A 55.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 480 kg · m^2 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? counterclockwise or clockwiseWith 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

Question

Angelina Walls

Physics

5 December, 01:34

A 55.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 480 kg · m^2 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? counterclockwise or clockwiseWith 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

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Answers (1)

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Aileen Love · Answer 1

Check the explanation

Explanation:

net angular momentum is conserved as there is no net external torque.

therefore. 0 = angular momentum of woman + angular momentum of wheel.

0 = 55x2x1.5 + 480xω

ω = - 0.3056 rad/sec,-ve sign shows that the direction is opposite to woman, ie counterclockwise.

work done by woman = total kinetic energy of the wheel and woman

=1/2 x 55 x 1.52 + 1/2 Iω2 = 1/2x55x1.52 + 1/2x540x0.30562 = 41.8+73.35 = 115.14J