Question

# A bead of mass m = 0.2 kg lies at a distance r from the axis of rotation on a rigid wire as shown in figure. Coefficient of friction between wire and bead is μ, and wire starts rotating from rest at an angular acceleration of α at t = 0. The whole arrangement is in gravity free region. r = 1m, μ=0.4 & α=0.1 r/s2

A
Relative motion between bead and wire starts at t = 2 sec
B
Power of friction force at t = 2 is 0.0016 watt
C
Power of normal contact at t = 2 is 0.004 watt
D
Power of net force on bead at t = 2 is 0.004 watt

Solution

## The correct options are A Relative motion between bead and wire starts at t = 2 sec C Power of normal contact at t = 2 is 0.004 watt D Power of net force on bead at t = 2 is 0.004 watt N=mat=mrα fr=μN=μmrα The bead starts sliding when friction (fr) is less than centrifugal force. ⇒fr=mω2r ⇒μmrα =mω2r μm/r/α=m/ω2r/ ω2=μα and we know, ω=αt α2t2=μα⇒t=√μα=√0.40.1=2 sec v=rω=rαt⇒ at t = 2, v=1×0.1×2=0.2m/s fl=μN=μmrα=0.4×0.2×1×0.1=0.008 N=mrα=0.2×1×0.1=0.02 Power of friction = 0 Power of normal = N×v=0.004 Watt  Physics

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