You stand on a frictional platform that is rotating at 1.9 rev/s. Your arms are

Question

You stand on a frictional platform that is rotating at 1.9 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 7.9 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 3.7 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.)

your internal energy

gravity

air

resistancemass of the weight

skinetic energy of the platform

Your work in question(s) 1 will also be submitted or saved.

Explanation / Answer

a)

I1*1 = I2*2

so 2 = I1*1/I2 = 7.9*1.9/3.7 = 4.149 rev/s

b)

K1 = 1/2*i*^2

K1 = 1/2*7.9*(1.9*2)^2 = 562J

K2 = 1/2*3.7*(4.149*2*)^2 = 1260J

So K = 1260 - 562 = 698J

c) your internal energy

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