A ring of diameter `2m` oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equibvalent length of the simple pendulum is

Derive an expression for moment of inertia of a thin circular ring about an axis passing through its centre and perpendicular to the plane of the ring.

Radius of gyration of a ring about a transverse axis passing through its centre is

A disc of radius R is pivoted at its rim. The period for small oscillations about an axis perpendicular to the plane of disc is

The moment of inertia of ring about an axis passing through its diameter is I . Then moment of inertia of that ring about an axis passing through its centre and perpendicular to its plane is

A ring-1 oscillate with period T_1 about tangential axis in the plane of ring and an another ring-2 oscillate with period T_2 about tangential axis tangential axis perpendicular to plane of ring T_1/T_2 ?

A ring is suspended at a point on its rim and it behaves as a second's pendulum when it oscillates such that its centre move in its own plane. The radius of the ring would be (g = pi^(2))

A ring is suspended at a point on its rim and it behaves as a second’s pendulum when it oscillates such that its centre move in its own plane. The radius of the ring would be (g = pi^(2))

A rod with rectangular cross section oscillates about a horizontal axis passing through one of its ends and it behaves like a seconds pendulum, its length will be

An annular ring of internal and outer radii r and R respectively oscillates in a vertical plane about a horizontal axis perpendicular to its plane and passing through a point on its outer edge. Calculate its time period and show that the length of an equivalent simple pendulum is (3R)/(2) as r rarr 0 and 2 R as r rarr R .