Worked exercise — Circle with a hole

Problem

A disc of radius $2\;\text{m}$ and surface density $2\;\text{kg/m}^2$ has a circular hole of radius $0{,}5\;\text{m}$ centred at $(1,1)$. Where is the CM located?

Solution

We use the anti-matter method: whole disc (positive mass) plus hole (negative mass).

Mass of the whole disc (radius $2\;\text{m}$, area $\pi R^2 = 4\pi$): $m_1 = 2\cdot\pi\cdot 4 = 25{,}1\;\text{kg}, \qquad \text{CM at } (0,0)$

Mass of the hole (radius $0{,}5\;\text{m}$, area $\pi (0{,}5)^2 = 0{,}25\pi$): $m_2 = 2\cdot\pi\cdot 0{,}25 = 1{,}57\;\text{kg}, \qquad \text{CM at } (1,1)$

Anti-matter formula (the hole enters with a minus sign): $\vec{s}_{\text{CM}} = \frac{m_1\,(0,0) - m_2\,(1,1)}{m_1 - m_2} = \frac{25{,}1\,(0,0) - 1{,}57\,(1,1)}{25{,}1 - 1{,}57} = \frac{(-1{,}57;\; -1{,}57)}{23{,}5}$

$\ev{\vec{s}_{\text{CM}} = (-0{,}07;\; -0{,}07)\;\text{m}}$

The CM shifts slightly in the direction opposite the hole: the hole is at $(1,1)$, so removing mass from that side pulls the centre of gravity toward the opposite quadrant.

Links

Topics: Centre of mass Concepts: Centre of mass Skills: Calculating the centre of mass