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