Question:

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

Updated On: Aug 1, 2022
  • $ \frac{16}{5} mr^2 $
  • $ 4 m r^2 $
  • $ \frac{11}{5} mr^2 $
  • $ 3mr^2 $
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The Correct Option is B

Solution and Explanation

Net moment of inertia of the system, $ I = I_1 + I_2 + I_3 $ The moment of inertia of a shell about its diameter, $ I_1 = \frac{2}{3} mr^2 $ The moment of inertia of a shell about its tangent is given by $ I_2 = I_3 = I_1 + mr^2 = \frac{2}{3} mr^2 + mr^2 = \frac{5}{3} mr^2 $ $ \therefore I = 2 \times \frac{5}{3} mr^2 + \frac{2}{3} mr^2 = \frac{12mr^2}{3} = 4mr^2 $
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Concepts Used:

Moment of Inertia

Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.

Moment of inertia mainly depends on the following three factors:

  1. The density of the material
  2. Shape and size of the body
  3. Axis of rotation

Formula:

In general form, the moment of inertia can be expressed as, 

I = m × r²

Where, 

I = Moment of inertia. 

m = sum of the product of the mass. 

r = distance from the axis of the rotation. 

M¹ L² T° is the dimensional formula of the moment of inertia. 

The equation for moment of inertia is given by,

I = I = ∑mi ri²

Methods to calculate Moment of Inertia:

To calculate the moment of inertia, we use two important theorems-

  • Perpendicular axis theorem
  • Parallel axis theorem