density of water,"},{"Start":"00:58.160 ","End":"01:00.320","Text":"one of our givens, times v,"},{"Start":"01:00.320 ","End":"01:01.760","Text":"the volume of the ball,"},{"Start":"01:01.760 ","End":"01:04.120","Text":"and g, the gravitational constant."},{"Start":"01:04.120 ","End":"01:05.950","Text":"Lastly, we have M_g,"},{"Start":"01:05.950 ","End":"01:07.310","Text":"the force of gravity."},{"Start":"01:07.310 ","End":"01:12.190","Text":"Now the mass for M_g is going to be Rho ball,"},{"Start":"01:12.190 ","End":"01:13.430","Text":"which is also given to us,"},{"Start":"01:13.430 ","End":"01:15.650","Text":"the ball\u0027s density, times v,"},{"Start":"01:15.650 ","End":"01:17.300","Text":"the volume of the ball,"},{"Start":"01:17.300 ","End":"01:22.655","Text":"and the volume of the ball we have the formula here as 4Pi r^3 over 3,"},{"Start":"01:22.655 ","End":"01:27.695","Text":"and we\u0027ll also be using that for the F_b, the buoyancy force."},{"Start":"01:27.695 ","End":"01:32.105","Text":"Knowing those forces, we did our sum of forces along the y-axis,"},{"Start":"01:32.105 ","End":"01:34.325","Text":"we got M_g minus F_b,"},{"Start":"01:34.325 ","End":"01:35.984","Text":"these are constants,"},{"Start":"01:35.984 ","End":"01:41.890","Text":"plus F_ky, which is the Stokes force along the y-axis."},{"Start":"01:41.890 ","End":"01:47.510","Text":"Along the x-axis, the only force that we had present is Stokes force."},{"Start":"01:47.510 ","End":"01:52.820","Text":"If you look here, our x-axis equation is negative kv_x,"},{"Start":"01:52.820 ","End":"01:57.200","Text":"remember k is the constant standing in for all of these figures here,"},{"Start":"01:57.200 ","End":"01:58.520","Text":"and both of these things,"},{"Start":"01:58.520 ","End":"02:00.665","Text":"the x equation and the y equation,"},{"Start":"02:00.665 ","End":"02:05.600","Text":"are in terms of v and a relative to their respective coordinates."},{"Start":"02:05.600 ","End":"02:11.705","Text":"The last note to remember from part A is that we call M_g minus F_b over c,"},{"Start":"02:11.705 ","End":"02:19.675","Text":"and we call the entire constant preceding v in our Stokes force k. Moving on to part B,"},{"Start":"02:19.675 ","End":"02:21.910","Text":"we\u0027re asked to find the terminal velocity."},{"Start":"02:21.910 ","End":"02:24.755","Text":"Now, we know that when we\u0027re looking at terminal velocity,"},{"Start":"02:24.755 ","End":"02:27.965","Text":"v is constant and therefore, a=0."},{"Start":"02:27.965 ","End":"02:34.125","Text":"So all we had to do was plug in a=0 to our 2 equations of motion and solve,"},{"Start":"02:34.125 ","End":"02:38.400","Text":"and what we get for y is c over k is the terminal velocity,"},{"Start":"02:38.400 ","End":"02:39.680","Text":"and on the x-axis,"},{"Start":"02:39.680 ","End":"02:42.005","Text":"the terminal velocity is 0."},{"Start":"02:42.005 ","End":"02:47.600","Text":"Now we saw that in regards to our terminal velocity along the y-axis that if M_g,"},{"Start":"02:47.600 ","End":"02:48.980","Text":"the force of gravity,"},{"Start":"02:48.980 ","End":"02:50.405","Text":"is greater than F_b,"},{"Start":"02:50.405 ","End":"02:51.800","Text":"the force of buoyancy,"},{"Start":"02:51.800 ","End":"02:54.290","Text":"then our terminal velocity will be positive,"},{"Start":"02:54.290 ","End":"02:56.510","Text":"meaning that our ball will sink."},{"Start":"02:56.510 ","End":"02:58.350","Text":"If the opposite is true,"},{"Start":"02:58.350 ","End":"03:00.750","Text":"if F_b is greater than M_g,"},{"Start":"03:00.750 ","End":"03:03.585","Text":"then our ball will float."},{"Start":"03:03.585 ","End":"03:05.500","Text":"Moving on to part C,"},{"Start":"03:05.500 ","End":"03:08.900","Text":"we\u0027re asked to find the maximal depth reached by the ball."},{"Start":"03:08.900 ","End":"03:10.985","Text":"Now, we know that to do that,"},{"Start":"03:10.985 ","End":"03:13.610","Text":"we need to find the position as a function of time,"},{"Start":"03:13.610 ","End":"03:16.250","Text":"and from there find the y maximum."},{"Start":"03:16.250 ","End":"03:18.695","Text":"In order to find y as a function of time,"},{"Start":"03:18.695 ","End":"03:22.510","Text":"we need to find the velocity of y as the function of time."},{"Start":"03:22.510 ","End":"03:26.000","Text":"The way that I get v_y or the velocity as a function of"},{"Start":"03:26.000 ","End":"03:29.150","Text":"time is by the method of separation of variables."},{"Start":"03:29.150 ","End":"03:34.955","Text":"I take my equation of motion and eventually get to a situation where on the left side,"},{"Start":"03:34.955 ","End":"03:36.640","Text":"I have some function, in our case,"},{"Start":"03:36.640 ","End":"03:40.640","Text":"it was the constant 1 times dt, and on the right side,"},{"Start":"03:40.640 ","End":"03:44.950","Text":"I have some function in terms of v multiplied by d_vy,"},{"Start":"03:44.950 ","End":"03:47.765","Text":"and then I can take an integral in each side."},{"Start":"03:47.765 ","End":"03:52.870","Text":"In this integral, I\u0027ll take my left side limit from t=0 to t=t,"},{"Start":"03:52.870 ","End":"03:54.164","Text":"and on the right side,"},{"Start":"03:54.164 ","End":"03:55.475","Text":"I\u0027ll take those same limit,"},{"Start":"03:55.475 ","End":"03:59.885","Text":"t=0 and t=t and plug them into my y function."},{"Start":"03:59.885 ","End":"04:04.895","Text":"So v_y(t)=0, once I use my given data from before,"},{"Start":"04:04.895 ","End":"04:07.970","Text":"equals v_0 sine Theta,"},{"Start":"04:07.970 ","End":"04:11.173","Text":"and v_y(t) equals v_y(t),"},{"Start":"04:11.173 ","End":"04:12.650","Text":"and this is what we\u0027re trying to find."},{"Start":"04:12.650 ","End":"04:14.270","Text":"I solve my integral,"},{"Start":"04:14.270 ","End":"04:15.785","Text":"do some creative algebra,"},{"Start":"04:15.785 ","End":"04:18.320","Text":"and what I end up with is v_y(t)."},{"Start":"04:18.320 ","End":"04:20.560","Text":"I also wrote out v_x here,"},{"Start":"04:20.560 ","End":"04:22.670","Text":"not because it\u0027s necessary for this problem,"},{"Start":"04:22.670 ","End":"04:25.580","Text":"but because it will be necessary for other problems,"},{"Start":"04:25.580 ","End":"04:29.360","Text":"but it is a very good exercise for anyone who\u0027s not yet comfortable"},{"Start":"04:29.360 ","End":"04:33.635","Text":"with the method of separation of variables and taking integrals that way."},{"Start":"04:33.635 ","End":"04:37.970","Text":"What I ultimately have here is v_y(t) and v_x(t)."},{"Start":"04:37.970 ","End":"04:40.820","Text":"This is my full velocity function in terms of time,"},{"Start":"04:40.820 ","End":"04:45.650","Text":"and I\u0027m going to use v_y(t) to find y(t) and then the y maximum."},{"Start":"04:45.650 ","End":"04:47.255","Text":"In the next stage,"},{"Start":"04:47.255 ","End":"04:51.800","Text":"I took an integral of v_y(t) in order to find y(t)."},{"Start":"04:51.800 ","End":"04:54.430","Text":"I did this using an indefinite integral,"},{"Start":"04:54.430 ","End":"04:58.495","Text":"so I had to add on c tag, some constant afterwards."},{"Start":"04:58.495 ","End":"05:01.750","Text":"Here I wrote my solution with c tag on the end,"},{"Start":"05:01.750 ","End":"05:05.405","Text":"and in order to find c tag, I put t=0."},{"Start":"05:05.405 ","End":"05:07.970","Text":"I know that this has to equal 0 for y,"},{"Start":"05:07.970 ","End":"05:10.685","Text":"and even though y=0 in this case,"},{"Start":"05:10.685 ","End":"05:13.535","Text":"our c tag ended up equaling negative 1,"},{"Start":"05:13.535 ","End":"05:15.080","Text":"so I added that in."},{"Start":"05:15.080 ","End":"05:19.550","Text":"Now we stand on the last stage where all we need to do is find y max."},{"Start":"05:19.550 ","End":"05:22.520","Text":"We said the best way to do that is take a derivative of"},{"Start":"05:22.520 ","End":"05:26.435","Text":"y relative to t and set that equal to 0."},{"Start":"05:26.435 ","End":"05:28.160","Text":"When that is equal to 0,"},{"Start":"05:28.160 ","End":"05:32.030","Text":"also the derivative of y to t is y. or v_y,"},{"Start":"05:32.030 ","End":"05:33.785","Text":"the velocity function,"},{"Start":"05:33.785 ","End":"05:35.420","Text":"then we\u0027re at a maximum."},{"Start":"05:35.420 ","End":"05:38.315","Text":"That also makes sense from a common sense point of view,"},{"Start":"05:38.315 ","End":"05:40.925","Text":"where your velocity equals 0,"},{"Start":"05:40.925 ","End":"05:42.130","Text":"it means you\u0027re turning around,"},{"Start":"05:42.130 ","End":"05:45.535","Text":"it means that you\u0027ve reached a maximum and you\u0027re going the other direction."},{"Start":"05:45.535 ","End":"05:48.545","Text":"We took v_y and set it equal to 0."},{"Start":"05:48.545 ","End":"05:50.210","Text":"When we find the answer to this,"},{"Start":"05:50.210 ","End":"05:53.240","Text":"the t time is going to be the time we need to plug back"},{"Start":"05:53.240 ","End":"05:56.710","Text":"into our y(t) equation to find y max."},{"Start":"05:56.710 ","End":"06:00.305","Text":"What I\u0027ll do now are 2 technical mathematical procedures."},{"Start":"06:00.305 ","End":"06:03.320","Text":"The first is I\u0027ll solve the integral from before for"},{"Start":"06:03.320 ","End":"06:06.560","Text":"those who still have trouble with that or would like to see it done,"},{"Start":"06:06.560 ","End":"06:12.155","Text":"and the second thing is do the algebra to find t and plug that back into our y equation."},{"Start":"06:12.155 ","End":"06:14.540","Text":"If you feel like you have control of these things,"},{"Start":"06:14.540 ","End":"06:16.310","Text":"if you feel comfortable with them already,"},{"Start":"06:16.310 ","End":"06:17.900","Text":"this part of the video is optional."},{"Start":"06:17.900 ","End":"06:19.670","Text":"However, if you still feel in any way"},{"Start":"06:19.670 ","End":"06:22.790","Text":"uncomfortable or that you don\u0027t yet understand all the material,"},{"Start":"06:22.790 ","End":"06:26.360","Text":"I would very highly suggest watching this."},{"Start":"06:26.360 ","End":"06:27.590","Text":"Let\u0027s look at the integral."},{"Start":"06:27.590 ","End":"06:29.330","Text":"If you recall when we first did this,"},{"Start":"06:29.330 ","End":"06:32.225","Text":"I just wrote down an answer without explaining how we got there."},{"Start":"06:32.225 ","End":"06:35.945","Text":"So now I want to take a step-by-step approach and move towards our answer."},{"Start":"06:35.945 ","End":"06:39.640","Text":"First of all, y(t), the result,"},{"Start":"06:39.640 ","End":"06:42.560","Text":"if we look at c over k, that\u0027s a constant,"},{"Start":"06:42.560 ","End":"06:43.746","Text":"it will stay the same,"},{"Start":"06:43.746 ","End":"06:45.860","Text":"multiplied by t,"},{"Start":"06:45.860 ","End":"06:53.430","Text":"and we need to subtract from that c minus v_0 sine Theta,"},{"Start":"06:53.430 ","End":"06:56.240","Text":"and we\u0027re going to multiply that by e. Now,"},{"Start":"06:56.240 ","End":"06:58.925","Text":"e to the power of something will stay the same,"},{"Start":"06:58.925 ","End":"07:02.630","Text":"except that we need to divide by the inner derivative."},{"Start":"07:02.630 ","End":"07:05.855","Text":"Instead of dividing by negative k over m,"},{"Start":"07:05.855 ","End":"07:07.610","Text":"we\u0027ll multiply by the inverse,"},{"Start":"07:07.610 ","End":"07:11.495","Text":"so multiply this by negative m over"},{"Start":"07:11.495 ","End":"07:16.670","Text":"k. Because we\u0027re multiplying this whole article by negative mk,"},{"Start":"07:16.670 ","End":"07:18.200","Text":"and initially we\u0027re subtracting it,"},{"Start":"07:18.200 ","End":"07:21.725","Text":"we can just multiply it by positive mk and add it."},{"Start":"07:21.725 ","End":"07:23.525","Text":"Now that I have my solution,"},{"Start":"07:23.525 ","End":"07:27.260","Text":"all I need to do is add my constant c and move on from there."},{"Start":"07:27.260 ","End":"07:30.650","Text":"Just to summarize what we did before, we set t=0."},{"Start":"07:30.650 ","End":"07:36.530","Text":"When t=0, e^negative k over m t equals 1,"},{"Start":"07:36.530 ","End":"07:39.200","Text":"and c over k t equals 0."},{"Start":"07:39.200 ","End":"07:43.530","Text":"So what we\u0027re left with is y(t) equals 0,"},{"Start":"07:43.530 ","End":"07:50.085","Text":"equals c minus v_0 sine Theta,"},{"Start":"07:50.085 ","End":"07:55.215","Text":"m over k plus c tag."},{"Start":"07:55.215 ","End":"07:57.755","Text":"If we take c tag to the other side,"},{"Start":"07:57.755 ","End":"08:02.120","Text":"will find that c tag equals negative c minus"},{"Start":"08:02.120 ","End":"08:08.175","Text":"v_0 sine Theta times m over k,"},{"Start":"08:08.175 ","End":"08:11.990","Text":"and ultimately, we incorporate that by taking negative 1 on"},{"Start":"08:11.990 ","End":"08:16.185","Text":"the end of e^negative k m, t,"},{"Start":"08:16.185 ","End":"08:20.105","Text":"and we incorporated that below with this negative 1 in the parentheses,"},{"Start":"08:20.105 ","End":"08:21.740","Text":"because as you see when we multiply that out,"},{"Start":"08:21.740 ","End":"08:27.085","Text":"it equals negative c minus v_0 sine Theta times mk."},{"Start":"08:27.085 ","End":"08:29.990","Text":"You can see that we incorporated this in"},{"Start":"08:29.990 ","End":"08:32.960","Text":"your equation on the left with a negative 1 here,"},{"Start":"08:32.960 ","End":"08:35.720","Text":"it ends up equaling the same thing in the end."},{"Start":"08:35.720 ","End":"08:38.675","Text":"Let\u0027s move on to solving for t here."},{"Start":"08:38.675 ","End":"08:48.500","Text":"We know that v_y(t) equals 0 and that equals c over k minus c minus"},{"Start":"08:48.500 ","End":"08:52.625","Text":"v_0 sine Theta times"},{"Start":"08:52.625 ","End":"08:59.120","Text":"e^negative k over m times t. First we want to isolate our t,"},{"Start":"08:59.120 ","End":"09:02.920","Text":"and the way we\u0027ll do that is by isolating the e on the other side."},{"Start":"09:02.920 ","End":"09:04.850","Text":"If we put the e on our left side,"},{"Start":"09:04.850 ","End":"09:08.420","Text":"we get e^negative k over m,"},{"Start":"09:08.420 ","End":"09:12.605","Text":"t equals c over"},{"Start":"09:12.605 ","End":"09:20.854","Text":"k times c minus v_0 sine Theta."},{"Start":"09:20.854 ","End":"09:25.670","Text":"The way I did this was by subtracting this whole section on the right from"},{"Start":"09:25.670 ","End":"09:30.665","Text":"both sides and then dividing by c minus v_0 sine Theta."},{"Start":"09:30.665 ","End":"09:34.055","Text":"The next step is to take the natural log of both sides."},{"Start":"09:34.055 ","End":"09:39.500","Text":"So the e will drop out on the left and we\u0027ll be left with negative k over m,"},{"Start":"09:39.500 ","End":"09:41.900","Text":"t equals, on the right,"},{"Start":"09:41.900 ","End":"09:44.720","Text":"the natural log of everything we had before,"},{"Start":"09:44.720 ","End":"09:51.905","Text":"c divided by k times c minus v_0 sine Theta."},{"Start":"09:51.905 ","End":"09:55.745","Text":"Now, if we multiply both sides by negative m over k,"},{"Start":"09:55.745 ","End":"09:58.460","Text":"what we\u0027re left with is an isolated t on the left side,"},{"Start":"09:58.460 ","End":"09:59.540","Text":"that\u0027s what we\u0027re looking for,"},{"Start":"09:59.540 ","End":"10:01.160","Text":"and on the right side,"},{"Start":"10:01.160 ","End":"10:06.695","Text":"negative m over k times the natural log of c"},{"Start":"10:06.695 ","End":"10:13.565","Text":"divided by k times c minus v_0 sine Theta."},{"Start":"10:13.565 ","End":"10:16.970","Text":"Now, don\u0027t worry about this negative mk here,"},{"Start":"10:16.970 ","End":"10:18.725","Text":"we\u0027re not going to get a negative time."},{"Start":"10:18.725 ","End":"10:21.140","Text":"Because we\u0027re taking the natural log of a fraction,"},{"Start":"10:21.140 ","End":"10:22.430","Text":"this will also be negative,"},{"Start":"10:22.430 ","End":"10:24.470","Text":"so our ultimate result will be positive."},{"Start":"10:24.470 ","End":"10:26.900","Text":"But now that we have an isolated t,"},{"Start":"10:26.900 ","End":"10:29.630","Text":"we can use this time negative m over k,"},{"Start":"10:29.630 ","End":"10:32.185","Text":"et cetera, to find our y max."},{"Start":"10:32.185 ","End":"10:34.135","Text":"So let\u0027s start writing that out."},{"Start":"10:34.135 ","End":"10:38.235","Text":"y max equals,"},{"Start":"10:38.235 ","End":"10:41.295","Text":"now remember we\u0027re going to take ck times t,"},{"Start":"10:41.295 ","End":"10:51.245","Text":"so what we end up with is cm over k^2 times the natural log of k times"},{"Start":"10:51.245 ","End":"10:56.030","Text":"c minus v_0 sine Theta over"},{"Start":"10:56.030 ","End":"11:00.560","Text":"c. What we\u0027ve done here is take the"},{"Start":"11:00.560 ","End":"11:02.540","Text":"negative from the outside and put it on"},{"Start":"11:02.540 ","End":"11:05.555","Text":"the inside so that we can switch these terms around."},{"Start":"11:05.555 ","End":"11:14.850","Text":"Now we can add on m over k times c minus v_0 sine Theta,"},{"Start":"11:14.850 ","End":"11:18.305","Text":"and instead of e^negative km, t,"},{"Start":"11:18.305 ","End":"11:20.390","Text":"we\u0027ll replace it to something that equals that so we"},{"Start":"11:20.390 ","End":"11:22.610","Text":"can keep things in terms of our constants."},{"Start":"11:22.610 ","End":"11:28.175","Text":"We multiply this by c over k,"},{"Start":"11:28.175 ","End":"11:33.310","Text":"c minus v_0 sine Theta,"},{"Start":"11:33.310 ","End":"11:36.510","Text":"and we\u0027re going to subtract from that 1,"},{"Start":"11:36.510 ","End":"11:38.475","Text":"that\u0027s our c tag,"},{"Start":"11:38.475 ","End":"11:39.905","Text":"and we have our answer,"},{"Start":"11:39.905 ","End":"11:41.855","Text":"this is our y maximum."},{"Start":"11:41.855 ","End":"11:45.515","Text":"I realize this may look long and unwieldy, but c, m,"},{"Start":"11:45.515 ","End":"11:47.195","Text":"k, and Theta,"},{"Start":"11:47.195 ","End":"11:49.460","Text":"as well as v_0 are all given to us."},{"Start":"11:49.460 ","End":"11:53.300","Text":"So we can actually find a numerical answer very simply by plugging in our values."},{"Start":"11:53.300 ","End":"11:56.730","Text":"Now we\u0027ve solved the entire problem."}],"ID":9303}],"Thumbnail":null,"ID":5337}]

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