Question

A disc of radius $R=10$cm oscillated as a physical pendulum about an axis perpendicular to the plane of the disc at a distance r from its centre. If $r=\frac{R}{4}$, the approximate period of oscillation is
(Take $g=10m/s^{-2})$

• A. $0.84s$
• B. $0.94s$
• C. $1.26s$
• D. $1.42s$

Answer 1

Answer: (B)

$0.94s$

Time period of a physical pendulum is
$T=2\pi \sqrt{\frac{I}{mgh}}$
where I is the moment of inertia of the pendulum about an axis through the pivot, m is the mass of the pendulum and h is the distance from the pivot to the centre of mass.
In this case, a solid disc of R oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance r from its centre.

$\therefore I=\frac{m{R}^2}{2}+m{r}^2=\frac{m{R}^2}{2}+m\left[\frac{m{R}^2}{4}\right]=\frac{m{R}^2}{2}+\frac{m{R}^2}{16}=\frac{9m{R}^2}{16}$

Here, R = $10$cm = $0.1m,h=\frac{R}{4}$

$\therefore T=2\pi \sqrt{\frac{\frac{9m{R}^2}{16}}{\frac{mgR}{4}}}=2\pi \sqrt{\frac{9R}{4g}}$

$=2\pi \sqrt{\frac{9\times 0.1}{4\times 10}}=2\pi \times \frac{3}{2}\times \frac{1}{10}=0.94s$