Two touching solid spheres A (mass M, radius R) and B (mass m, radius r) have their centres separated by R + r. Let $I_A$ and $I_B$ be moments of inertia of the pair about parallel vertical axes through the centres of A and B, respectively. Find $I_A$ - $I_B.$ Image
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Question

Two touching solid spheres A (mass M, radius R) and B (mass m, radius r) have their centres separated by R + r. Let $I_A$ and $I_B$ be moments of inertia of the pair about parallel vertical axes through the centres of A and B, respectively. Find $I_A$ - $I_B.$ Image

Options

G1 Physics · Accepted Answer

Step 1: Let d = R + r be the centre separation. Step 2: About axis through A: I_A = (2/5)MR² + [(2/5)mr² + md²]. Step 3: About axis through B: I_B = [(2/5)MR² + Md²] + (2/5)mr². Step 4: Subtracting cancels each sphere's central moment of inertia: I_A − I_B = md² − Md². Step 5: Substitute d = R + r. Final answer: (m − M)(R + r)² (Option B).

Two touching solid spheres A (mass M, radius R) and B (mass m, radius r) have their centres separated by R + r. Let I_A and I_B be moments of inertia of the pair about parallel vertical axes through the centres of A and B, respectively. Find I_A − I_B. Image