A rod of length 50 c m is pivoted at one end. It is raised such that it makes an angle of 30 o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in r a d s - 1 ) will be g = 10 m s - 2

Step-by-step solution

From conservation of mechanical energy Loss in gravitational P E = gain in K E m g l 2 sin ⁡ 30 o = 1 2 I ω 2 m g l 2 sin ⁡ 30 o = 1 2 m l 2 3 ω 2 ω = 3 g 2 l = 3 × 10 2 × 50 × 10 - 2 = 30   r a d / s