Question

Find the time period of a uniform disc of mass $m$ and radius $r$ suspended through a point at a distance $r/2$ from the centre, oscillating in a plane parallel to its plane.

• A. $2\pi \sqrt{\frac{2r}{3g}}$
• B. $4\pi \sqrt{\frac{r}{2g}}$
• C. $2\pi \sqrt{\frac{3r}{2g}}$
• D. None

Answer 1

The correct option is C $2\pi \sqrt{\frac{3r}{2g}}$

Time period of a physical pendulum is given by
$T=2\pi \sqrt{\frac{I}{mgd}}$
where $I$ is the moment of inertia and $d$ is the perpendicular distance of its COM from the axis of rotation.

Since it is given that the disc oscillates in a plane parallel to its plane, the axis of rotation must be perpendicular to its plane.

Hence, MOI of disc about the axis of rotation
$I=I_c+m(r/2{)}^2$ [using parallel axis theorem]
$I=m\frac{r^2}{2}+m\frac{r^2}{4}$

and distance of COM of disc from the axis of rotation, $d=\frac{r}{2}$

$\Rightarrow T=2\pi \sqrt{\frac{\frac{1}{2}m{r}^2+m(r/2{)}^2}{mg\left(r/2\right)}}$
$=2\pi \sqrt{\frac{3r^2/4}{gr/2}}=2\pi \sqrt{\frac{3r}{2g}}$