Question

The ratio of radii of gyration of a thin circular ring and a circular disc of same radius about a tangential axis in their own planes is $\sqrt{12}:\sqrt{K}$. The value of K is

• A. 10
• B. 24
• C. 5
• D. 12

Hint:

To find the moment of inertia about an axis, first identify a standard axis (like one through the center of mass). Then, use the parallel axis theorem ($I = I_{cm} + Md^2$) or the perpendicular axis theorem ($I_z = I_x + I_y$) to shift to the required axis.

Answer 1

The Correct Option is A

Let $k_g$ be the radius of gyration and $I$ be the moment of inertia. The relationship is $I = Mk_g^2$.
We need to find the moment of inertia for both a ring and a disc about a tangential axis in their plane.
Step 1: Moment of Inertia of a thin circular ring.
The moment of inertia of a ring about a diameter is $I_{dia, ring} = \frac{1}{2}MR^2$.
Using the parallel axis theorem, the moment of inertia about a tangential axis in its plane (which is parallel to a diameter and at a distance R) is:
$I_{tan, ring} = I_{dia, ring} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
The radius of gyration of the ring is $k_{ring}^2 = \frac{I_{tan, ring}}{M} = \frac{3}{2}R^2 \implies k_{ring} = R\sqrt{\frac{3}{2}}$.
Step 2: Moment of Inertia of a circular disc.
The moment of inertia of a disc about a diameter is $I_{dia, disc} = \frac{1}{4}MR^2$.
Using the parallel axis theorem, the moment of inertia about a tangential axis in its plane is:
$I_{tan, disc} = I_{dia, disc} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.
The radius of gyration of the disc is $k_{disc}^2 = \frac{I_{tan, disc}}{M} = \frac{5}{4}R^2 \implies k_{disc} = R\sqrt{\frac{5}{4}}$.
Step 3: Find the ratio of the radii of gyration.
$\frac{k_{ring}}{k_{disc}} = \frac{R\sqrt{3/2}}{R\sqrt{5/4}} = \sqrt{\frac{3/2}{5/4}} = \sqrt{\frac{3}{2} \times \frac{4}{5}} = \sqrt{\frac{6}{5}}$.
We are given the ratio as $\sqrt{12}:\sqrt{K}$, which is $\frac{\sqrt{12}}{\sqrt{K}}$.
Equating the two ratios: $\frac{\sqrt{6}}{\sqrt{5}} = \frac{\sqrt{12}}{\sqrt{K}}$.
Squaring both sides: $\frac{6}{5} = \frac{12}{K}$.
Solving for K: $6K = 12 \times 5 \implies 6K = 60 \implies K = 10$.