Question

A mass slides down an inclined plane and reaches the bottom with a velocity $v$. If the same mass is in the form of a disc and rolls down the same inclined plane, what will be its velocity at the bottom?

• A. $v$
• B. $\sqrt{\frac{2}{3}}v$
• C. $\sqrt{2}v$
• D. $\frac{v}{\sqrt{2}}$

Answer 1

The correct option is B $\sqrt{\frac{2}{3}}v$

$\therefore PE=\frac{1}{2}m{v}^2$

$KE=\frac{1}{2}m{u}^2+\frac{1}{2}I{\omega }^2$

But, $KE=PE$

Or, $\frac{1}{2}m{u}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2$

Now for disc $I=\frac{m{R}^2}{2}$

$\therefore \frac{1}{2}m{u}^2+\frac{1}{2}\frac{m{R}^2}{2}\frac{u^2}{R^2}=\frac{1}{2}m{v}^2$

$u^2+\frac{u^2}{2}=v^2$

$3u^2=2v^2$

$\therefore u=\sqrt{\frac{2}{3}}v$