A uniform rod AB of mass 2 kg and length 1 m is placed on a sharp support O , such that AO = a = 25 cm and OB = b = 75 cm . A spring of force constant 600 N m - 1 is attached to the end B , as shown. Initially, the spring is extended by 2 cm at equilibrium. As soon as the thread is burnt, find the normal force (in N ) exerted by the support at O .

Step-by-step solution

I about O is m L 2 12 + m L 2 16 = 7 m L 2 48 Spring force ⇒ K Δ x = 600 × 2 100 = 12 N Torque, just after burning the string, τ = 2 × 10 × 1 4 + 12 × 3 4 ⇒ τ = 5 + 9 = 14 N m ∴ α = 14 × 48 7 × 2 × 1 = 48 rad s - 2 ∴ Acceleration of COM, a = 48 × 1 4 = 12 m s - 2 Now, total vertical force 12 + 20 – N = 2 × 12 ⇒ N = 32 – 24 = 8 N