Question

A solid cylinder of mass $m$ and radius $r$ starts rolling down on inclined plane of inclination $\theta$. Friction is sufficient enough to prevent slipping. Find the speed of centre of mass of solid cylinder, when it's centre of mass has fallen through a height of $h=9\text{m}$. (Take $g=10{\text{m/s}}^2$)

• A. $\sqrt{120}\text{m/s}$
• B. $\sqrt{125}\text{m/s}$
• C. $\sqrt{180}\text{m/s}$
• D. $\sqrt{90}\text{m/s}$

Answer 1

The correct option is A $\sqrt{120}\text{m/s}$

In case of pure rolling, total mechanical energy remains conserved as work done by dissipative forces are zero i.e $W_f=0,$ mechanical energy is conserved.

Since sphere 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 position $\left(1\right)$:
$E_1=PE+{KE}_{Trans}+{KE}_{Rot}$
$\therefore E_1=mgh+0+0=mgh$
Taking reference of $PE=0$ at position $\left(2\right)$
Total mechanical Energy at position $\left(2\right)$:
$E_2=PE+{KE}_{Trans}+{KE}_{Rot}$
$E_2=0+\frac{1}{2}m{v}_{\text{CM}}^2+\frac{1}{2}{I}_{CM}{\omega }^2$
$\because \frac{v_{\text{CM}}}{r}=\omega$ 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 energy,
$\Rightarrow E_1=E_2$
$mgh=\frac{3}{4}m{v}_{\text{CM}}^2$
$v_{\text{CM}}=\sqrt{\frac{4}{3}gh}$
$v_{\text{CM}}=\sqrt{\frac{4\times 10\times 9}{3}}=\sqrt{120}\text{m/s}$
$\therefore v_{\text{CM}}=\sqrt{120}\text{m/s}$