Question

The ratio of the radii of gyration of a hollow cone and a hollow cylinder of the same mass and radius about a tangential axis parallel to their central axis is:

• A. \(\sqrt{3} : \sqrt{2}\)
• B. \(\sqrt{3} : 2\)
• C. \(1 : \sqrt{2}\)
• D. \(\sqrt{5} : \sqrt{6}\)

Hint:

For a thin hollow cylinder about its own axis: \[ I = MR^2 \] For a hollow cone about its own axis: \[ I = \frac{1}{2}MR^2 \] Always apply the Parallel Axis Theorem carefully when shifting to a tangential axis.

Answer 1

The Correct Option is B

Concept: The radius of gyration \(k\) is related to moment of inertia \(I\) by: \[ I = Mk^2 \] or \[ k = \sqrt{\frac{I}{M}} \] To find the moment of inertia about a tangential axis parallel to the central axis, we use the Parallel Axis Theorem: \[ I = I_{\text{cm}} + MR^2 \] where \(R\) is the perpendicular distance between the two axes. Step 1: Find the radius of gyration of the hollow cone.
For a hollow cone about its central axis: \[ I_{\text{cm, cone}} = \frac{1}{2}MR^2 \] Using the Parallel Axis Theorem: \[ I_{\text{tangent, cone}} = \frac{1}{2}MR^2 + MR^2 \] \[ = \frac{3}{2}MR^2 \] Now, \[ k_1 = \sqrt{\frac{I}{M}} \] \[ = \sqrt{\frac{\frac{3}{2}MR^2}{M}} \] \[ = \sqrt{\frac{3}{2}}R \]

Step 2: Find the radius of gyration of the hollow cylinder.
For a thin hollow cylinder about its central axis: \[ I_{\text{cm, cylinder}} = MR^2 \] Applying the Parallel Axis Theorem: \[ I_{\text{tangent, cylinder}} = MR^2 + MR^2 \] \[ = 2MR^2 \] Hence, \[ k_2 = \sqrt{\frac{2MR^2}{M}} \] \[ = \sqrt{2}R \]

Step 3: Calculate the required ratio.
\[ \frac{k_1}{k_2} = \frac{\sqrt{\frac{3}{2}}R}{\sqrt{2}R} \] \[ = \frac{\sqrt{3}}{2} \] Therefore, \[ k_1 : k_2 = \sqrt{3} : 2 \] Hence, the correct answer is: \[ \boxed{\text{(B)}} \]