A small object placed on a rotating horizontal turn table just slips when it is placed at a distance 4cm 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=4cm,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 4cmR2=(2ωω)2 ∴R2=1cm