Question

The moment of inertia of a solid sphere of mass 'm' and radius 'R' about its diametric axis is 'I'. Its moment of inertia about a tangent in the plane is ______.

• A. 2.5I
• B. 3.0I
• C. 3.5I
• D. 4I

Hint:

Be careful to read whether the object is a solid sphere ($I = \frac{2}{5}mR^2$) or a hollow sphere ($I = \frac{2}{3}mR^2$). A hollow sphere tangent would yield $I_t = \frac{5}{3}mR^2$, which would equal $2.5 I$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5} mR^2$. To find the moment of inertia about a tangent, we use the Parallel Axis Theorem.

Step 2: Formula Application:
Parallel Axis Theorem: $I_{tangent} = I_{cm} + md^2$. Here, $I_{cm} = I = \frac{2}{5} mR^2$ and the distance $d$ between the diameter and the tangent is $R$.

Step 3: Explanation:
$I_{tangent} = \frac{2}{5} mR^2 + mR^2 = \frac{7}{5} mR^2$. Since $I = \frac{2}{5} mR^2$, we can write $mR^2 = \frac{5}{2} I$. Substituting this: $I_{tangent} = \frac{7}{5} \times \left( \frac{5}{2} I \right) = \frac{7}{2} I = 3.5I$.

Step 4: Final Answer:
The moment of inertia about the tangent is 3.5I.