NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
For the arrangement in the figure, the particle M 1 attached to one end of a string which moves on a horizontal table in a circle of radius = l 2 (where l is the length of the string) with the constant angular speed ω . The other end of the string attached to mass M 2 which rests on a vertical rod. When the rod collapse, the acceleration of mass M 2 at that instant
Options
- Ag
- Bω 2 l 2
- C2 M 2 g - M 1 l ω 2 2 M 1 + M 2
- DM 2 g + M 1 l ω 2 M 1 + M 2
Correct answer
C. 2 M 2 g - M 1 l ω 2 2 M 1 + M 2
Step-by-step solution
From Newton's second law T - M 1 ω 2 l 2 = M 1 a And M 2   g - T = M 2 a After solving above equations, we get a = 2 M 2 g - M 1 l ω 2 2 M 1 + M 2