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 $3 \,kg$ and its radius is $20\, cm$ . If the ring is slightly displaced, the time of resulting oscillations will be

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

Answer 1

Answer: (B)

$1.3 \, s$

Answer 2

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

Moment of inertia of the ring about XX'
$I=I_{0}+mr^{2}=mr^{2}+mr^{2}=2mr^{2}$
$=2 \times 3\times (0.2)^{2}=0.24\,kg-m^{2}$
Putting values in Eq (i)
$T=2\pi \sqrt{\frac{0.24}{3\times10\times0.2}}=2\pi \sqrt{0.04}$
$=2\pi\times0.2=1.2566\,s$
$T=1.3\,s$