A ring with mass m and radius r is suspended from a point on its periphery. Find the time period of its small oscillations.

Step-by-step solution

The time period of a physical pendulum is given by T = 2 I mgd , where I is the moment of inertia about the point of suspension and d is the distance from the point of suspension to the center of mass. For a ring of mass m and radius r , the moment of inertia about its center of mass is I_ cm = mr^2 . Using the parallel axis theorem, the moment of inertia about a point on its periphery is I = I_ cm + md^2 = mr^2 + mr^2 = 2mr^2 . The distance from the point of suspension to the center of mass is d = r . Substituting