Question

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

• A. 2:1
• B. \(\sqrt2:1\)
• C. 4:1
• D. \(1:\sqrt2\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The radius of gyration (\(k\)) of a body about a given axis is the distance from the axis at which the entire mass of the body can be assumed to be concentrated such that the moment of inertia remains the same.
The relation is given by \(I = Mk^2\), which implies \(k = \sqrt{\frac{I}{M}}\).
Key Formula or Approach:
For a thin uniform disc of mass \(M\) and radius \(R\):
1. Moment of inertia about an axis passing through the centre and perpendicular to the plane (\(I_c\)) is \(\frac{1}{2}MR^2\).
2. Moment of inertia about a diameter (\(I_d\)) is \(\frac{1}{4}MR^2\).
Step 2: Detailed Explanation:
Let \(k_1\) be the radius of gyration about the axis through the centre and normal to the plane:
\[ Mk_1^2 = I_c = \frac{1}{2}MR^2 \implies k_1 = \sqrt{\frac{R^2}{2}} = \frac{R}{\sqrt{2}} \]
Let \(k_2\) be the radius of gyration about the diameter:
\[ Mk_2^2 = I_d = \frac{1}{4}MR^2 \implies k_2 = \sqrt{\frac{R^2}{4}} = \frac{R}{2} \]
The required ratio is:
\[ \frac{k_1}{k_2} = \frac{R/\sqrt{2}}{R/2} = \frac{2}{\sqrt{2}} = \frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}} = \sqrt{2} \]
Thus, the ratio is \(\sqrt{2} : 1\).
Step 3: Final Answer:
The ratio of the radii of gyration is \(\sqrt{2} : 1\).