MCQ
Moment of inertia of a semicircular disc of mass $M$ and radius $R$ about the shown axis is
- A$\frac{{M{R^2}}}{2}$
- ✓$\frac{{M{R^2}}}{4}$
- C$\frac{{M{R^2}}}{2}{\sin ^2}\theta $
- D$\frac{{M{R^2}}}{2}{\cos ^2}\theta $
✓
Answer
Correct option: B.
$\frac{{M{R^2}}}{4}$
b
angular momentum $=I w$
angular momentum $=I w$
Momentum of inertia $I_{x}=M R^{2} / 4$
$I_{X}=M R^{2} / 4$
$w_{x}=w \cos \theta$
$w_{y}=w \sin \theta$
angular momentum $L=I_{x} w_{x}+I_{y} w_{y}$
$L=\frac{M R^{2}}{4} w \cos ^{2} \theta+\frac{M R^{2}}{4} w \sin ^{2} \theta$
$T w=w\left[\frac{m R^{2}}{4} \cos ^{2} \theta+\frac{m R^{2}}{4} \sin ^{2} \theta\right]$
$I=\frac{m R^{2}}{4}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$
$I=\frac{m R^{2}}{4}$
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