Question

The moment of inertia of semicircular plate of radius R and mass M about axis AA' in its plane passing through its center is :

• A. $\frac{{MR}^2}{2}$
• B. $\frac{{MR}^2}{4}{cos}^2\theta$
• C. $\frac{{MR}^2}{4}{sin}^2\theta$
• D. $\frac{{MR}^2}{4}$

Answer 1

The correct option is D $\frac{{MR}^2}{4}$

angular momentum $=Iw$

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 $L=I_x{w}_x+I_y{w}_y$

$L=\frac{M{R}^2}{4}w{\cos }^2\theta +\frac{M{R}^2}{4}w{\sin }^2\theta$

$Tw=w[\frac{m{R}^2}{4}{\cos }^2\theta +\frac{m{R}^2}{4}{\sin }^2\theta ]$

$I=\frac{m{R}^2}{4}({\cos }^2\theta +{\sin }^2\theta )$

$I=\frac{m{R}^2}{4}$