NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
A solid sphere of radius r is gently placed on a rough horizontal surface with an initial angular speed ω 0 but no linear velocity. If the coefficient of friction is μ , then the time t when the slipping will stop is
Options
- A2 7 r ω 0 μ g
- B3 7 r ω 0 μ g
- C4 7 r ω 0 μ g
- Dr ω 0 μ g
Correct answer
A. 2 7 r ω 0 μ g
Step-by-step solution
Let m be the mass of the sphere. Since, it is a case of backward slipping, force of friction is in forward direction. Limiting friction will act in this case. Linear acceleration a = f m = μ mg m = μ g Angular retardation α = τ I = f · r 2 5 mr 2 = 5 2 μ g r Slipping will be stopped when v = rω at = r ( ω 0 - αt ) μ gt = r ( ω 0 - 5 2 μ g t r ) 7 2 μ gt = r ω 0 ∴ t = 2 7 r ω 0 μ g