Question

A uniform solid disc of radius $R$ and mass $M$ is free to rotate on a frictionless pivot through a point on its rim. If the disk is released from rest in the position shown in figure. The speed of the lowest point on the disc in the dashed circle is

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

Answer 1

The correct option is A $4\sqrt{\frac{gR}{3}}$

To identify the change in gravitational potential energy, think of the height through which the center of mass falls. From the parallel axis theorem, the moment of inertia of the disc about the pivot point on the circumference is

$I=I_{CM}+M{d}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

The pivot point is fixed, so the kinetic energy is entirely rotational around the pivot.

Applying conservation of energy,

$K_f+U_f=K_i+U_i$

$\frac{1}{2}{I}_{\text{axis}}{\omega }^2=mgR$

$\frac{1}{2}\times \frac{3}{2}m{R}^2\times {\omega }^2=mgR$

$\therefore \omega =\sqrt{\frac{4g}{3R}}=2\sqrt{\frac{g}{3R}}$

Velocity of the lowest point $=\omega \times 2R=4\sqrt{\frac{gR}{3}}$