Question

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

• A. $\frac{M R^{2}}{2}$
• B. $\frac{M R^{2}}{4}\left(cos\right)^{2}\left(\text{\theta }\right)$
• C. $\frac{M R^{2}}{4}\left(sin\right)^{2}\left(\text{\theta }\right)$
• D. $\frac{M R^{2}}{4}$

Answer 1

Answer: (D)

Moment of inertia of disk w.r.t. principle axis (1) = $\frac{M R^{2}}{2}$
Moment of inertia of disk w.r.t. diametrical axis (2) -

From perpendicular axes theorem $I_{z}=I_{x}+I_{y}$ $...\left(i\right)$
here $I_{x}=I_{y}$ , because both are diametrical axis,
So here $I_{z}=\frac{M R^{2}}{2}$ and $I_{x}=I_{y}=I\left(L e t\right)$
So from eq. $\left(i\right)$
$\Rightarrow $ $\frac{M R^{2}}{2}=I+I\Rightarrow I=\frac{M R^{2}}{4}$
For half disk also, it is applicable, So
$I=\frac{MR^{2}}{4}$
(It does not depend on $\theta $ given in diagram, because given axis is also similar to diametrical axis)