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

The problem involves understanding the concept of the radius of gyration for different axes of a uniform disc. The radius of gyration \( k \) is defined as \( k = \sqrt{\frac{I}{m}} \), where \( I \) is the moment of inertia and \( m \) is the mass of the object. 

For a thin uniform disc of radius \( R \), the moment of inertia \( I \) about an axis through its center and perpendicular to its plane is \( I_c = \frac{1}{2}mR^2 \). Hence, the radius of gyration \( k_c \) about this axis is:

\( k_c = \sqrt{\frac{\frac{1}{2}mR^2}{m}} = \sqrt{\frac{1}{2}}R \)

The moment of inertia \( I_d \) about the diameter of the disc is \( I_d = \frac{1}{4}mR^2 \). Therefore, the radius of gyration \( k_d \) about the diameter is:

\( k_d = \sqrt{\frac{\frac{1}{4}mR^2}{m}} = \frac{R}{2} \)

Thus, the ratio of the radius of gyration about the center perpendicular axis to that about the diameter is given by:

\( \frac{k_c}{k_d} = \frac{\sqrt{\frac{1}{2}}R}{\frac{R}{2}} = \sqrt{2} \)

The correct ratio is therefore \( \sqrt{2}:1 \).