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 0.570{ m m} from the axis of rotation of the stool. She is given an angular velocity of 2.95{ m rad}/{ m s} , after which she pulls the dumbbells in until they are only 0.230{ m m} distant from the axis. The woman's moment of inertia about the axis of rotation is 4.80{ m{ { m kg}}} cdot { m{{ m m}}}^{ m{2}} and may be considered constant. Each dumbbell has a mass of 5.45{ m kg} and may be considered a point mass. Neglect friction.

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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