NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
A uniform ball of radius r rolls without slipping down from the top of a sphere of radius R . The spin angular velocity of the ball when it breaks away from the sphere is (Assume initial velocity negligible)
Options
- A1 0 g R + r 1 7 r 2
- B1 0 g R - r 1 7 r 2
- C1 0 g R + r 1 7
- D1 0 R + r 17 r 2
Correct answer
A. 1 0 g R + r 1 7 r 2
Step-by-step solution
mv 2 R + r = mg cos θ ; mgh = 1 2 mv 2 + 1 2 I ω 2 mg R + r 1 - cos θ = 1 2 mv 2 + 1 5 mv 2 = 7 1 0 mv 2 1 0 7 mg 1 - cos θ = mg cos θ mv 2 = 1 0 7 mg R + r 1 - cos θ 1 0 7 = 1 7 7 cos θ or cos θ = 1 0 1 7 v = g R + r cos θ = 1 0 1 7 g R + r and ω = v r = 1 0 g R + r 1 7 r 2