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 A disc of mass ' m ' and radius 'r ' is made to rotate with an angular velocity of \omega° and then it is placed between two smooth walls on a rough horizontal surface. If the distance between the walls is slightly greater than the diameter of the disc. If coefficient of friction between the ground and the surface is \mu, then the time after which the disc will stop rotating is given by -

Option: 1

\frac{\omega_{0}}{\mu g}


Option: 2

\frac{2 \omega_{0}}{\mu g}


Option: 3

\frac{\omega_{0}}{2 \mu g}


Option: 4

\frac{2}{3} \frac{\omega_{0}}{\mu g}


Answers (1)

best_answer

$$ \begin{aligned} & f=\mu N=\mu m g \\ & \tau=\gamma f=\mu m g r \\ & \alpha=\frac{\mu m g r}{I} \\ & \omega=\omega_{0}-\alpha t \\ & t=\frac{\omega_{0}}{\alpha}=\frac{\omega_{0} I}{\mu m g r} \\ & t=\frac{\omega_{0}\left(\frac{1}{2} m \gamma^{\gamma}\right)}{\mu m g \gamma}=\frac{\omega_{0}}{2 \mu g} \end{aligned}

 

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Divya Prakash Singh

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