Question

A cylinder is given angular velocity ${\omega }_0$ and kept on a horizontal rough surface, the initial velocity is zero. Find out distance travelled by the cylinder before it performs pure rolling and work done by friction force.

• A. $\frac{{\omega }_0^2}{12}\frac{R^2}{\mu g},-\frac{m{\omega }_0^2{R}^2}{18}$
• B. $\frac{{\omega }_0^2}{18}\frac{R^2}{\mu g},\frac{m{\omega }_0^2{R}^2}{18}$
• C. $\frac{{\omega }_0^2}{18}\frac{R^2}{\mu g},-\frac{m{\omega }_0^2{R}^2}{6}$
• D. $\frac{{\omega }_0^2}{6}\frac{R^2}{\mu g},\frac{m{\omega }_0^2{R}^2}{6}$

Answer 1

The correct option is C $\frac{{\omega }_0^2}{18}\frac{R^2}{\mu g},-\frac{m{\omega }_0^2{R}^2}{6}$

Let the given situation be as shown below

Torque about lowest point of the cylinder
$f_{k}R=I\alpha$
$\Rightarrow \mu mgR=\frac{m{R}^2\alpha }{2}$

$\Rightarrow \alpha =\frac{2\mu g}{R}...\left(1\right)$
Initial velocity, $u=0$
So, we have
$v^2=u^2+2as$
or, $v^2=2as...\left(2\right)$
Also we have
$f_k=ma$
$\Rightarrow a=\mu g...\left(3\right)$

Again we know that
$\omega ={\omega }_0-\alpha t$
or, $\omega ={\omega }_0-\frac{2\mu g}{R}t$ $[$from equation $\left(1\right)]$
Also, $v=u+at$
or, $v=\mu gt$ $[$from equation $\left(3\right)]$
Combining above obtained equations we get
$\omega ={\omega }_0-\frac{2v}{R}$
or, $\omega ={\omega }_0-2\omega$
or, $\omega =\frac{{\omega }_0}{3}$

Now from equation $\left(2\right)$
${\left(\frac{{\omega }_{0}R}{3}\right)}^2=\left(2as\right)=2\mu gs$
or, $s=\left(\frac{{\omega }_0^2}{18}\frac{R^2}{\mu g}\right)$

work done by the friction force
$w=(-f_{k}Rd\theta +f_k\Delta s)$
$-\mu mgR\Delta \theta +\frac{\mu mg\times {\omega }_0^2{R}^2}{18\mu g}$
As we know that,
$\Delta \theta ={\omega }_0\times t-\frac{1}{2}\alpha t^2={\omega }_0\times \left(\frac{{\omega }_{0}R}{3\mu g}\right)-\frac{1}{2}\times \frac{2\mu g}{R}{\left(\frac{{\omega }_{0}R}{3\mu g}\right)}^2$
$=\frac{{\omega }_0^{2}R}{3\mu g}-\frac{{\omega }_0^{2}R}{9\mu g}=\frac{2{\omega }_0^{2}R}{9\mu g}$

Putting the value in the equation obtained for work done, we get
$w=-\mu mg\times R\times \frac{2{\omega }_0^{2}R}{9\mu g}+\mu mg\times \frac{{\omega }_0^2{R}^2}{18\mu g}$
or, $w=-\frac{2m{\omega }_0^2{R}^2}{9}+\frac{m{\omega }_0^2{R}^2}{18}$
or, $w=\frac{-3m{\omega }_0^2{R}^2}{18}=-\frac{m{\omega }_0^2{R}^2}{6}$


Alternative solution:

Using work energy theorem $w_g+w_a+w_{f_k}=\Delta K$
$w_{f_k}=[\frac{1}{2}m{\left(\frac{{\omega }_{0}R}{3}\right)}^2+\frac{1}{2}\frac{m{R}^2}{2}\times {\left(\frac{{\omega }_0}{3}\right)}^2]-[\frac{1}{2}\frac{m{R}^2}{2}\times {\omega }_0^2]=(-\frac{m{\omega }_0^2{R}^2}{6})$