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$\cdot$ On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.60 m from the axis of rotation of the stool. She is given an angular velocity of 3.00 rad/s, after which she pulls the dumbbells in until they are only 0.20 m distant from the axis. The woman's moment of inertia about the axis of rotation is 5.00 $\mathrm{kg} \cdot \mathrm{m}^{2}$ and may beconsidered constant. Each dumbbell has a mass of 5.00 $\mathrm{kg}$ and may be considered a point mass. Neglect friction. (a) What is the initial angular momentum of the system? (b) What is the angular velocity of the system after the dumbbells are pulled in toward the axis? (c) Compute the kinetic energy of the system before and after the dumbbells are pulled in. Account for the difference, if any.

a) 25.8 $\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$b) 4.78 $\mathrm{rad} / \mathrm{s}$c) 22.9 $\mathrm{J}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

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So here we're trying to find the initial angular momentum, the final angular velocity and the ah change and kinetic energy. So we have an initial radius equal in 0.600 meters. We're going to say that the final radius is going to be equal to point 200 meters. We know that thie Ah, angular. The initial angular velocity is equaling 3.0 radiance per second and we know that the mass is going to be 5.0 kilogram. And these these are for the dumb bills. So there's going to be two dumbbells, so we can say that the we also we also know that the moment of inertia of the woman it's going to be equal to 5.0 kilograms meters squared. So for party, when they want us to find the initial angular momentum, this will be equal to the initial moment of inertia times the initial angular velocity. This is equaling the moment of inertia of the woman, plus two dumbbells times the mass of one dumb bell times the radius initial squared Time's omega initial and this is equaling three times five plus two times five Times point fix squared, and the initial angular momentum is equaling 25.8 kilogram meters squared per second for Part B. We know that the angular momentum is conserved, so we can say that I initial omega initial equals I Final times Omega final. We're trying to find Omega Final. So mega final is going to be equal to a moment of inertia. Initial omega initial divided by moment of inertia Final. And this will equal moments of a rather angular the initial angular velocity times the angular moment of inertia for the woman plus two em are initial squared, divided by the moment of inertia of the woman once again plus two em are final squared, and at this point, we can solve. So Omega Final is going to be equal 23 times, five plus two times, five times 0.6 squared And this is going to be divided by five plus two times five times 0.3 squared. Another 0.2 squared my apologies, and we find that the final angular velocity is going to be 4.78 radiance per second. So this will be the final answer for Part B. This is your final answer for part A. Let's get a new workbook for part. See report. See, they want the change in kinetic energy so the kinetic energy initial will simply be equal to 1/2. I initial Omega initial squared, and this is simply equal to 1/2 times 8.6 there o times three squared, and this is giving us 38.7 jewels. Kinetic energy final will be equal to 1/2 the moment of inertia Final Omega final squared. So this will be 1/2 times. Ah, 5.40 times 4.78 squared. And this is giving us 61.7 jewels. So the change in kinetic energy will simply be the final kinetic energy. Minus the initial kinetic energy. This will be 61.7, minus 38.7, and this is giving us 23.0 jewels. So this will be the changing kinetic energy that is the end of the solution. Thank you for watching

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