Question

Four spheres of diameter 2a and mass m are placed on corners of square of side b. The moment of inertia of system about an axis through one side is

• A. \(\frac{4}{5}Ma^2+2Mb^2\)
• B. \(\frac{8}{5}Ma^2+8Mb^2\)
• C. \(\frac{8}{5}Ma^2\)
• D. \(\frac{5}{2}Ma^2+4Mb^2\)

Hint:

Parallel axis theorem: \(I=I_{\text{cm}}+mr^2\).

Answer 1

The Correct Option is B

Step 1: M.I. of one sphere about its own centre: \[ I_0=\frac25 m a^2. \]
Step 2: Shift to axis through side using parallel axis theorem. Two spheres lie at distance \(0\); two at distance \(b\).
Step 3: \[ I = 4\left(\frac25 ma^2\right) + 2(mb^2)+2(mb^2). \]
Step 4: \[ I=\frac{8}{5}ma^2+4mb^2. \] Total mass \(M=4m\). Converting: \[ I=\frac{8}{5}Ma^2+8Mb^2. \] Hence option (B).