Question

A disc of radius R and mass M is pivoted at the rim and is set 't' small oscillations. It' simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be:

• A. $3R/2$
• B. $2R/2$
• C. $3R$
• D. $3R/4$

Answer 1

The correct option is A $3R/2$

Moment of lnertia of the disc $ab$ at the axis of oscillations

$R=I_o+M{R}^2$

$=\frac{1}{2}M{R}^2+M{R}^2$

$R=\frac{3}{2}M{R}^2$

Time period of the oscillation of the disc in

$T=2\pi \sqrt{\frac{I}{mgrcm}}$

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

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

Let a simple pendulam of the effective length 'l' has the same time period

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

$L=\frac{3}{2}R$