A hanging object cannot be considered as a point mass and it oscillates about a fixed axis which does not pass through its centre of mass uniform rod of mass M and length L. The rod is pivoted at one end and hangs vertically in equilibrium with its centre of mass vertically below the point of suspension. The rod is slightly at the lower end and released. It then oscillates in a vertical plane in a simple harmonic manner, at any instant angular displacement of the rod from its vertical position is θ assuming small θ, equation of motion of the rod can be expressed $$as(Here$$ I is the moment of inertia of the rod about the axis about which the rod oscillates $$)$$

Question

A hanging object cannot be considered as a point mass and it oscillates about a  fixed axis which does not pass through its centre of mass uniform rod of mass M  and length L. The rod is pivoted at one end and hangs vertically in equilibrium with  its centre of mass vertically below the point of suspension.  The rod is slightly at  the lower end and  released. It then oscillates in a vertical plane in a simple  harmonic manner, at any instant angular displacement of the rod from its vertical  position is θ assuming small θ, equation of motion of the rod can be expressed  $$as(Here$$ I is the moment of inertia of the rod about the axis about which the rod  oscillates $$)$$

Infinity Learn · Accepted Answer

For an object executing angular simple harmonic motion. τ$$=c$$θ  Where, τ$$=I$$α, Therefore,  Iα$$=$$−cθWhere $$d2$$θ$$dt2=$$α and Restoring 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  , for small angles $$sin$$θ≈θ  Therefore,  τ$$=$$−cθ$$=$$−$$Mgl2$$θ……………….$$(1)Therefore$$ $$Id2$$θ$$dt2=$$−$$Mgl2$$θ  is the equation of angular simple harmonic motion.

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