On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbel

Question

On an old-fashioned rotating piano stool, a woman sits holding a pair of dumbbells at a distance of 0.590 from the axis of rotation of the stool. She is given an angular velocity of 3.05 , after which she pulls the dumbbells in until they are only 0.210 distant from the axis. The woman's moment of inertia about the axis of rotation is 5.30 and may be considered constant. Each dumbbell has a mass of 5.40 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 the dumbbells are pulled in. D. Compute the kinetic energy of the system after the dumbbells are pulled in.

Explanation / Answer

What is the initial angular momentum of the system? --->add the woman's I to I of dumblells (m*r^2 = 10.0 * (0.600)^2) then momentum = I * omega (3.00 rad/s) What is the angular velocity of the system after the dumbbells are pulled in toward the axis? --->angular momentum is conserved so momentum before (see above) = momentum after = I * omega I = moment of woman + (10.0 *(0.200)^2) then solve for omega (it will be greater than 3) Compute the kinetic energy of the system before the dumbbells are pulled in. KE = 1/2*I*omega^2 you have I and omega from parts 1 and 2 above solve for KE Compute the kinetic energy of the system after the dumbbells are pulled in.f --->you can repeat the energy equation for the new condition but energy is also conserved so this KE is the same as above

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