The moment of inertia of thin spherical shell of mass M and radius R about a diameter is `(2)/(3)`MR. Its radius of gyration K about a tangent will be

Options

A. `sqrt((2)/(3))R`
B. `(2)/(3)R`
C. `(5)/(3)R`
D. `sqrt((5)/(3))R`

Answer

To find the radius of gyration \( K \) of a thin spherical shell about a tangent, we can use the parallel axis theorem. Let's go through the solution step by step:

### Step 1: Understand the moment of inertia about the diameter
The moment of inertia \( I \) of a thin spherical shell of mass \( M \) and radius \( R \) about a diameter is given by:
\[ I_{CM} = \frac{2}{3} M R^2 \]

### Step 2: Apply the parallel axis theorem
The parallel axis theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is given by:
\[ I = I_{CM} + M d^2 \]
where \( d \) is the distance between the two axes.

### Step 3: Determine the distance \( d \)
In our case, the distance \( d \) from the center of the sphere to the tangent is equal to the radius \( R \) of the sphere.

### Step 4: Substitute values into the parallel axis theorem
Now, substituting the values into the equation:
\[ I_{tangent} = I_{CM} + M R^2 \]
\[ I_{tangent} = \frac{2}{3} M R^2 + M R^2 \]

### Step 5: Simplify the equation
Combine the terms:
\[ I_{tangent} = \frac{2}{3} M R^2 + \frac{3}{3} M R^2 = \frac{5}{3} M R^2 \]

### Step 6: Relate moment of inertia to radius of gyration
The moment of inertia can also be expressed in terms of the radius of gyration \( K \):
\[ I = M K^2 \]
Setting the two expressions for \( I \) equal gives:
\[ M K^2 = \frac{5}{3} M R^2 \]

### Step 7: Solve for \( K^2 \)
Dividing both sides by \( M \) (assuming \( M \neq 0 \)):
\[ K^2 = \frac{5}{3} R^2 \]

### Step 8: Take the square root to find \( K \)
Taking the square root of both sides:
\[ K = \sqrt{\frac{5}{3}} R \]

### Final Answer
Thus, the radius of gyration \( K \) about a tangent is:
\[ K = R \sqrt{\frac{5}{3}} \]

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