A uniform solid cylinder of mass m and radius | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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: $\mu gt = R(\omega_0 - \frac{2\mu gt}{R}) = \omega_0 R - 2\mu gt$.$3\mu gt = \omega_0 R \Rightarrow t = \frac{\omega_0 R}{3\mu g}$.

Similar Questions

$A$ solid sphere is moving on a horizontal plane. The ratio of its translational kinetic energy to its rotational kinetic energy is:

$A$ force $F$ is applied at the centre of a disc of mass $M$. The minimum value of the coefficient of friction of the surface for pure rolling is:

$A$ solid sphere spinning about a horizontal axis with an angular velocity $\omega$ is placed on a horizontal surface. Subsequently,it rolls without slipping with an angular velocity of

Vedclass Test Series

Exam Paper Generator

Online Exam Module