**Question:** A circular pizza of radius R has a circular piece of radius R/4 removed from one side, as shown in the figure below. Where is the centre of mass of the pizza with the hole in it?

**Answer:**

Let the centre of mass of the pizza with the hole in it be x.

From the definition of centre of mass,

$$x_{CM} = \frac{1}{M} \sum \, mx$$

Since the centre of mass of a WHOLE pizza is at the centre, x_{CM} is 0.

$$0 = \frac{1}{M} \left\{ \sigma \left[ \pi \left( \frac{R}{4} \right)^{2} \left( – \frac{3}{4} R \right) \right] + \sigma x \left[ \pi \left( R \right)^{2} – \pi \left( \frac{R}{4} \right)^{2} \right] \right\}$$

Basically, the above equation is saying that the whole pizza is made up of two portions: the cut out piece and the left over piece. The centre of mass of the cut out piece is $-\left( \frac{3}{4} \right) R$ while the centre of mass of the left over piece is x.

Re-arranging and solving,

$$x = \left( \frac{1}{20} \right) R$$