A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is a/3 R²ω. The value of a will be :

Solution

Physics sir · 4 steps

Split the angular momentum into spin about the centre plus the centre's own motion about O, then add.

1. Two parts. For a body that moves and spins at once, L_O = I_(cm)ω + M v_(cm) d. Here d is the perpendicular distance from O to the line along which the centre moves.
2. Spin part. For a thin spherical shell, I_(cm) = 2/3MR². With M = 1 kg this gives 2/3R²ω, clockwise, which points into the page.
3. Orbital part. Rolling without slipping gives v_(cm) = ω R towards +x. The centre moves along the line y = R, so d = R wherever the shell is. This gives M v_(cm) R = R²ω, also into the page.

   L_O = 2/3R²ω + R²ω = 5/3R²ω

4. So. Both parts point the same way, so they add directly. Comparing with a/3R²ω gives a = 5, which is option (c).

The idea to keep. Angular momentum about any point = spin about the centre of mass + M v_(cm) times the perpendicular distance. Check both directions before you add them.

Answer from the NTA final answer key. Solution by PrepWiser.