JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Rotational Mechanics

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

\frac{a}{3} R^{2} ω

. The value of a will be:

(A)2
(B)3
(C)5✓ Correct
(D)4

Answer: (C)

Solution

For a rolling spherical shell, the angular momentum about point O is

L_{O} = m v_{c m} R + I ω

, where

v_{c m} = ω R

(rolling condition) and

I = \frac{2}{3} m R^{2}

for a spherical shell. Substituting:

L_{O} = m (ω R) R + \frac{2}{3} m R^{2} ω = m R^{2} ω + \frac{2}{3} m R^{2} ω = \frac{5}{3} m R^{2} ω

. With

m = 1

kg,

L_{O} = \frac{5}{3} R^{2} ω

, so

a = 5

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Single correct option

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