Question

A cylinder initially kept at rest on a horizontal rough surface is given an angular velocity $10\text{rad/s}$. What is the distance travelled by the centre of mass of the cylinder till it begins pure rolling ?
Given: Coefficient of kinetic and static frcition are equal $(\mu =0.5)$ and radius of cylinder $R=3\text{m}$. Also, take $g=10{\text{m/s}}^2$.

• A. $5\text{m}$
• B. $10\text{m}$
• C. $1\text{m}$
• D. $20\text{m}$

Answer 1

The correct option is B $10\text{m}$


Initially, linear velocity $v_i=0$
Friction force will oppose the rotational motion, hence kinetic friction will act towards right as shown in figure

Here $a=a_{\text{CM}}$
$\Rightarrow f_k=ma$
$\Rightarrow a=\frac{f_k}{m}...\left(i\right)$
From equation of torque,
${\tau }_{com}=I_{com}\alpha$
$\Rightarrow f_k.R=\frac{m{R}^2}{2}.\alpha ...\left(ii\right)$
From eq. $\left(i\right)\&\left(ii\right)$
$ma.R=\frac{m{R}^2}{2}.\alpha$
$\Rightarrow a=\frac{R}{2}\alpha ...\left(iii\right)$
Conserving angular momentum about point of contact, since ${\tau }_{f_k}=0$
$L_i=L_f$
$\Rightarrow 0+(-I\omega )=-mvR+(-I{\omega }^{\prime })$
$\Rightarrow \frac{m{R}^2}{2}.\omega =mvR+\frac{m{R}^2}{2}\left(\frac{v}{R}\right)$
where $v$ is the final velocity of COM.
[$\because$ for achieving pure rolling, $v={\omega }^{\prime }R$]
$\Rightarrow \frac{\omega R^2}{2}=\frac{3vR}{2}$
$\therefore v=\frac{\omega R}{3}...\left(iii\right)$

Since friction will be kinetic till pure rolling starts,
$f_k=\mu N=\mu mg$
$\therefore a=\frac{f_k}{m}=\mu g...\left(iv\right)$
From equation $v=u+at$
$u=0\Rightarrow v=at$
Substitute from eq. $\left(iii\right)\&\left(iv\right)$,
$\frac{\omega R}{3}=\mu gt$
$\Rightarrow t=\frac{\omega R}{3\mu g}$
Now, distance travelled till pure rolling is achieved:
$S=\frac{1}{2}a{t}^2$
$S=\frac{1}{2}\times \left(\mu g\right)\times {\left(\frac{\omega R}{3\mu g}\right)}^2=\frac{{\omega }^2{R}^2}{18\mu g}$
$\therefore S=\frac{(10{)}^2\times (3{)}^2}{18\times 0.5\times 10}=10\text{m}$