Question

A solid cylinder P rolls without slipping from rest down an inclined plane attaining a speed $v_p$ at the bottom. Another smooth solid cylinder Q of same mass and dimensions slides without friction from rest down the inclined plane attaining a speed $v_q$ at the bottom. The ratio of the speeds $\left(\frac{v_p}{v_q}\right)$ is$?$

• A. $\left(\sqrt{\frac{3}{4}}\right)$
• B. $\left(\sqrt{\frac{3}{2}}\right)$
• C. $\left(\sqrt{\frac{2}{3}}\right)$
• D. $\left(\sqrt{\frac{4}{3}}\right)$

Answer 1

The correct option is C $\left(\sqrt{\frac{2}{3}}\right)$

Case 1 :
Solid cylinder rolls on the inclined plane.
Initial speed of the cylinder is zero i.e. $K.E_i=0$
Work-energy theorem $W=\Delta K.E=K.E_f-K.E_i$
$\therefore$ $mgh=\frac{1}{2}I{w}^2+\frac{1}{2}m{v}_p^2-0$
where $I=\frac{1}{2}m{r}^2$ and $v_p=rw$
We get $mgh=\frac{4}{3}m{v}_p^2$
$⟹$ $v_p=\sqrt{\frac{4}{3}gh}$
Case 2 :
Solid cylinder slides on the inclined plane.
Initial speed of the cylinder is zero i.e. $K.E_i=0$
Work-energy theorem $W=\Delta K.E=K.E_f-K.E_i$
$\therefore$ $mgh=\frac{1}{2}m{v}_q^2-0$
We get $mgh=\frac{1}{2}m{v}_q^2$
$⟹$ $v_q=\sqrt{2gh}$
Thus we get $\frac{v_p}{v_q}=\sqrt{\frac{2}{3}}$