Star Moment

We can consider the star to be made of five rods, each with mass $\frac{M}{5}$ . Each rod will have the same moment of inertia about the center of the star, since the distance of each rod about the center is the same.

We will first find the length of one rod. Consider the right triangle formed by joining the center of the circle, the midpoint of a rod, and an endpoint of the rod.

The hypotenuse of the triangle is $1$ , and the other sides have the length $\sin 72^\circ$ and $\cos 72^\circ$ . The length of the rod is $\ell = 2 \sin 72^\circ$ . Thus, the moment of inertia of each rod about its center is $\dfrac{m \ell ^2 }{12} = \dfrac{M}{5} \dfrac{ (2\sin 72^\circ)^2}{ 12}$ .

We want to find the moment of inertia of the rod about an axis which is at a distance $d = \cos 72^\circ$ away from the center of the rod. Using the parallel axis theorem, the moment of inertia of the rod about the center of the circle $I'$ is given by

$\begin{aligned} I' &= I + md^{2} \\ &= I + \frac{M}{5} (\cos 72^{\circ})^{2} \\ &= \frac{M}{5} \frac{(2\sin 72^{\circ} )^{2}}{12} + \frac{M}{5} (\cos 72^{\circ} )^{2} \\ \end{aligned}$

The moment of inertia of the complete star is

$\begin{aligned}5 I' & = M \dfrac{ (2\sin 72^\circ)^2}{12} + M (\cos 72^\circ)^2 \\ & = M (\frac{\sin^2 72^\circ}{3} + \cos^2 72^\circ) \\ & \approx 0.397 M \end{aligned}$

Use the expression for a rod about the center, along with the parallel axis theorem. From the figure, notice that the length of each of the 5 lines is $2 \cos(18 ^\circ)$ , and that the distance of each line from the origin is $\sin(18 ^\circ)$ .

$I = 5(\frac{M}{5}\frac{(2\cos(18 ^\circ))^{2}}{12} + \frac{M}{5}(\sin(18 ^\circ))^{2}) \approx \, 0.397$