NTA Abhyas JEE Main2020PhysicsRotational MotionPractice
A solid sphere and hollow sphere of the same mass and radius are given a spin about their centre of mass and then, they are placed on a rough horizontal surface. The spin angular velocity is the same for both the spheres and it is equal to ω 0 . Once the pure rolling starts; let v 1 and v 2 be the linear speeds of their centres of mass, then
Options
- Av 1 = v 2
- Bv 1 > v 2
- Cv 1 < v 2
- DData is insufficient
Correct answer
C. v 1 < v 2
Step-by-step solution
From conservation of angular momentum about point of contact : I ω 0 = I ω + mRv or I ω 0 = I v R + mRv or v = I ω 0 I R + mR or or v = ω 0 1 R + mR I Now I solid sphere < I hollow ∴ v solid < v hollow ∴ v 1 < v 2