Moment of inertia of semi circular disc with hole

Question

Suppose we have semi circular disc of radius $R$ with semi circular hole of radius $R/2$ , how can I find moment of inertia from axis thru $H$ I fail to see any symmetry so I cannot integrate , help?

Answer 1

If the mass is homogeneous with density $\rho$ then the angular inertia of a shape $\Omega$ around the origin is given by the integral

$$I = \rho\iint_{\Omega}(x^2+y^2)dx\wedge dy=\frac{\rho}{3}\oint_{\partial \Omega}\left(x^3 dy -y^3 dx\right).$$

Your shape is a difference of two half disks. Let's compute the above for $\Omega$ a half disk with radius $r$ and its diameter along the $x$-axis and a corner at the origin. I will use

$$\gamma(t)=r (\cos(t)+1, \sin(t))$$

for $t \in [0, \pi]$ as the circular part of the contour $\partial \Omega$. The linear part along the $x$-axis does not contribute to the integral. The angular inertia of this half disk is

$$I_r=\frac{\rho\, r^4}{3}\int_0^{\pi}\left((\cos(t)+1)^3\cos(t)+\sin(t)^4\right)dt = \frac{3}{4}\pi\rho\, r^4.$$

The inertia for the entire shape is the difference of the inertia of two half disks:

$$I=I_r-I_{r/2}=\frac{45}{64}\pi\rho\,r^4=\frac{15}{8}M r^2$$

where $M$ is the mass of the shape.

Answer 2

Choose a more comfortable point to compute the integral and then use Huygens-Steiner theorem.