Question

As shown in the figure, radius of gyration about the axis shown in \(\sqrt{n}\) cm for a solid sphere. Find 'n'. 

Hint:

For radius of gyration problems, mass usually cancels out.
Focus on the geometric terms: \(k^2 = k_{cm}^2 + d^2\).

Answer 1

Correct Answer: 265

Step 1: Understanding the Concept:
The radius of gyration \(k\) is defined by the relation \(I = Mk^2\).
For an axis not passing through the center of mass, we use the parallel axis theorem.
Step 2: Key Formula or Approach:
1. Moment of inertia of a solid sphere about its center: \(I_{cm} = \frac{2}{5}MR^2\).
2. Parallel axis theorem: \(I = I_{cm} + Md^2\).
3. Radius of gyration: \(k^2 = \frac{I}{M}\).
Step 3: Detailed Explanation:
Given: Radius \(R = 10 \text{ cm}\) and distance to the axis \(d = 15 \text{ cm}\).
The total moment of inertia about the given axis is:
\[ I = \frac{2}{5}MR^2 + Md^2 \]
Since \(I = Mk^2\), we can write:
\[ Mk^2 = M \left( \frac{2}{5}R^2 + d^2 \right) \]
\[ k^2 = \frac{2}{5}R^2 + d^2 \]
Substitute the given values into the equation:
\[ k^2 = \frac{2}{5}(10)^2 + (15)^2 \]
\[ k^2 = \frac{2}{5}(100) + 225 \]
\[ k^2 = 40 + 225 = 265 \]
Therefore, \(k = \sqrt{265} \text{ cm}\).
Comparing this with \(\sqrt{n}\), we find \(n = 265\).
Step 4: Final Answer:
The value of \(n\) is 265.