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A physical pendulum pivoted at a point executes angular oscillations. Its mass is m, has its centre of mass at distance r from the point of suspension. If its moment  of Inertia is  I, then its angular frequency is

a
ω=MgrI2
b
ω=I2Mgr
c
ω=MgrI
d
ω=IMgr

detailed solution

Correct option is C

For a body executing simple harmonic motion, Restoring torque acting on it after a  small displacement θ, about an axis,torque τ=−cθ  here c=couple acting; θ is angular displacement  We know that this results in angular oscillations which can be also related by the equation  torque=τ=Iα  ; here I=moment of inertia; α=angular accelerationEquating the above two we get  −cθ=IαWhere for angular oscillations  α=−ω2θHence −cθ=−Iω2θ Simplifying this c=Iω2 Hence  ω=cIFor angular oscillation ω=cI , where  ω=2πTRestoring torque acting on a rod after a small displacement  θ, about an axis  passing through point of contact O with the curved path,τ=−Mgrsinθ=-r×F=force×perpendicular distance  , for small angles   sinθ≈θTherefore,  τ=−cθ=−MgrθHence  c=Mgr. Substituting this is (1) we get ω=MgrI .

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