Question

A ring of mass m and Radius R oscillate about point O as shown in figure, then it's time period is​

Answer 1

Answer:

T=2π√I/mgd

T= 2π√2R/g

Answer 2

IT's time period is​T=21TV2R/g.

I=MR^2+MR^2

=MR^2

I=Given BY The

T=2π√(1/mgr)

T=2π√(2MR^2/Mgr)

T=2π√(2r/g)

*the point of the axis of rotation is present at a periphery of the circle so the moment of inertia of the point o will be I=MR^2 + MR^2=2MR^2. Since, the moment of inertia of a ring is MR^2. Now to get the time period of the ring we know that the time period of a body . *moment of inertia I is given by the T=217(1/MgR). Now, substituting the values of the I we will get that T=217V(2MR^2/MgR). Which on solving we will get that

T=211V2R/g.