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Dizzy physics student(fill in name) is spinning on our near-frictionless blue thing while holding weights. The weights each have a mass of 10 kg: We can assume that...

Question

Dizzy physics student(fill in name) is spinning on our near-frictionless blue thing while holding weights. The weights each have a mass of 10 kg: We can assume that the student measures meter from hand to sternum; and shoulders are 5 meters wide: The weights can be considered particles: the student is spun at an initial angular speed of 2 revolutions per second with the hands extended, what is the new angular speed when the weights are pulled into the shoulders?metermeters

dizzy physics student (fill in name) is spinning on our near-frictionless blue thing while holding weights. The weights each have a mass of 10 kg: We can assume that the student measures meter from hand to sternum; and shoulders are 5 meters wide: The weights can be considered particles: the student is spun at an initial angular speed of 2 revolutions per second with the hands extended, what is the new angular speed when the weights are pulled into the shoulders? meter meters



Answers

A professor doing a lecture demonstration stands at the center of a frictionless turntable, holding 5.00 -kg masses in each hand with arms extended so that each mass is $1.20 \mathrm{~m}$ from his centerline. A (carefully selected!) student spins the professor up to a rotational frequency of 1.00 rpm. If he then pulls his arms in by his sides so that each mass is $0.300 \mathrm{~m}$ from his centerline, what is his new angular speed? Assume that his rotational inertia without the masses is $2.80 \mathrm{~kg} \mathrm{~m}^{2}$, and neglect the effect on the rotational inertia of the position of his arms, since their mass is small compared to the mass of the body.

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.

In this question. Uh, we have a situation whereby a student hold bicycle view a spinning view, and then initially, it has an angular momentum pointing up after that, he has angular momentum down. And because of that, the students and, uh, to, uh, system. Uh, rotating. Uh, Okay. So, um, we want to find out, uh, students finally, angular speed after the turning and then in the direction of the final students final. Angular Momentum. Okay. Okay. So, uh, in this question, Okay, it is about conservation of angular momentum. Okay, so okay, using, uh, conservation. Okay, uh, angular momentum. So Okay, so the system that we are dealing with is bicycle view plus student plus stew. Okay, so we have, uh I be, uh, angular. This moment of inertia of bicycle view. Oh, my God. Be I It goes to I b *** B f s i b plus I s This is just, um hi. Scooping class to my guys King. Mhm. All right, So, um, initially that we take up to be the positive direction. Okay, uh, to be positive, uh, before after. So, um, Andy is even to be zero from one high from the Gabby. I will be 17 5. Okay. And then, uh, at the end after that is, uh oh, my God, we have is negative because it's pointing downward, then three and oh, my God s okay. So Oh, my God s is, uh, 0.3 times 17.5 by three. And you get 1.75 to get a second thing. And from the diagram, uh, you can tell that, uh, the direction of the Omega s here or the direction of China. Angular momentum of the students is pointing up. Okay, so that's how we use a conservation of angular momentum to solve the problem, that's all.

In this solution given that it is that is there is a gender platform given in the shape of circular this. So consider that this is the circular disk and this circular disk is rotating around a access vertical axis which is passing through. Yeah the center of this circular this so consider that this is the rotational axis. And for the information is given that is math of this circular. This is given that is M. Is equal to 1 50 g and sorry, 1 50 kg. And radius is given that is r is equal to two m. And rotational inertia is given. That is equal to 300 G.m. Square. So for the information is given that is I consider this is the radius of this circular disk. So in the question is saying that at the rim of this circular disk there is a student is assume that there is a student is Present, whose weight is given that is 60 Kg and no in the question it is ended. and this in this condition the Angular is well of system is given that is equal to 1.5 radiant per second. Further in the question is saying that now this a student is walking slowly from the rim to the was the center. And afraid when she is 0.50 m from the center, then we have to find the angular is created. So considered that the distances, if I consider that this is the equal to value that is from the respect to center of this healthier, then I can see that It is 0.5 km and at this point when it is reaching year, after working slowly slowly from the rim towards the center, then we have to find the angular speed. So so this is a fusion. That means if I consider that this is the omega Desh, then we have to find the anglers with off that a student. So this execution now let's start answering discussion. So in the answered first first I will write the rotational inertia here and initial condition, rotational inertia of total system. That means when the student is at this point at the dream of this circular disk and then afraid teaching that at this point from distances that is 0.50 m from center of the circular disk. So I can say that the initially the rotational inertia of the system will be written as. Yeah that is I want will be equal to total value. Is the dissertation inertia of disk. Less rotational inertia of a student with respect to rotational axis which is passing through this center of this circular this so value will be, I can say that the rotational inertia of this will be written is that is equal to in the Russian given There is 300 Kg metarie square plus no rotational energy of a student will be written as that is about the rotational axis, about this vertical axis which is passing through the center of this circular this. So I can say that value will be, there is M. R square. So I will be, I can say that that is initially it is resting at the rim of this circular disk. So I considered are will be witness that is equal to areas of this circular disk. So that is given in the question that is to metal and me's that is mass of this student. That is given in the question that is 60 Kg. So it will be witness that is 300 plus 60 multiplied by two is square. So or I can say that value will be That is a rotational inertia of the system I one will be that that is at initial portion. It will be equal to 540 kg meters square. Now I will write the rotational inertia of system after when the student is at this point. So I considered in this condition that are will be euthan,ized that is 0.50 m And the M. is constant. That means massive student. That was written by city 60 Kg. So use this expression again. Then I can say that the rotational industry of system I do after reaching the student at this point where it is from 0.5 m from the center of this circular disk. So where you will be internalized, that is rotational inertia of this. That is given 300 plus rotational inertia of a student about the vertical axis which is passing through the center of the circular days. So where you will be that is M. R. Square and will be that is 60 kg multiplied by our will be written is that is Do you know .50 square. So No simplify it. Then I will get the value of protesters in Austria I. two. That is equal to I consider 300, multiplied by 60 but declared by. Sorry, total value will be that is 315 KK m square. So no nexus to bees. Yeah. Here I will write the angular momentum conservation equation, Social bitterness. That is Ivan Omega one is equal to I two. Omega 2. So here you can see that Omega one is uh angular speed of the system at initial portion and Omega two is an angular speed of system after after when a student is reaching at this point. That means at this point. So I considered in the question Omega one is given yeah, That is 1.5 radian per second and omega dishes That is equal to Omega two so which we have to find out. So No next step is here. Put the value of I one and I do. Which I had get in the above step. So I wan is that is 540 kg meter square and I do is 3300 and 15 kg meter square. So after securing it will be organized. That is Ivan will be 30 540 multiplied by Omega one is given. That is mm 1.5 multiplied by I do is that is 315 multiplied by. Omega two will be that is Omega desh. So simplify it. Then I will get the answer for omega desk that is equal to After solving final value is there is 2.6 radiant per second. Yes. So this is the sort of given question. So I can see that The Anglo speed when the student is 0.5 km from the center, that will be equal to 2.6 radian per second. So I have saw the vision completely. Thank you.

Problem. 11.58. We have a discus rotating horizontal plane that has a massive 150 kilograms a radius, two meters and the given moment of inertia. Leo person walking from the rim thio half a meter away from the center. And given the initial frequency, your angular speed with which it's moving, we'd like to find what its final angular speed is. So this is ah, and conservation of momentum. Angular momentum problem. Yeah, make a final. And so, um, the total moment of inertia is just the sub of the that of the disc. So this is the initial so, um, r squared, invited by my our final squared. So the moment of inertia is just the sum of the moment of inertia the disc, plus the point, the point mass that the this person represent Erupt that represents his person. Uh huh. And so then we get 2.6 radiance per second is the final angular speed


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