Conservation of angular momentum during a jump to his...

Conservation of Angular Momentum During a jump to his partner, an aerialist is to make a quadruple somersault lasting a time $t=1.87$ s. For the first and last quarter-revolution, he is in the extended orientation shown in Fig. $11-55,$ with rotational inertia $I_{1}=19.9 \mathrm{kg} \cdot \mathrm{m}^{2}$ around his center of mass (the dot). During the rest of the flight he is in a tight tuck, with rotational inertia $I_{2}=3.93 \mathrm{kg} \cdot \mathrm{m}^{2} .$ What must be his angular speed $\omega_{2}$ around his center of mass during the tuck?

Transcript

00:02 So in this video we're trying to figure out how fast an aerialist must spin in order to complete a trick while jumping to their partner.

00:13 So the trick that this aerialist is trying to perform is a quadruple somersault.

00:19 So that means that the total amount of rotation that we're going to go through is four rotations which is eight pi radians, two pi for one rotation times four.

00:34 And we know that the total time that they have in the air is 1 .87 seconds.

00:46 Now, it also tells us that the first and last quarter of the revolution, they are extended outwards with a moment of inertia of 19 .9 kilograms times meter square.

01:08 And then the remaining three and a half rotations, they are tucked at a moment of inertia of, 3 .93 kilograms meters squared.

01:27 So where do we start? so first we need to figure out, all right, well what is the ratio of how fast they're going to spin in position one versus position two? you can do this with angular momentum.

01:41 So conservation of angular momentum, the angular momentum in position one has to equal the angular momentum in position two.

01:48 So that means that the moment of inertia times the angular velocity in position 1 has to equal the moment of inertia times the angular velocity in position 2.

02:04 So this gives us...

02:07 I'm going to leave out units just for room.

02:09 19 .9 omega -1 equals 3 .93 omega -2.

02:21 And what we want to do, since we want to figure out specifically just the angular momentum in position 2, i want to solve for position.

02:30 Omega 1.

02:32 So i'm going to divide both sides by my 19 .9.

02:35 And what that's going to give me is omega 1 equals omega 2 divided by 5 .06.

02:49 And we'll round the zigfigs up here.

02:52 Okay.

02:55 So what we know is that angular velocity is equal to the angular distance theta over t.

03:09 So, but what we do know is.

03:11 Is that in position one, our angular velocity is omega 2 over, omega 2 over 5, 5, 06, 4.

03:35 This is all position 1.

03:42 We know that in position 1, if you go through a total of 1 half of a rotation, so pi radiance, over some amount of time.

03:56 So if we cross -multiplied divide, we end up with t equals 5 .064 times pi over omega 2.

04:14 Get in somewhere.

04:16 So that tells us the amount of time that is taken up by position 1...

Question

Please give Ace some feedback