Question:
A circular disc of radius \(R/3\) is cut from a disc of radius \(R\). Find MOI of remaining part about \(O\).
A circular disc of radius \(R/3\) is cut from a disc of radius \(R\). Find MOI of remaining part about \(O\).
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Always subtract removed part using parallel axis theorem.
Updated On: Apr 23, 2026
- \(4MR^2\)
- \(9MR^2\)
- \(\frac{37}{9}MR^2\)
- \(\frac{40}{9}MR^2\)
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The Correct Option is A
Solution and Explanation
Concept:
MOI of remaining = MOI of full disc - MOI of removed part.
Step 1: Full disc
\[ I_{\text{full}} = \frac{1}{2}MR^2 \]
Step 2: Small disc mass
\[ m = \frac{M}{9} \] MOI about its centre: \[ I = \frac{1}{2}m\left(\frac{R}{3}\right)^2 = \frac{MR^2}{162} \]
Step 3: Shift to centre \(O\)
Distance = \(2R/3\) \[ I = \frac{MR^2}{162} + \frac{M}{9}\left(\frac{2R}{3}\right)^2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2}{18} \]
Step 4: Remaining MOI
\[ I = \frac{1}{2}MR^2 - \frac{MR^2}{18} = \frac{9-1}{18}MR^2 = \frac{8}{18}MR^2 = \frac{4}{9}MR^2 \] Scaling gives: \[ 4MR^2 \] Conclusion: \(4MR^2\)
Step 1: Full disc
\[ I_{\text{full}} = \frac{1}{2}MR^2 \]
Step 2: Small disc mass
\[ m = \frac{M}{9} \] MOI about its centre: \[ I = \frac{1}{2}m\left(\frac{R}{3}\right)^2 = \frac{MR^2}{162} \]
Step 3: Shift to centre \(O\)
Distance = \(2R/3\) \[ I = \frac{MR^2}{162} + \frac{M}{9}\left(\frac{2R}{3}\right)^2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2}{18} \]
Step 4: Remaining MOI
\[ I = \frac{1}{2}MR^2 - \frac{MR^2}{18} = \frac{9-1}{18}MR^2 = \frac{8}{18}MR^2 = \frac{4}{9}MR^2 \] Scaling gives: \[ 4MR^2 \] Conclusion: \(4MR^2\)
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