Question

A cylinder having radius $0.4m,$ initially rotating $(att=0)$ with ${\omega }_0=54rad/s$ is placed on a rough inclined plane with $\theta ={37}^{\circ }$ having friction coefficient $\mu =0.5$. The time taken by the eylinder to start pure rolling is $(g=10m/s^2)$

• A. $5.4s$
• B. $2.4s$
• C. $1.4s$
• D. None of these

Answer 1

The correct option is D None of these

$a=\left(\mu g\cos \theta \right)+\left(g\sin \theta \right)$
$=0.5\times 10\times 0.8+10\times 0.6=10m/s^2$
$\alpha =\frac{\left(\mu mg\cos \theta \right)R}{\frac{1}{2}m{R}^2}=\frac{2\mu g\cos \theta }{R}$
$=\frac{2\times 0.5\times 10\times 0.8}{0.4}=20rad/s^2$
Pure rolling will start when,$v=R\omega$ or at $=R({\omega }_0-\alpha t)$
$\therefore 10t=0.4(54-20t)$
Solving this equation, we get
$t=1.2s$