Stool (initially motionless) while holding bicycle wheel A student is sitting on radius of 75 cm. Assume the wheel to be thin which has mass of 1.2 kg and hoop: The...

A student holds a spinning bicycle wheel while sitting motionless on a stool that is free to rotate about a vertical axis through its center (Fig. P8.67). The wheel spins with an angular speed of 17.5 rad/s and its initial angular momentum is directed up. The wheel's moment of inertia is 0.150 $\mathrm{kg} \cdot \mathrm{m}^{2}$ and the moment of inertia for the student plus stool is $3.00 \mathrm{kg} \cdot \mathrm{m}^{2}$ . (a) Find the student's final angular speed after he turns the wheel over so that it spins at the same speed but with its angular momentum directed down. (b) Will the student's final angular momentum be directed up or down?

In this exercise, we have a student that is sitting in a rotating rotating stool here in green, and the students starts rotating with his arms extended about a rotation access that I drew here as this vertical black line. And if the student is holding in his hands to weights off 3 kg each and when his arms are extended, the distance between each hey, each weight and the rotation axis is a go to 1 m. Then after that, they student starts to close, closes his arms until the distance between each weight and the deportation access is your 0.3 m, okay, and when he does this, the angler velocity off the student changes from W I to w f great eso I wrote here on the right, the mass, the quantities that were given to us by the exercise. So the mess off each weight, the initial distance off each weight and rotational axis the final distance, whether is to close his arms between the weight and the rotational axis, I hear is the moment off. Inertia off the students and the to and this moment of inertia way assume that this does not change and Also, we have the initial angler velocity off the student in the before scenario. Okay, so question a asks us to evaluate what is the final angular velocity wf when they shouldn't closes his arms. Okay, so to do this to calculate this, we must notice that in our system there is no external force been applied on the student and on the weights. So we can and infer that the angler, because off no external force, the angular momentum off the student is the angular momentum off the system student. Plus weight is conserved, so the England momentum is conserved. Okay, so we can So by that we can say that the final angular momentum when the student closes his arms is equal to the initial angular momentum when the students is with the arms extended. So Okay, so, um, we have that The expression for angular momentum is l is equal to the moment of inertia times the angler velocity. So we have that we have the final moment, Offner Sha times The final angular velocity shall be equal to the initial moment of inertia times the initial angular velocity. We know what the initial angular velocity iss okay. And we can find what is initial and final moment of inertia, and we can calculate this by the following. So if we treat each off the weights as a point particle and knowing that the moment of nursing off the student is constant in both scenarios, we have that. So, for instance, the initial is equal to the moment of inertia off the student, plus the moment of inertia off each weight. Okay, so the moment off nurse off the student is given it is three plus the moment of inertia off the weight. So treating the weights as, uh, two point particles we have that this is equal to two times mg. Sorry. Uh, two times. Uh, m r. Sorry, I m d I squared on which the eye is a decent between each weight and the rotation axis. So we can to be stupid, The values and we have that this is equal to three plus two times, three times one squared. So we have that the total initial moment of inertia is equal to, uh, 9 kg meter squared. Furthermore, we can do a similar thing to calculate the final moment of inertia. So it's the final moment of inertia off the student, plus the final moment of inertia off the weights. Okay, so let me put this here. Um, So again, the moment of inertia off their student, uh, is the same is conserved as stated by the exercise, plus two times the moment of inertia off each. Wait. So M, which is three And now the distance between each, uh, wait to the rotational access, it changes to 0.3. So now we have time, 0.3 squared. And by doing so, we find that the final moment of inertia let me write it down here. So you have the initial and the final changes to 3.54 kilograms meter squared. So in the final scenario is much easier to rotate the system than in the beginning. Due to the due to the change on the moment of inertia. If the moment of inertia is greater, it's easier. It's harder to rotate the system. If the moment of inertia is small, it's easier to rotate the system. Um, okay. So we can substitute this value. So we have that, uh, I s Omega s. It goes to I I Omega I And this leads to, uh, nine time 0.75 is equal to sorry. This is the initial. Let me right here. So we don't confuse make, uh, any mistakes. So the final moments of inertia 3.54 times omega f is equal to now nine times 0.75. And we can isolate Omega F and find it to be equal to 1.91 radiance per second. Okay, Now, question be asks us to calculate what is the connect the kinetic energy off the system before and after he pulls the weights, uh, inwards towards the rotational axis. So we want to find what is the kinetic energy ke I on the before scenario and k f off the after scenario. Okay, so that's right here. We know that the definition off kinetic energy for our rotation system is you go to I Omega squared. Okay, so for the before scenario, okay, I is equal to I Omega. I squares and we have all these values, so we just Oh, sorry. It's I only get squared over two so over to Andi will just be suit the values and this is nine times 0.75 squared over two. So we find, um, que i the initial kinetic energy to be equal to two point 53 Jews. Yeah, And for the after scenario, the final kinetic energy It's just is just i f times omega f squared over two, which is, uh, 3.54 times 1.91 b squared. Oh, virtue. And we find that the final kinetic energy is equal to 6.44 jewels. And with this, we conclude the exercise.

Here. We're going to use Newton's second law of rotational analog of that. And actually more of an analog of the impulse momentum theorem, in order to figure out the force a person is applying to a bicycle wheel to change its angular momentum to a standstill. So impulse angular momentum is a rotational analog of the impulse momentum theorem for linear motion. And so basically we have that torque or some of torques is equal to change in angular momentum in time. And if we move the time over to the torque, we have something that looks like impulse angular impulse. Okay, so the torque a couple things to break out. We're going to have to use the rigid body formula for angular momentum, which is equal to moment of inertia of the object time, It's angular velocity, and our initial amount is the moment of inertia of the who times its angular velocity. Which we will want to get in terms of radiance per second, and the final amount is zero, and our delta L. Is el final minus L. Initial And that is -1 times omega initial. And we'll drop the negative sign. But what that negative sign tells us is in my picture, I already have incorporated that in order to slow the wheel down. The force must be creating a torque that's opposite the motion. So our torque will also be negative um And will be in the amount F types the radius of the wheel so that negative sign will disappear very nicely. And we are trying to figure out the force given the time it takes to slow down from a certain initial speed. And now it's a matter of putting in for the moment of inertia of the wheel which we are going to treat like a hoop with all of its mass concentrated on the rim. That's a little bit of a approximation. But we'll go ahead and do that anyway. Um It's usually good enough to get us started anyway. Um So let's see. We've got a 3.25 kg wheel With a radius .41 m. And putting that all together gives us a moment of inertia of less than one kilogram metre squared and the mega initial. We want to make sure that we Take our two ribs per seconds, multiply by two pi radiance per revolution to get a uh angular speed. That's in an acceptable form of units. And that is 12 points six. So we can now put that all together if we use them. Okay. S or System International Units we should wind up with a force in newtons. Let's see. The time was 3.50 so we'll wind up with a force in newtons And we get about 4.8 newtons. Mhm. Which is not too much. Person could easily do that. Mhm.

For this problem on the topic of angular momentum, we're told that a student is rotating on a stool whilst holding two weights. Each of these weights have a mass of three kgs. And when the students arms accident horizontally, the way it's a meter from the axis of rotation And rotates an angular speed of 0.75 radiance per second. We are told at the moment of inertia of the student plus two is constant at three kg meter squared. The student then pulls the weights inward horizontally to a position that is 0.3 m from the rotation axis. We want to firstly find the new angle angular speed of the student and then find the kinetic energy of the rotating system before and after the weights are pulled inward. Now the total angular momentum of the system of the student, the stool and the weights about the axis of rotation is given by I total and this is equal to the moment of finisher of the weights I w plus the moment of inertia of the student is so this is equal to two into em r squared plus three kg meet a squid. So Before the weights are brought in word, R is equal to one m and the initial moment of inertia II is equal to two into three kg times one meter squared plus three kg meter squared, which gives the initial moment of inertia to be nine kg meter squared. Afterwards, R is equal to 0.3 m when the weights are brought in word. And so the final moment of initial system I. F. Is equal to two into three kg time, zero point three m squared plus three kg meter squared. And so the final moment of Russia is three 0.54 kg meter squared. Now we can use the conservation of angular momentum and this tells us that I. F. Omega F is equal to I I. Omega I. And so this means that the final angular speed of the student omega F is equal to the initial moment of inertia over the final moment of inertia times the initial speed of rotation. And so this is nine divided by three 0.54 Times the initial angular speed of 0.75 radiance the second, and so the angular speed of rotation after the weights are brought in word Is 1.91 radiance for second. Next you want to find the kinetic energy of the entire system before and after the weights are pulled inward. So the initial kinetic energy K. I. Is a half I I omega I squared, which is a half times nine kg meters squared Times 0.75 gradients, the second squared. So the initial kinetic energy Is 2.53 jules. Now the final kinetic energy k f is equal to a half i f. Omega F squared, Which is a half times the final moment of Inertia, k g meter squared Times the final angular speed of 1.91 radiance per second squared, Which gives us the final Kinetic energy to be 6.44 jewels.

Okay. And this problem, you have a student rotating, holding dumbbells, and then the arms extended and the dumbbells aired a certain distance. Um, and the student pulls a dumb around dumbbells in, and we want the new angular speed, and we want to find the kinetic energy before and after. So let's write down the givens. Then we can discuss how to do the problem. So, um, the masses three of each dumb bell. So I'll put, um, put times to whose 3.0 kilograms. And, um, the radius the distance they are is one meter from the axis of rotation and the, um Omega is 0.75 radiance per second. And, um, I is 3.0 kilogram meters squared, so our final is 0.3. So I'll call this our initial, and then this Our final equals 0.3 oh meters. And our goal is to get oh, may go final. So this was also Omega initial. So there is no external torques acting on the system of the student and the dumbbells. So you can say angular momentum is conserved. How initials you go out final. The initial angular momentum is the moment of inertia of the system, Um, which consists of the moment of inertia of a student plus and and stool, plus the moment of inertia of the dumbbells. Initially so put a little I hopes. And then we want to multiply by, uh, initial And the final moment of angular momentum is the, um, a moment of inertia of the student and stool, plus the moment of inertia of the dumbbells finally all multiplied by Omega final. And so we can use this to sell for Omega Final. So I have the students, um, in stool that's given, and then I of the dumbbells, um, we can model them. It's point masses located a distance r from the axis of rotation. So then that makes them, um m r squared. And then there's two of them, so it's gonna be to our initial squared times homemade. Uh oh, my gosh. What happened to Omega Times? Omega initial and then I asked, is gonna be the same. But now they're at their new radius. So it's two AM our final squared times, Omega final and then Omega final. What we want to do is we want to divide this by this to get a mega final, and we got so that becomes, um, this. And now I'm gonna put back into a calculator to perfectly align on my tabs. So, um, e got three plus two times. Um, three times one times, 0.75 And then we want to divide this whole seeing by, um, almost the exact same thing. Oh, I need to multiply the whole thing by that omega with me. Fix my parentheses later. Um, hey, so we want to do three plus two times, three times 30.3 squared. Um, and now I got 1.91 radiance per second, and we, um so next we want to find the, um, kinetic energy before and after. So, um que initial We actually label this? Oh, yeah. I said I would put parentheses here. Okay. So OK, initial is 1/2 I omega initial squared, and then Okay, you final. And then I boobs me, Raymond, I s, um, Eyes, Of course. I s plus two. Um, our initial squared and, um uh, and then times omega initial squared. And then k final is one have times basically the same thing, except for everything. Has a final subscript if it had an initial subscript initially. So let's plug all that into a calculator. Um, so I got Do you want half times High Times Square, my omega. And with that, I get, um, 2.5 three. All right, in a second. And then finally, um, basically everything the same, except for a change, are 2.3, and then I can who make up final to, um, 1.91 So, um, the final kinetic energy is six point for six, Jules. And in the initial whoa is 2.53 jewels and