A man stands at the center of a platform that rotates withoutfriction with an an

Question

A man stands at the center of a platform that rotates withoutfriction with an angular speed of 1.8 rev/s. His arms areoutstretched, and he holds a heavy weight in each hand. The momentof inertia of the man, the extended weights, and the platform is6.4 kg·m2. When the manpulls the weights inward toward his body, the moment of inertiadecreases to 2.2 kg·m2. (a) What is the resultingangular speed of the platform? (b) What is the change inkinetic energy of the system? (c) Where did this increase inenergy come from? (a) What is the resultingangular speed of the platform? (b) What is the change inkinetic energy of the system? (c) Where did this increase inenergy come from?

Explanation / Answer

(a) What is the resulting angularspeed of the platform?


L before=L after
Ibb = Iaa

a =b [Ib/Ia]=1.8x2 [ 6.4 / 2.2] = 32.9 rad/s

(b) What is the change in kineticenergy of the system?
Kebefore=0.5Ibb2
Ke after= 0.5Iaa2= 0.5Ia[b [Ib/Ia] ]2 =0.5 b2 Ib2/Ia
Ke= Ke before - Keafter = 0.5Ibb2 - 0.5 b2Ib 2/Ia
Ke=0.5b2Ib[ 1 -   Ib/Ia]

Ke= 0.5(1.8x2)2 (6.4 ) [ 1 - 6.4 /2.2 ] = -780J
(c)Where did this increase in energy come from?

Conservation of angularmomentum.

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