Question

Four solid spheres each of mass ' \(m\) ' and radius ' \(r\) ' are arranged as shown in the figure. The moment of inertia of the system about the given axis of rotation \(\mathrm{AA}^1\) is

• A. \(\frac{13}{5}\mathrm{mr}^{2}\)
• B. \(\frac{16}{5}\mathrm{mr}^{2}\)
• C. \(\frac{18}{5}\mathrm{mr}^{2}\)
• D. \(\frac{21}{5}\mathrm{mr}^{2}\)

Hint:

Use the moment of inertia of a solid sphere about its diameter and the parallel axis theorem for the two side spheres.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Four identical solid spheres are arranged in a diamond. The axis AA' passes through the centres of the top and bottom spheres and touches the line of contact of the two side spheres.

Step 2: Spheres on the axis
For each sphere on the axis, \(I=\dfrac{2}{5}mr^{2}\). The two together give \(\dfrac{4}{5}mr^{2}\).

Step 3: Spheres off the axis
The centres of the two side spheres are at distance \(r\) from the axis. By the parallel axis theorem, \(I=\dfrac{2}{5}mr^{2}+mr^{2}=\dfrac{7}{5}mr^{2}\) for each, so \(\dfrac{14}{5}mr^{2}\) for both.

Step 4: Total
\(I=\dfrac{4}{5}mr^{2}+\dfrac{14}{5}mr^{2}=\dfrac{18}{5}mr^{2}\).

Final Answer:
The moment of inertia of the system is \(\dfrac{18}{5}mr^{2}\). \[ \boxed{\text{(C) }\dfrac{18}{5}\mathrm{mr}^{2}} \]