Question

A disk of mass m and radius R rolls down an inclined plane of height h without slipping. What is the velocity of the disk at the bottom of the incline? The moment of inertia for a disk is $1/2m{R}^2$.

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

Answer 1

The correct option is B $\sqrt{\frac{4}{3}gh}$

For no slipping condition, $v=R\omega$

From conservation of energy,

Loss in gravitational potential energy=Gain in translational kinetic energy+Gain in rotational kinetic energy

$⟹mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

$⟹v=\sqrt{\frac{4}{3}gh}$