Question

For the same total mass which of the following will have the largest moment of inertia passing through its centre of mass and perpendicular to the plane of the body

• A. a disc of radius a
• B. a ring of radius a
• C. a square lamina of side 2a
• D. four rods forming a square of side a

Answer 1

Answer: (B)

B

Answer 2

To determine which object has the largest moment of inertia, we need to calculate the moment of inertia for each option about an axis passing through its center of mass and perpendicular to its plane. Let M be the total mass for all objects.

1. A disc of radius a: The moment of inertia of a disc of mass M and radius 'a' about an axis passing through its center and perpendicular to its plane is: $I_A = \frac{1}{2} M a^2 = 0.5 M a^2$

2. A ring of radius a: The moment of inertia of a ring of mass M and radius 'a' about an axis passing through its center and perpendicular to its plane is: $I_B = M a^2 = 1.0 M a^2$

3. A square lamina of side 2a: For a square lamina of mass M and side length L, the moment of inertia about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{6} M L^2$. Here, the side length L = 2a. $I_C = \frac{1}{6} M (2a)^2 = \frac{1}{6} M (4a^2) = \frac{2}{3} M a^2 \approx 0.667 M a^2$

4. Four rods forming a square of side a: The total mass of the system is M. Since there are four rods, the mass of each rod is $m = \frac{M}{4}$. Each rod has a length L = a. The axis of rotation passes through the center of the square and is perpendicular to its plane. Consider one rod. Its center of mass is at its midpoint. The distance from the center of mass of a rod to the central axis of the square is $d = \frac{a}{2}$. The moment of inertia of a single rod of mass 'm' and length 'a' about an axis perpendicular to its length and passing through its center of mass is $I_{cm} = \frac{1}{12} m a^2$. Using the parallel axis theorem, the moment of inertia of one rod about the central axis of the square is: $I_{rod} = I_{cm} + m d^2 = \frac{1}{12} m a^2 + m \left(\frac{a}{2}\right)^2 = \frac{1}{12} m a^2 + \frac{1}{4} m a^2 = \frac{1}{12} m a^2 + \frac{3}{12} m a^2 = \frac{4}{12} m a^2 = \frac{1}{3} m a^2$ Since there are four such rods, the total moment of inertia for the system is: $I_D = 4 \times I_{rod} = 4 \times \frac{1}{3} m a^2 = \frac{4}{3} m a^2$ Substitute $m = \frac{M}{4}$: $I_D = \frac{4}{3} \left(\frac{M}{4}\right) a^2 = \frac{1}{3} M a^2 \approx 0.333 M a^2$

Comparing the calculated moments of inertia: $I_A = 0.5 M a^2$ $I_B = 1.0 M a^2$ $I_C = 0.667 M a^2$ $I_D = 0.333 M a^2$

The largest moment of inertia is $I_B = M a^2$, which corresponds to the ring of radius 'a'.