1. Tardigrade
  2. Question
  3. Physics
  4. The time-period of a physical pendulum is 2π√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

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Q. 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

AMUAMU 2018Oscillations

Solution:

Given, $T=2\pi \sqrt{\frac{I}{mg. d}} \ldots\left(i\right)$
$m=3\,kg, d$ or $r=20\,cm=0.2\, m$
image
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)
$T=2\pi \sqrt{\frac{0.24}{3\times10\times0.2}}=2\pi \sqrt{0.04}$
$=2\pi\times0.2=1.2566\,s$
$T=1.3\,s$