A uniform solid cylinder of mass m and radius | vedclass

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Question

(C) When the cylinder is placed on the surface,the friction force $f = \mu mg$ acts in the direction opposite to the velocity of the point of contact.

Vedclass · Accepted Answer

$\frac{\omega_0 R}{3\mu g}$

Vedclass · Answer

(C) When the cylinder is placed on the surface,the friction force $f = \mu mg$ acts in the direction opposite to the velocity of the point of contact.This friction force provides a linear acceleration $a = \frac{f}{m} = \mu g$ and a retarding angular acceleration $\alpha = \frac{	au}{I} = \frac{fR}{\frac{1}{2}mR^2} = \frac{2\mu g}{R}$.At time $t$,the linear velocity is $v = at = \mu gt$.The angular velocity is $\omega = \omega_0 - \alpha t = \omega_0 - \frac{2\mu g t}{R}$.Rolling without slipping starts when $v = R\omega$.Substituting the expressions: $\mu gt = R(\omega_0 - \frac{2\mu gt}{R}) = \omega_0 R - 2\mu gt$.$3\mu gt = \omega_0 R \Rightarrow t = \frac{\omega_0 R}{3\mu g}$.

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