Question

Find the speed of uniform solid sphere after rolling down (without sliding) an inclined plane of vertical height $h=0.14\text{m}$ from rest is (Take $g=9.8{\text{m/s}}^2$)

• A. $1.4\text{m/s}$
• B. $1.2\text{m/s}$
• C. $1\text{m/s}$
• D. $1.3\text{m/s}$

Answer 1

The correct option is A $1.4\text{m/s}$

For pure rolling, $\omega =\frac{v}{r}...\left(i\right)$
where, $v=v_{CM}$
Applying mechanical energy conservation:
$\text{Loss in}P{E}_{\text{grav}}=\text{Gain in}K{E}_{\text{Trans}}+\text{Gain in}K{E}_{\text{Rot}}$
$mgh=(\frac{1}{2}m{v}^2-0)+(\frac{1}{2}I{\omega }^2-0)...\left(ii\right)$
$\because \text{Initially}K{E}_{\text{Trans}}=0,K{E}_{\text{Rot}}=0$
For solid sphere,
$I_{\text{CM}}\Rightarrow I=\frac{2}{5}m{r}^2$
From Eq $\left(i\right)\&\left(ii\right)$,
$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2\times {\left(\frac{v}{r}\right)}^2$
$mgh=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2$
$mgh=\frac{7}{10}m{v}^2$
$\Rightarrow v=\sqrt{\frac{10gh}{7}}$
$\therefore v=\sqrt{\frac{10\times 9.8\times 0.14}{7}}=1.4\text{m/s}$