Question

In a free space, a thin rod carrying uniformly distributed negative charge $-q$ is placed symmetrically along the axis of a thin ring of a radius $R$ carrying uniformly distributed charge $Q$. Mass of the rod is m and length is l=2R.The ring is fixed and the rod is free to move.The rod is displaced slightly along the axis of the ring and then released, Find the period T of the small amplitude oscillations of the rod.

Answer 1

In a free space, a thin rod carrying uniformly distributed negative charge $=-q$

Radius $=R$

Uniform distributed charge$=Q$

Mass of the rod $=m$

Length $l=2R$

Then find the period of Time $T$ of the small amplitude oscillations of the rod.

For a rigid body oscillating about horizontal axis passing through the point O the time period is given by

Where $g=$ moment of the rigid body

$OG$ the distance of the point of pilot to the centre of gravity G and I is the moment of inertia of the rod.

$g=\frac{1}{3}m{e}^{2}OG=\frac{l}{2}=\frac{2R}{2}=RT=2\pi \sqrt{\frac{\frac{1}{3}m{l}^{2}mgR}{mgR}}=2\pi \sqrt{\frac{2l}{3g}}$