Rotation Ques 34
- From a uniform circular disc of radius $R$ and mass $9 M$, a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of
disk is
(2018 Main)
(a) $\frac{37}{9} M R^{2}$
(b) $4 M R^{2}$
(c) $\frac{40}{9} M R^{2}$
(d) $10 M R^{2}$
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Answer:
Correct Answer: 34.(b)
Solution:
Formula:
Two Important Theorems on Moment of Inertia :
- $I _{\text {Remaining }}=I _{\text {Total }}-I _{\text {Cavity }}$
$$ \Rightarrow \quad I=\frac{9 M R^{2}}{2}-\frac{M R^{3}}{6}+M \frac{2 R^{2}}{3}=4 M R^{2} $$
BETA
