NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
A weightless rod of length l carries two equal masses m one fixed at the end and other in the middle of the rod. The rod can revolve in a vertical plane about A . Then horizontal velocity which must be imparted to end C of the rod to deflect it to the horizontal position is
Options
- A12 5 g l
- B3 g l
- C16 5 g l
- D2 g l
Correct answer
A. 12 5 g l
Step-by-step solution
Loss in kinetic energy = gain in potential energy ∴ 1 2 I ω 2 = m g l 2 + m g l ...(i) Here, I = m l 2 4 + m l 2 = 5 4 m l 2 From (i), 1 2 5 4 m l 2 ω 2 = m g l 2 + m g l = 3 2 m g l m l 2 ω 2 = 12 5 m g l ∴ v c = 12 g l 5