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A hollow spherical shell of mass 1 kg and radius R rolls without slipping on a horizontal floor with angular speed omega. The magnitude of the angular momentum of the shell about the contact point O on the floor is (a/3) * R² * omega. Find the value of a.

1. 2
2. 3
3. 5
4. 4

Correct answer: 5

Solution

For a hollow spherical shell: I_cm = (2/3)mR². For rolling: v_cm = omega*R. Angular momentum about origin O (contact point on floor): L = I_cm*omega + m*v_cm*R = (2/3)*1*R²*omega + 1*(omega*R)*R = (2/3 + 1)*R²*omega = (5/3)*R²*omega. Comparing with (a/3)*R²*omega gives a = 5.

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