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 $θ$ , time period of oscillation of the rod is (Here I is the moment of inertia of the rod about the axis about which the rod oscillates )

Moderate

Solution

For an object executing angular simple harmonic motion.
$τ = c θ$
Where, $τ = I α$ , Therefore, $I α = - c θ$
Where $\frac{d^{2} θ}{d t^{2}} = α$ 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,
$τ = - M g r \sin θ$ , for small angles $\sin θ \approx θ$
Therefore, $τ = - c θ = - M g \frac{l}{2} θ$ ………………….(1)
Therefore $I \frac{d^{2} θ}{d t^{2}} = - M g \frac{l}{2} θ$ is the equation of angular simple harmonic motion.
Here, $α = \frac{d^{2} θ}{d t^{2}} = - ω^{2} θ$
Substituting this we get $- I ω^{2} θ = - M g \frac{l}{2} θ$
i.e. $- I ω^{2} = - M g \frac{l}{2}$
Moment of inertia of a rod oscillating about one of its end is given by $I = \frac{M l^{2}}{3}$
Substituting this we get $\frac{M l^{2}}{3} ω^{2} = M g \frac{l}{2}$
i.e. $\frac{l}{3} ω^{2} = g \frac{1}{2}$
simplifying this we get $ω^{2} = \frac{3 g}{2 l} \Rightarrow ω = \sqrt{\frac{3 g}{2 l}}$
We know that $ω = \frac{2 π}{T}$
Using this we get $\frac{2 π}{T} = \sqrt{\frac{3 g}{2 l}}$
Hence, $T = 2 π \sqrt{\frac{2 l}{3 g}}$

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