Question

From a uniform circular disc of radius $R$ , a circular disc of radius $R/6$ and having a center at a distance $R/2$ from the centre of the disc is removed. Determine the center of mass of the remaining portion of the disc.

Answer 1

The center of mass of a disc after removing the potion can be calculated by assuming the removed portion has negative mass and then calculating the center of mass due to the combination of the removed portion and the full disc without the removed portion.
${x_{COM}} = \sum\limits_{i = 1}^n {{m_i}{x_i}}$ where ${x_{COM}}$ is the $x$ coordinate of the center of mass, ${x_i}$ and ${m_i}$ are the x-coordinate and mass of the ith object respectively.

Complete step by step answer:
To solve such problems, a convenient way to find the center of mass is to assume the removed portion has negative mass and then use the combination of the removed mass and the complete disc. Let’s assume that the complete disc without the removed portion has mass $M$ . Thus the density of the disc, which is uniform, can be calculated as:
$\rho = \dfrac{M}{{4\pi {R^2}}}$
The mass of the removed portion $(m)$ then can be calculated as:
m = \rho \times 4\pi {\left( {\dfrac{R}{6}} \right)^2} \\\
= \dfrac{{\rho \times 4\pi {R^2}}}{{36}} \\\
We can write the mass of the removed portion in terms of the total mass as
$m = \dfrac{M}{{36}} \Rightarrow M = 36m$
Now assuming that the mass of the removed disc is negative and it is located $\dfrac{R}{2}$ away from the centre of the disc, we can use the formula for determining the centre of mass as:
${x_{COM}} = \dfrac{{M \times 0 - m \times \left( {\dfrac{R}{2}} \right)}}{{M - m}}$
Substituting $M = 36m$ , we can write:
{x_{COM}} = \dfrac{{m \times \left( {\dfrac{R}{2}} \right)}}{{36m - m}} \\\
$\therefore {x_{COM}} = - \dfrac{R}{{70}}$
Hence the center of mass of the remaining disc is located at a distance $R/70$ on the left side of the centre of the disc.

Note:
Since we’re removing mass on the right side of the disc, we can intuitively expect the center of mass to shift to the left side of the center which can be used as a consistency check for our final answer. While we’re assuming the mass of the removed portion as negative, it is a trick to solve such problems and in reality, there is no concept as a negative mass.