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Question

Physics Question on Moment Of Inertia

Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis X X' which is touching to two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about XX' axis is

A

165mr2\frac{16}{5} mr^2

B

4mr24 m r^2

C

115mr2\frac{11}{5} mr^2

D

3mr23mr^2

Answer

4mr24 m r^2

Explanation

Solution

Net moment of inertia of the system, I=I1+I2+I3I = I_1 + I_2 + I_3 The moment of inertia of a shell about its diameter, I1=23mr2I_1 = \frac{2}{3} mr^2 The moment of inertia of a shell about its tangent is given by I2=I3=I1+mr2=23mr2+mr2=53mr2I_2 = I_3 = I_1 + mr^2 = \frac{2}{3} mr^2 + mr^2 = \frac{5}{3} mr^2 I=2×53mr2+23mr2=12mr23=4mr2\therefore I = 2 \times \frac{5}{3} mr^2 + \frac{2}{3} mr^2 = \frac{12mr^2}{3} = 4mr^2