Question

A solid cylinder is released from rest from the top of an inclined plane of inclination $\theta$ and length ${}^{\prime }{l}^{\prime }$. If the cylinder rolls without slipping, then find it's speed when it reaches the bottom of inclined plane.

• A. $\sqrt{\frac{4gl\sin \theta }{3}}$
• B. $\sqrt{\frac{3gl\sin \theta }{2}}$
• C. $\sqrt{\frac{4gl}{3\sin \theta }}$
• D. $\sqrt{\frac{4g\sin \theta }{3l}}$

Answer 1

The correct option is A $\sqrt{\frac{4gl\sin \theta }{3}}$

In case of pure rolling, total mechanical energy remains conserved i.e $W_f=0,$
$v_{\text{CM}}=r\omega$ for pure rolling
Since solid cyliinder is starting from rest, ${\omega }_i=0,v_i=0$
${KE}_{Trans}=\frac{1}{2}m{v}_{\text{CM}}^2$
${KE}_{Rot}=\frac{1}{2}{I}_{CM}{\omega }^2$
$\Rightarrow$ Total mechanical energy at initial position:
$E_1=PE+{KE}_{Trans}+{KE}_{Rot}$
$\therefore E_1=mgh+0+0=mgh$
Taking reference of $PE=0$ at final position i.e at bottom of inclined plane
$\Rightarrow$Total mechanical energy at final position:
$E_2=PE+{KE}_{Trans}+{KE}_{Rot}$
$E_2=0+\frac{1}{2}m{v}_{\text{CM}}^2+\frac{1}{2}{I}_{CM}{\omega }^2$
$\because \frac{v_{\text{CM}}}{r}=\omega$ and $I_{CM}=\frac{m{r}^2}{2}$
$\Rightarrow E_2=\frac{1}{2}m{v}_{\text{CM}}^2+\frac{1}{2}\left(\frac{m{r}^2}{2}\right)\left(\frac{v_{\text{CM}}^2}{r^2}\right)$
$\therefore E_2=\frac{3}{4}m{v}_{\text{CM}}^2$
From conservation of total mechanical energy,
$\Rightarrow E_1=E_2$
$mgh=\frac{3}{4}m{v}_{\text{CM}}^2$
$v_{\text{CM}}=\sqrt{\frac{4}{3}gh}$


$\therefore v_{\text{CM}}=\sqrt{\frac{4gl\sin \theta }{3}}$