# Moment of inertia of a sector about its center point?

What is the moment of inertia of a pizza slice that has a radius r, an angle (radians) of theta, and a height of h about the center point perpendicular to the cheese plane?

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It could possibly be like a cylinder? if $$I = (mr^2) / 2$$ for a cylander than maybe for a sector: $$I = (\theta m r^2) / 2$$ – Sam Mar 4 '12 at 21:42
Do you mean the entire moment of inertia tensor, or only about the axis through the "center point" (I assume this means center of mass) and perpendicular to the plane of the cheese/crust? – Mark Eichenlaub Mar 4 '12 at 21:42
I think, OP meant The moment of inertia about the axis through the "center point" (center of complete circle) and perpendicular to the plane. Using the forlmulae $I=\int r^2 dm$, you are receive the same formulae, as for disk $I=\frac{mr^2}{2}$, where $m$ - is the mass of pizza slice/ – Sergio Mar 4 '12 at 21:52

I assume you're talking about spinning the pizza around the axis a pizza-dough spinner would spin it.

The moment of inertia of a solid disk is $\frac{mr^2}{2}$. If it spins, each slice contributes angular momentum proportional to $\theta$. The slice's mass is also proportional to $\theta$ so the moment of inertia of a slice of pizza about its tip is also $\frac{mr^2}{2}$ where $m$ is now the mass of the slice.

We can use the parallel axis theorem to find the moment of inertia through the slice's center of mass. The center of mass is displaced a distance $\frac{2}{3}r\mathrm{sinc}\frac{\theta}{2}$ from the tip, so the moment of inertia through this axis is

$$I = \left(\frac{1}{2} + (\frac{2}{3}\mathrm{sinc}\frac{\theta}{2})^2\right)mr^2$$