Question

A solid sphere is rolling without slipping on a frictionless surface with a translational velocity $v\text{m/s}$ as shown in the figure. If it has to climb the inclined surface then $v$ in $\text{m/s}$ should be:

• A. $\ge \sqrt{\frac{10gh}{7}}$
  
• B. $\le \sqrt{2gh}$
  
• C. $\ge \sqrt{\frac{5gh}{7}}$
  
• D. $\le \frac{17}{7}\sqrt{gh}$
  
  

Answer 1

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


Considering that sphere just climbs the inclined surface, it will stop at it's final position hence ${\omega }_f=0,v_f=0$
$\Rightarrow$In absence of any dissipative external forces, using conservation of mechanical energy for sphere:
$\text{Loss in}{KE}_{Trans}+\text{Loss in}{KE}_{Rot}=\text{Gain in PE}$
$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh...\left(i\right)$
$\because MI$ for solid sphere about axis passing through it's centre is $I=\frac{2}{5}m{r}^2$
Using condition for pure rolling $v=r\omega$ in Eq.$\left(i\right)$ gives,
$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\frac{2}{5}m{r}^2\times \frac{v^2}{r^2}=mgh$

$\frac{v^2}{2}+\frac{2v^2}{10}=gh$

$\frac{5v^2+2v^2}{10}=gh\Rightarrow v^2=\frac{10}{7}gh$
$\Rightarrow v=\sqrt{\frac{10}{7}gh}$
This is the required minimum speed to just climb the inclined surface.
Hence,
$\therefore v\ge \sqrt{\frac{10}{7}gh}\text{m/s}$