You stand on a frictional platform that is rotating at 1.3 rev/s. Your arms are

Question

You stand on a frictional platform that is rotating at 1.3 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.6 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 4.4 kg · m2.
(a) What is the resulting angular speed of the platform?
rev/s

(b) What is the change in kinetic energy of the system?
J
(c) Where did this increase in energy come from? (Select all that apply.)
gravity
your internal energy
kinetic energy of the platform
air resistance
mass of the weights

Explanation / Answer

a) First, we convert the initial speed to radians per second,

1.3 rev/s = 2*1.3 rad/s = 8.2 rad/s

Next, we apply conservation of angular momentum

I11 = I22

2 = I11/I2 = (8.6 kg*m2)(8.2 rad/s)/(4.4 kg*m2) = 16.0 rad/s

b) Then, we compute the change in kinetic energy, which is:

I222/2 - I121/2 = (4.4 kg*m2)(16.0 rad/s)2/2 - (8.6 kg*m2)(8.2 rad/s)2/2 = 274 J

This change in energy comes from the effort that you put in in order to pull the weights closer to yourself, so the answer is "your internal energy."

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