Question

Three bodies have equal mass m. Body A is solid cylinder of radius R, body B is a square lamina of side R, and body C is a solid sphere of radius R. Which body has the smallest moment of inertia about an axis passing through their centre of mass and perpendicular to the plane (in case of lamina)

• A. A
• B. B
• C. C
• D. A and C both

Answer 1

Answer: (B)

B

Answer 2

To determine which body has the smallest moment of inertia, we need to calculate the moment of inertia for each body about the specified axis. All bodies have equal mass 'm' and radius 'R' (or side 'R' for the square lamina).

1. Body A: Solid cylinder of radius R.
   The axis passes through its center of mass and is perpendicular to its plane (i.e., along its central axis).
   The moment of inertia of a solid cylinder about its central axis is given by:
   $I_A = \frac{1}{2} m R^2$
   In decimal form: $I_A = 0.5 m R^2$

2. Body B: Square lamina of side R.
   The axis passes through its center of mass and is perpendicular to its plane.
   For a square lamina of mass 'm' and side 'R', the moment of inertia about an axis passing through its center and perpendicular to its plane is given by:
   $I_B = \frac{1}{6} m R^2$
   In decimal form: $I_B \approx 0.1667 m R^2$

3. Body C: Solid sphere of radius R.
   The axis passes through its center of mass.
   The moment of inertia of a solid sphere about an axis passing through its center is given by:
   $I_C = \frac{2}{5} m R^2$
   In decimal form: $I_C = 0.4 m R^2$

Now, let's compare the numerical coefficients of $mR^2$ for each body:

• For Body A: 0.5
• For Body B: 0.1667
• For Body C: 0.4

Comparing these values: $0.1667 < 0.4 < 0.5$.
Therefore, $I_B$ is the smallest moment of inertia.

The body with the smallest moment of inertia is Body B, the square lamina.