A man of mass 100. kg stands at the rim of a turntable of radius 2.00 meters and

Question

A man of mass 100. kg stands at the rim of a turntable of radius 2.00 meters and moment of inertia 4000. kg-m2, mounted on a frictionless vertical shaft at its center. The whole system is initially at rest. The man now walks along the outer edge of the turntable with a velocity of 1.00 meter/second relative to the earth in a clockwise direction. (a) With what angular velocity and in what direction does the turntable rotate? (b) Through what angle will it have rotated when the man first returns to his initial positionon the turntable? (c) Through what angle will it have rotated when the man returns to his initial position relative to the earth?

Explanation / Answer

(A) Applying angular momentum conservation,

0 = (100 x 2 x 1) + (4000 w)

w = 0.05 rad/s ......Counter clockwise direction

(B) v of man wrt turnable = vman - v turnable

v = 1 - (-0.05 x 2) = 1.1 m/s

t = 2 x pi x 2 / 1.1 = 11.4 sec

theta = 0.05 x 11.4 = 0.57 rad ......Ans

(c) t = 2 pi 2 / 1 = 12.57 sec

angle = 0.628 rad ........Ans

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