A 55.0-kg woman stands at the rim of a horizontal turntable having a moment of i

Question

A 55.0-kg woman stands at the rim of a horizontal turntable having a moment of inertia of 420 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?. With what angular speed does the turntable rotate? (b) How much work does the woman do to set herself and the turntable into motion?

Explanation / Answer

since there is no net external torque on the system of woman and turntable

so angular momentum will be conserved

MOI of woman = MR^2 = 55*2^2 =220kgm2

I1w1 - I2w2 = 0

220*(1.5/2) = 420 * w

w = 0.39285714285 rad/s

it will rotate in counter clockwise direction so that total angular momentum will be zero

b) since all final kinetic energy will be due to work done by woman

work done = 0.5* I1 *w1^2 + 0.5 * I2 *w2^2

=0.5*420*(0.39285714285)^2 +0.5* (220) (1.5/2)^2

= 94.2857142845 J

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