Question

A cylinder is given angular velocity ${\omega }_0$ and kept on a horizontal rough surface. The initial velocity is zero. The distance travelled by it before it starts performing pure rolling is: [Assume radius of cylinder $=R$]

• A. $\frac{{\omega }_0^2{R}^2}{18\mu g}$
• B. $\frac{{\omega }_0^2{R}^2}{\mu g}$
• C. $\frac{{\omega }_0^2{R}^2}{2\mu g}$
• D. $\frac{{\omega }_0^2{R}^2}{6\mu g}$

Answer 1

The correct option is A $\frac{{\omega }_0^2{R}^2}{18\mu g}$

From torque equation of cylinder,
$fR=\mu MgR=I\alpha =\frac{M{R}^2\alpha }{2}$
$\Rightarrow \alpha =\frac{2\mu g}{R}$ ... (1)


From force balance equation,
$f_k=\mu Mg=Ma\Rightarrow a=\mu g$ ... (2)

Initial velocity, $u=0$
Using $v^2=u^2+2as$
$v^2=2as$ ... (3)
Also,
$\omega ={\omega }_0-\alpha t$ ... (4)
From equation (1), $\omega ={\omega }_0-\frac{2\mu g}{R}t$
& $v=u+at$
From equation (2), $v=\mu gt$
$\Rightarrow \omega ={\omega }_0-\frac{2v}{R}$
$\Rightarrow \omega ={\omega }_0-2\omega$
[for pure rolling, $v=R\omega$]
$\Rightarrow \omega =\frac{{\omega }_0}{3}$
From equation (3),
${\left(\frac{{\omega }_{0}R}{3}\right)}^2=\left(2as\right)=2\mu gs$
$\therefore s=\left(\frac{{\omega }_0^2{R}^2}{18\mu g}\right)$ is the distance covered before pure rolling begins.