Question

A uniform disc of mass $M$ and radius $R$ is suspended in vertical plane from a point on its periphery. Its time period of oscillation is

• A. $2\pi \sqrt{\frac{3R}{g}}$
• B. $2\pi \sqrt{\frac{R}{3g}}$
• C. $2\pi \sqrt{\frac{2R}{3g}}$
• D. $2\pi \sqrt{\frac{3R}{2g}}$

Answer 1

Answer: (D)

$2\pi \sqrt{\frac{3R}{2g}}$

Let's consider,

$M=$ mass of a uniform disc

$R=$ radius of the suspended uniform disc

The time period of oscillation is given by

$T=2\pi \sqrt{\frac{I}{MgR}}$. . . . . . . .(1)

where $I=$ moment of inertia froma point in its periphery

Using paarallel axis theorem,

$I=M{R}^2+\frac{1}{2}M{R}^2=\frac{3}{2}M{R}^2$. . . . .(2)

Substitute equation (2) in equation (1), we get

$T=2\pi \sqrt{\frac{3M{R}^2/2}{MgR}}$

$T=2\pi \sqrt{\frac{3R}{2g}}$

The correct option is D.