A 70.0-kg woman stands at the rim of a horizontal turntable having a moment of i

Question

A 70.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 560 kg · m2 and a radius of 2.00 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 1.50 m/s relative to the Earth.
(a) In what direction does the turntable rotate?
clockwise counterclockwise

With what angular speed does the turntable rotate?
rad/s

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

Explanation / Answer

(a) When the women walk with angular velocity
= -(v/r)k = -((1.5 m/s)/(2 m))k = -(0.75/s)k
her angular momentum is
L = I = (-70 kg * 4 m2 * 0.75/s)k = -(210 kgm2/s)k.
The angular momentum of the turntable will be
L = (210kgm2/s)k
and its angular velocity
= (210kgm2/s)k/(560 kgm2) = (0.375/s)k.
The turntable turns counterclockwise.

b) The kinetic energy of the women is
(1/2)I2 = (1/2) 280 kgm2(0.75 s)2 = 78.75 J.
The kinetic energy of the turntable is
(1/2)I2 = (1/2) 560 kgm2(0.375s)2 = 39.37 J.
The work done by the women is W = (78.75 + 39.37)J = 118.12J

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