Question

A solid sphere is rolling on a surface as shown in the figure with a translational velocity $v$. If it is to climb the inclined surface, then $v$ should be

• A. $\ge \sqrt{\frac{10}{7}gh}$
• B. $\ge \sqrt{\frac{15}{7}gh}$
• C. $\ge \sqrt{\frac{13}{7}gh}$
• D. $\ge \sqrt{\frac{11}{7}gh}$

Answer 1

The correct option is A $\ge \sqrt{\frac{10}{7}gh}$

Condition for pure rolling, $v=r\omega ,$
$v\to$ translational velocity of center of mass $(C.O.M)$.
$\omega \to$ angular velocity of sphere about $C.O.M$
Let the solid sphere just climbs the inclined surface.
The Condition for it is, ${\omega }_f=0$ and $v_f=0$.
On applying the law of conservation of mechanical energy,
$\Rightarrow K{E}_i+P{E}_i=K{E}_f+P{E}_f$
$\Rightarrow (KE{}_{rot}+KE{}_{trans}{)}_i+P{E}_i=(K{E}_{rot}+K{E}_{trans}{)}_f+P{E}_f$
$\Rightarrow \frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2+0=0+0+mgh$
[considering reference of $PE=0$ at the bottom surface]


$\Rightarrow \frac{1}{2}\times \left(\frac{2}{5}m{r}^2\right)\times {\left(\frac{v}{r}\right)}^2+\frac{1}{2}m{v}^2=mgh$
$\Rightarrow \frac{7m{v}^2}{10}=mgh$
$\Rightarrow v=\sqrt{\frac{10}{7}gh}$
This is the minimum translational velocity required to just climb the inclined surface.
Hence, $v\ge \sqrt{\frac{10}{7}gh}$ for solid sphere to climb without fail.