Question

A uniform disc of mass m and radius r is suspended in vertical plane from a point on its periphery. Its time period of oscillation is

Answer 1

The time period of the disc is 2π√(3r/2g)

We know that the time period of an object,

T = 2π√(I/mgL)

where ,

I  = moment of inertia from the suspended point

L = distance of its centre from suspended point = r

we know that, the moment of inertia of disc about its centre = mr²/2

using parallel axis theoram the moment of inertia from a point in its periphery,

I = mr² + mr²/2 = 3mr²/2

putting the values in the above equation we get,

T = 2π√(3mr²/2mgr)

= 2π√(3r/2g)

Hence the time period of the disc is 2π√(3r/2g)

Answer 2

This way you can solve