Solid Sphere Radius of Gyration Calculation

This solution determines the value of $n$ based on the radius of gyration ($k$) of a solid sphere rotating about a specified axis.

Key Physics Concepts

• Moment of Inertia (MI): A measure of a body's resistance to changes in its rotation.
• Radius of Gyration ($k$): The effective distance from the axis of rotation where the object's mass can be assumed concentrated. It relates MI ($I$) and mass ($M$) as $I = Mk^2$.
• Parallel Axis Theorem: Relates the MI about a central axis ($I_{cm}$) to the MI about a parallel axis ($I$) at distance $d$: $I = I_{cm} + Md^2$.

Moment of Inertia Calculation

The moment of inertia for a solid sphere of radius $R$ about an axis through its center is:

$ I_{cm} = \frac{2}{5}MR^2 $

Using the Parallel Axis Theorem, the moment of inertia ($I$) about an axis at distance $d$ from the center is:

$ I = I_{cm} + Md^2 = \frac{2}{5}MR^2 + Md^2 $

Radius of Gyration Calculation

We relate the moment of inertia $I$ to the radius of gyration $k$ using $I = Mk^2$. Equating this to the parallel axis result:

$ Mk^2 = \frac{2}{5}MR^2 + Md^2 $

Cancelling mass $M$ gives the square of the radius of gyration:

$ k^2 = \frac{2}{5}R^2 + d^2 $

Substitute the given values:

• Sphere radius $R = 10 \text{ cm}$
• Distance from center $d = 15 \text{ cm}$

Calculation:

$ k^2 = \frac{2}{5}(10 \text{ cm})^2 + (15 \text{ cm})^2 $

$ k^2 = \frac{2}{5}(100 \text{ cm}^2) + 225 \text{ cm}^2 $

$ k^2 = 40 \text{ cm}^2 + 225 \text{ cm}^2 $

$ k^2 = 265 \text{ cm}^2 $

Determining the Value of n

The problem states the radius of gyration is $k = \sqrt{n} \text{ cm}$. Squaring this gives:

$ k^2 = n \text{ cm}^2 $

Comparing $k^2 = 265 \text{ cm}^2$ with $k^2 = n \text{ cm}^2$, we find:

$ n = 265 $

Final Answer

The value of $n$ is 265.