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Question

A small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation is (in cm)

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Solution

Given:
R1=4 cm,ω1=ω
ω2=2ω,R2=?

The object will slip if centripetal force equals to force of friction
mrω2=μmg
rω2=μg
rω2=constant

As ω changes from ω1 to ω2
(R1R2)=(ω2ω1)2
4 cmR2=(2ωω)2
R2=1 cm

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