A ring of mass 2 π kg and of the radius 0 . 25 m is making 300 rpm about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is:

Step-by-step solution

As 2 t sin θ 2 = dm ω 2 r (for small angle sin θ 2 → θ 2 ) but dm = m ℓ θ r As ℓ = 2 π r ∴    T = m ω 2 r 2 π Put m = 2 π kg ,  ω = 1 0 π rad   s - 1 and r = 0 . 25   m ∴    T = 2 5 0  N