Question

Four identical uniform solid spheres each of same mass '$M$' and radius '$R$' are placed touching each other as shown in figure, with centres $A$, $B$, $C$, $D$. $I_A$, $I_B$, $I_C$ and $I_D$ are the moment of inertia of these spheres respectively about an axis passing through centre and perpendicular to the plane. The difference in $I_A$, and $I_B$ is

• A. 24 MR^2
• B. 32 MR^2
• C. 56 MR^2
• D. 80 MR^2

Hint:

The internal moments of inertia ($I_{cm}$) cancel out in the subtraction, so the difference is just based on the squared distances: $M[(4R^2 + 16R^2 + 36R^2) - (4R^2 + 4R^2 + 16R^2)]$.

Answer 1

The Correct Option is A

Step 1: Use the parallel axis theorem, $I = I_{cm} + Md^2$, to find the total moment of inertia of the system. For a solid sphere, $I_{cm} = \frac{2}{5} MR^2$.
Step 2: Calculate the total moment of inertia $I_A$ about an axis through center $A$. The distances from $A$ to the centers of spheres $B, C, D$ are $2R, 4R, 6R$ respectively. \[ I_A = 4\left( \frac{2}{5}MR^2 \right) + M(2R)^2 + M(4R)^2 + M(6R)^2 \] \[ I_A = 1.6MR^2 + 4MR^2 + 16MR^2 + 36MR^2 = 57.6MR^2 \]
Step 3: Calculate the total moment of inertia $I_B$ about an axis through center $B$. The distances from $B$ to centers of spheres $A, C, D$ are $2R, 2R, 4R$ respectively. \[ I_B = 4\left( \frac{2}{5}MR^2 \right) + M(2R)^2 + M(2R)^2 + M(4R)^2 \] \[ I_B = 1.6MR^2 + 4MR^2 + 4MR^2 + 16MR^2 = 25.6MR^2 \]
Step 4: Find the difference: \[ I_A - I_B = 57.6MR^2 - 25.6MR^2 = 32MR^2 \]