A man stands on a platform that is rotating (without friction) with an angular s

Question

A man stands on a platform that is rotating (without friction) with an angular speed of 9.42 rad/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 7.50 kg middot m^2. By moving the bricks the man decreases the rotational inertia of the system to 2.00 kg middot m^2. What is the resulting angular speed of the platform? What is the ratio of the new kinetic energy of the system to the original kinetic energy?

Explanation / Answer

apply angular momentum conservation.

L = I*w

Li = Lf

7.50* 9.42 = 2*w

w = 35.325 rad/s

original KE = 0.5 *I1*w1 ^2

new Ke = 0.5 *I2*w2^2

ratio = I2*w2^2 / I1*w1 ^2

= 2*35.325 ^2 / ( 7.5 * 9.42^2) =3.75

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