Question

A disc of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping. Find out the velocity of the disc when it reaches the bottom of the incline? The moment of inertia for a disk is $\left(1/2\right)m{R}^2$.

• A. $\sqrt{gh}$
• B. $\sqrt{\frac{4}{3}gh}$
• C. $\sqrt{2gh}$
• D. $2\sqrt{gh}$
• E. $2\sqrt{2gh}$

Answer 1

Answer: (B)

$\sqrt{\frac{4}{3}gh}$

Initial kinetic energy of the system is zero as the system is at rest initially.

Moment of inertia for a disk about its center $I=\frac{1}{2}m{R}^2$

As there is no slipping, thus $v=Rw$ .................(1)

Using work-energy theorem : $W=\Delta K.E$

$\therefore$ $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2-0$

OR $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{m{R}^2}{2}\times \frac{v^2}{R^2}$ (using (1))

OR $mgh=\frac{3}{4}m{v}^2$ $⟹v=\sqrt{\frac{4}{3}gh}$