Question

The time-period of a physical pendulum is $2\pi \sqrt{I/mgd}$ , where $I$ is the moment of inertia of the pendulum about the axis of rotation and $d$ is perpendicular distance between the axis of rotation and the centre of mass of the pendulum
A circular ring hangs from a nail on a wall. The mass of the ring is $3kg$ and its radius is $20cm$ . If the ring is slightly displaced, the time of resulting oscillations will be

• A. $1.0s$
• B. $1.3s$
• C. $1.8s$
• D. $2.1s$

Answer 1

Answer: (B)

Given, $T=2\pi \sqrt{\frac{I}{mg.d}}\dots \left(i\right)$
$m=3kg,d$ or $r=20cm=0.2m$

Moment of inertia of the ring about XX'
$I=I_0+m{r}^2=m{r}^2+m{r}^2=2m{r}^2$
$=2\times 3\times (0.2{)}^2=0.24kg-m^2$
Putting values in Eq (i)
$T=2\pi \sqrt{\frac{0.24}{3\times 10\times 0.2}}=2\pi \sqrt{0.04}$
$=2\pi \times 0.2=1.2566s$
$T=1.3s$