A solid sphere $A$ of radius $R$ and mass $M$ is attached to a smaller solid sphere $B$ of radius $r$ ($r\lt R$) and mass $m$ ($m\lt M$). The centres lie on the same horizontal line. The moments of inertia about the vertical axes passing through the centres of $A$ and $B$ are $I_A$ and $I_B$, respectively. The value of $I_A-I_B$ is:

Question

A thin circular disc of mass $ M $ and radius $ R $ is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with angular velocity $ \omega $. If another disc of the same dimensions but of mass $ M/2 $ is placed gently on the first disc co-axially, then the new angular velocity of the system is:

Answer 1

Step 1: Recall the tools.
A solid sphere about its own diameter has $I = \tfrac25 M R^2$, and the parallel axis theorem adds $Md^2$ when the axis is shifted by distance $d$. The two centres are separated by $R + r$.
Step 2: Moment of inertia about A's axis.
Sphere $A$ contributes $\tfrac25 M R^2$ about its own centre. Sphere $B$ is at distance $R+r$, so it contributes $\tfrac25 m r^2 + m(R+r)^2$. Adding: $I_A = \tfrac25 M R^2 + \tfrac25 m r^2 + m(R+r)^2$.
Step 3: Moment of inertia about B's axis.
Sphere $B$ contributes $\tfrac25 m r^2$ about its own centre. Sphere $A$ is at distance $R+r$, so it contributes $\tfrac25 M R^2 + M(R+r)^2$. Adding: $I_B = \tfrac25 m r^2 + \tfrac25 M R^2 + M(R+r)^2$.
Step 4: Subtract the two.
The $\tfrac25 M R^2$ and $\tfrac25 m r^2$ terms cancel, leaving $I_A - I_B = m(R+r)^2 - M(R+r)^2$.
Step 5: Factor the result.
\[ I_A - I_B = (m - M)(R+r)^2 \]
Step 6: Match the option.
Since $m < M$, this difference is negative, and it equals option C.
\[ \boxed{ I_A - I_B = (m - M)(R+r)^2 } \]

Show Hint

The Correct Option is C

Solution and Explanation

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