NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping then the value of F is:
Options
- A3 2 m a
- B2 m a
- C5 3 m a
- Dm a
Correct answer
A. 3 2 m a
Step-by-step solution
From free body diagram of cylinder F - f s = m a .....(1) ∵       ∑   f e x t = m a c m a l s o   ∑ τ e x t = I c m α ⟹ f s   R = I c m α ⟹ f s   R = 1 2   m R 2 α ..... (2) For rolling without slipping a = R α ...... (3) ⟹   α = a R ∴       f s   R = 1 2 m R 2 a R ⟹   f s = 1 2 m a Put in (1) F - 1 2 m a = m a ⟹ F = 3 2 m a