Moment of inertia of a semicircle by simple integration.

Question

The problem asks to calculate the moment of inertia of a semicircle:

By simple integration using polar coordinates:

$I_x=\int^{\pi/2}_{-\pi/2} y^2 dA$

Choosing a differential element with area $dA$ , $dA$ = $\frac{r_0^2}{2}d\theta$

Where $r_0$ is the radius of the cricle.

y=$r_0sin\theta$

$I_x=\int^{\pi/2}_{-\pi/2} {r_0^2sin^2\theta} \frac{r_0^2}{2}d\theta$

which evaluates to $\frac{\pi r_0^4}{4}$ while the solution should be $\frac{\pi r_0^4}{8}$

Where did i go wrong ?

Thanks in advance!

Answer 1

You can also calculate the moment of inertia by using double integrals. Considering the semicircle you have is from -π/2 to π/2 the moment of inertia you re looking for with respect to the x axis will be: integrate y^2 with polar coordinates: the inside integral will be bounded by -π/2, π/2 (with respect to θ --> the angle) and the other one will be bounded by 0 to r--> radius of the circle (with respect to p). Don't forget the jacobian( it will be p) and obviously y^2 = p^2 * (sinθ)^2. (You will transfer (x,y) to (p,θ) ).