Disc with a tangent hole

Problem

From a uniform disc of radius $R$, a smaller disc of radius $R/2$, internally tangent to the edge, is cut out. Using the anti-matter method, determine the position of the centre of mass of the remaining figure relative to the centre of the original disc.

Solution

Masses proportional to areas (uniform density). Whole disc: $m_1 \propto \pi R^2$, with CM at $x = 0$. Hole of radius $R/2$: $m_2 \propto \pi (R/2)^2 = \tfrac{1}{4}\pi R^2$, so $m_2 = \tfrac{1}{4}m_1$.

Position of the hole. Internally tangent to the edge: its centre is at a distance $R - \tfrac{R}{2} = \tfrac{R}{2}$ from the centre of the disc. We place the hole at $x_2 = +R/2$.

Anti-matter method (the hole enters with negative mass): $x_\text{CM} = \frac{m_1\cdot 0 - m_2\cdot \tfrac{R}{2}}{m_1 - m_2} = \frac{-\tfrac{1}{4}m_1\cdot \tfrac{R}{2}}{m_1 - \tfrac{1}{4}m_1} = \frac{-\tfrac{1}{8}R}{\tfrac{3}{4}}$ $\ev{x_\text{CM} = -\frac{R}{6} \approx -0{,}17\,R}$ The minus sign indicates that the CM shifts to the side opposite the hole.

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