A 65 kg woman stands at the rim of a horizontal turntable having a moment of ine

Question

A 65 kg woman stands at the rim of a horizontal turntable having a moment of inertia of 520 kg·m2 and a radius of 2.0 m. The turntable is initially at rest and is free to rotate about a frictionless vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.5 m/s relative to the Earth.

(a) In what direction and with what angular speed does the turntable rotate?

counterclockwise clockwise    

2 rad/s

(b) How much work does the woman do to set herself and the turntable into motion?
3 J

Explanation / Answer

Set the initial angular momentum of the system equal to the final angular momentum of the system. Call the angular momentum of the turntable L and that of the woman - L, then:

(L + L)(i) = (L + L)(f)

L is equal to the product of moment of inertia (I) and angular velocity (), and since initially the system is at rest, we write:

0 = I + I
= - I / I

You are given the woman’s speed as 1.5m/s( it should be -1.5m/s as she is moving clockwise), but angular speed is:

= v / r

So replacing with v / r and I with mr², we have:

(a) = -mr²v/r / I
= -mrv / I
= -(65kg)(2.0m)(-1.5m/s) / 520kgm²
= 0.375 rad/s (that is 0.375 rad/s counter-clockwise)

The work she does to get the turntable to this speed is found from the work-energy theorem, where the net work is equal to the change in kinetic energy:

(b), W = KE
= KE(f) - KE(i)
= [0.5I² + 0.5mv²] - 0
= [0.5(520kgm²)(0.375rad/s)² + 0.5(65kg)(-1.5m/s)²]
= 109.69 J

Hope this helps.

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