Question #88380
A Ring of radius R is first rotated with an angular velocity ∆ and then carefully placed on a horizontal rough surface. The coefficient of friction between the surface and the ring is ¶. Time after which its angular speed is reduced to half is
1
Expert's answer
2019-04-24T10:10:36-0400

Let μ\mu be the coefficient of friction, ω\omega be the angular speed and α\alpha the angular acceleration. Consider a torque of force of friction:

τf=μmgR,\tau_\text{f}=\mu mg\cdot R,

and according to Newton's second law for rotational motion, the net torque is


τnet=mR2α=mR2ωω/2t,\tau_\text{net}=mR^2\alpha=mR^2\cdot \frac{\omega-\omega/2}{t},

equate them:


μmgR=mR2ω2t,\mu mg\cdot R=mR^2\cdot \frac{\omega}{2t},

t=ωR2μg.t=\frac{\omega R}{2\mu g}.


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