NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
AB is a light rigid rod, which is rotating about a vertical axis passing through end A . A spring of force constant k and natural length l is attached at A and its other end is attached to a small bead of mass m . The bead can slide without friction on the rod. At the initial moment, the bead is at rest (with respect to the rod) and the spring is unstretched. The maximum speed of the bead with respect to the rod duri
Options
- AV max = m ω 2 l 2 k - m ω 2
- BV max = m ω 4 l 2 k - m ω 2
- CV max = m ω 4 l 2 m ω 2 - k
- DV max = m ω 2 l 2 m ω 2 - k
Correct answer
B. V max = m ω 4 l 2 k - m ω 2
Step-by-step solution
Velocity will be maximum at the equilibrium position ⇒ k x = m ω 2 l + x ⇒ x = m ω 2 l k - m ω 2 Now, using work-energy theorem ∆ K E = Work done by all the forces 1 2 m V max 2 = ∫ 0 x m ω 2 l + x d x - 1 2 k x 2 ⇒ V max 2 = 2 m ω 2 l x + m ω 2 x 2 - k x 2 m V max 2 = m ω 2 l + m ω 2 l + x - k x x m = m ω 2 l x m ⇒ V max 2 = ω 2 l x = m ω 4 l 2 k - m ω 2 V max = m ω 4 l 2 k - m ω 2