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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

  1. Av 1 = v 2
  2. Bv 1 > v 2
  3. Cv 1 < v 2
  4. 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

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