Question

The moment of inertia of a uniform semi circular disc about an axis passing through its centre of mass and perpendicular to its plane is (Mass of this disc is $M$ and radius is $R$)

• A. $\frac{M{R}^2}{2}+M{\left(\frac{4R}{3\pi }\right)}^2$
• B. $\frac{M{R}^2}{2}-M{\left(\frac{4R}{3\pi }\right)}^2$
• C. $\frac{M{R}^2}{2}-M{\left(\frac{2R}{\pi }\right)}^2$
• D. $\frac{M{R}^2}{2}+M{\left(\frac{2R}{\pi }\right)}^2$

Answer 1

The correct option is B $\frac{M{R}^2}{2}-M{\left(\frac{4R}{3\pi }\right)}^2$

Using parallel axis theorem:
$I_o=I_{cm}+m(O{O}^{\prime }{)}^2$ where OO' is the distance between the com, O' and the centre of the semicircular disc
$O{O}^{\prime }=\frac{4R}{3\pi }$ since com of the semi circular disc is at a distance $\frac{4R}{3\pi }$ from O. as shown in the figure.
$I_o=\frac{1}{2}\times I_{circulardisc}=\frac{1}{2}\left(\frac{1}{2}{M}_{circulardisc}{R}^2\right)=\frac{1}{2}{M}_{semicirculardisc}{R}^2$
$\therefore I_{cm}=I_o-M_{semicirculardisk}(\frac{4R}{3\pi }{)}^2$
$\therefore I_{cm}=\frac{1}{2}{M}_{semicirculardisk}{R}^2-M_{semicirculardisk}(\frac{4R}{3\pi }{)}^2$

$M=M_{semicirculardisc}$

$I_{cm}=[\frac{M{R}^2}{2}-M{\left(\frac{4R}{3\pi }\right)}^2]$