Question

A disc of radius $R=10cm$ oscillates 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)

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
$\therefore I=\frac{m{R}^2}{2}+m{r}^2=\frac{m{R}^2}{2}+m{\left(\frac{R}{4}\right)}^2=\frac{m{R}^2}{2}+\frac{m{R}^2}{16}$
$=\frac{9m{R}^2}{16}\left(\because r=\frac{R}{4}\right)$
Here, $R=10cm=0.1m,h=\frac{R}{4}$
$\therefore T=2\pi \sqrt{\frac{\frac{9m{R}^2}{16}}{\frac{mgR}{4}}}=2\pi \sqrt{\frac{9R}{4g}}$
$=2\pi \sqrt{\frac{9\times 0.1}{4\times 10}}$
$2\pi \times \frac{3}{2}\times \frac{1}{10}$
$=0.94s$