Question

The centre of mass of given system is at a distance x from geometrical centre of bigger body.

• A. x = 0 or x > R
• B. Centre of mass lies outside the bigger body
• C. 0 < x < R
• D. Centre of mass lies within bigger body
• E. None

Answer 1

Answer: (C)

0 < x < R

Answer 2

The center of mass of a system can be calculated using the formula $\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$. For a continuous body with mass density $\rho(\vec{r})$, the formula is $\vec{R}_{CM} = \frac{\int \vec{r} dm}{\int dm} = \frac{\int \vec{r} \rho(\vec{r}) dV}{\int \rho(\vec{r}) dV}$.

When dealing with a body with a cavity, we can use the concept of negative mass. The system is treated as a complete body of positive mass minus the mass of the cavity. If the density is uniform, $\rho = \sigma$ (mass per unit area for a thin disk).

Let the geometrical center of the bigger disk be the origin (0, 0). The radius of the bigger disk is R.

(I) Disk with one circular cavity: A circular cavity of radius r is removed, with its center at a distance d from the origin. The figure shows the cavity is off-center, so $d > 0$. The cavity is inside the disk, so $d+r \le R$. Mass of the complete disk $M = \sigma \pi R^2$. Its center of mass is at (0, 0). Mass of the cavity $m = \sigma \pi r^2$. Its center of mass is at a position $\vec{d}$ from the origin, with $|\vec{d}| = d$. The center of mass of the system with the cavity is $\vec{X}_{CM} = \frac{M \cdot \vec{0} - m \vec{d}}{M - m} = \frac{-m \vec{d}}{M - m}$. The distance of the center of mass from the origin is $x = |\vec{X}_{CM}| = \frac{m d}{M - m} = \frac{\sigma \pi r^2 d}{\sigma \pi R^2 - \sigma \pi r^2} = \frac{r^2 d}{R^2 - r^2}$. Since the cavity is off-center, $d > 0$, so $x > 0$. Since the cavity is inside the disk, $d \le R-r$. So $x \le \frac{r^2 (R-r)}{R^2 - r^2} = \frac{r^2 (R-r)}{(R-r)(R+r)} = \frac{r^2}{R+r}$. Since $r < R$, $R+r > 2r$, so $\frac{r^2}{R+r} < \frac{r^2}{2r} = \frac{r}{2}$. Also $r < R$. So $x < R$. Thus, for case (I), $0 < x < R$. This corresponds to option (R). Since $x < R$, the center of mass is within the bigger body, so it also corresponds to option (S). However, (R) is a more specific range for x.