Q. A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius r < R and mass m < M. Assume that the line joining their centres lies along the horizontal. The moment of inertia of the system calculated about a vertical axis passing through the centre of A is $I_A$ and that about a vertical axis passing through the centre of B is $I_B$. The difference $I_A-I_B$ is:
A) 0
B) $(M-m)(R+r)^2$
C) $(m-M)(R+r)^2$
D) $(m-M)(R-r)^2$
Explanation
[Solution from source] Centres are R + r apart. $I_A=\frac25MR^2+\frac25mr^2+m(R+r)^2$ and $I_B=\frac25MR^2+M(R+r)^2+\frac25mr^2$, so $I_A-I_B=(m-M)(R+r)^2$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.