A uniform rod A B of mass 2 kg and Length 30 cm at rest on a smooth horizontal surface. An impulse of force 0 . 2 N s is applied to end B. The time taken by the rod to turn through at right angles will be π x s , where x = ____.

Step-by-step solution

Impulse J = 0 . 2 N s ⇒ J = ∫ F d t = 0 . 2 N s Now, angular impulse ( M → ) will be M c = ∫ τ d t = ∫ F L 2 d t = L 2 ∫ F d t = L 2 × J = 0 . 3 2 × 0 . 2 = 0 . 03 Moment of inertia of the rod about the centre of mass, I cm = M L 2 12 = 2 × ( 0 . 3 ) 2 12 = 0 . 09 6 Angular impulse will be equal to the change in angular momentum. M = I cm ω f - ω i ⇒ 0 . 03 = 0 . 09 6 ω f ⇒ ω f = 2 rad s - 1 For angular displacement, we can write θ = ω t ⇒ t = θ ω = π 2 × 2 = π 4 s . Therefore, x = 4 .