The ratio of the radii of gyration of a hollow sphere and a solid sphere of the same radii about a tangential axis

Options

A. `sqrt((7)/(3))`
B. `(5)/(sqrt(21))`
C. `sqrt((21)/(5))`
D. `(25)/(9)`

Explanation

To find the ratio of the radii of gyration of a hollow sphere and a solid sphere of the same radius about a tangential axis, we can follow these steps:

### Step 1: Understand the Moment of Inertia
The moment of inertia (I) of an object about an axis is a measure of how difficult it is to change its rotational motion about that axis. For our case, we need to find the moment of inertia about a tangential axis for both the hollow sphere and the solid sphere.

### Step 2: Use the Parallel Axis Theorem
The parallel axis theorem states that:
\[ I = I_{cm} + Md^2 \]
where:
- \( I \) is the moment of inertia about the new axis,
- \( I_{cm} \) is the moment of inertia about the center of mass,
- \( M \) is the mass of the object,
- \( d \) is the distance from the center of mass to the new axis.

### Step 3: Moment of Inertia for Hollow Sphere
For a hollow sphere, the moment of inertia about its center of mass is:
\[ I_{cm} = \frac{2}{3} M_h R^2 \]
where \( M_h \) is the mass of the hollow sphere and \( R \) is its radius.

Using the parallel axis theorem to find the moment of inertia about the tangential axis:
\[ I_h = I_{cm} + M_h R^2 = \frac{2}{3} M_h R^2 + M_h R^2 = \frac{2}{3} M_h R^2 + \frac{3}{3} M_h R^2 = \frac{5}{3} M_h R^2 \]

### Step 4: Moment of Inertia for Solid Sphere
For a solid sphere, the moment of inertia about its center of mass is:
\[ I_{cm} = \frac{2}{5} M_s R^2 \]
where \( M_s \) is the mass of the solid sphere.

Using the parallel axis theorem:
\[ I_s = I_{cm} + M_s R^2 = \frac{2}{5} M_s R^2 + M_s R^2 = \frac{2}{5} M_s R^2 + \frac{5}{5} M_s R^2 = \frac{7}{5} M_s R^2 \]

### Step 5: Calculate the Radii of Gyration
The radius of gyration \( k \) is defined as:
\[ k = \sqrt{\frac{I}{M}} \]

For the hollow sphere:
\[ k_h = \sqrt{\frac{I_h}{M_h}} = \sqrt{\frac{\frac{5}{3} M_h R^2}{M_h}} = R \sqrt{\frac{5}{3}} \]

For the solid sphere:
\[ k_s = \sqrt{\frac{I_s}{M_s}} = \sqrt{\frac{\frac{7}{5} M_s R^2}{M_s}} = R \sqrt{\frac{7}{5}} \]

### Step 6: Find the Ratio of the Radii of Gyration
Now, we can find the ratio of the radii of gyration of the hollow sphere to that of the solid sphere:
\[ \frac{k_h}{k_s} = \frac{R \sqrt{\frac{5}{3}}}{R \sqrt{\frac{7}{5}}} = \frac{\sqrt{\frac{5}{3}}}{\sqrt{\frac{7}{5}}} = \sqrt{\frac{5}{3} \cdot \frac{5}{7}} = \sqrt{\frac{25}{21}} \]

### Step 7: Simplify the Ratio
This simplifies to:
\[ \frac{k_h}{k_s} = \frac{5}{\sqrt{21}} \]

### Final Answer
Thus, the ratio of the radii of gyration of a hollow sphere to a solid sphere about a tangential axis is:
\[ \frac{k_h}{k_s} = \frac{5}{\sqrt{21}} \]