A solid sphere A of radius R and mass M is at | vedclass

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(B) The distance between the centres of the two spheres is $d = R + r$.For the axis passing through the centre of $A$:The moment of inertia of sphere

Vedclass · Accepted Answer

$(m - M)(R + r)^2$

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(B) The distance between the centres of the two spheres is $d = R + r$.For the axis passing through the centre of $A$:The moment of inertia of sphere $A$ about its own centre is $I_{A,cm} = \frac{2}{5}MR^2$.The moment of inertia of sphere $B$ about the axis passing through the centre of $A$ is given by the parallel axis theorem: $I_{B,A} = I_{B,cm} + md^2 = \frac{2}{5}mr^2 + m(R + r)^2$.Thus, $I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R + r)^2$.For the axis passing through the centre of $B$:The moment of inertia of sphere $B$ about its own centre is $I_{B,cm} = \frac{2}{5}mr^2$.The moment of inertia of sphere $A$ about the axis passing through the centre of $B$ is given by the parallel axis theorem: $I_{A,B} = I_{A,cm} + Md^2 = \frac{2}{5}MR^2 + M(R + r)^2$.Thus, $I_B = \frac{2}{5}mr^2 + \frac{2}{5}MR^2 + M(R + r)^2$.Now, calculating the difference $I_A - I_B$:$I_A - I_B = [\frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R + r)^2] - [\frac{2}{5}mr^2 + \frac{2}{5}MR^2 + M(R + r)^2]$$I_A - I_B = m(R + r)^2 - M(R + r)^2$$I_A - I_B = (m - M)(R + r)^2$.

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