System of Particles and Rotational Motion

Concept. The centre of mass stays where it is only if the mass-weighted shifts cancel: $m_1\,\Delta x_1+m_2\,\Delta x_2=0$. So the two particles must move in opposite directions along the line joining them.

Step 1. Write the condition. Take the line through the particles as the $x$-axis. Moving $m_1$ towards the centre of mass by $2$ cm gives $\Delta x_1=+2$ cm (say). For the centre of mass not to move, $$m_1\Delta x_1+m_2\Delta x_2=0\ \Rightarrow\ 3(2)+2\,\Delta x_2=0\ \Rightarrow\ \Delta x_2=-3\ \text{cm}$$

Step 2. Read the sign. $\Delta x_2$ is opposite to $\Delta x_1$. Since $m_2$ lies on the other side of the centre of mass, moving opposite to $m_1$ means moving towards the centre of mass.

Answer. $m_2$ moves $3$ cm towards the centre of mass. The lighter particle has to move farther: shifts are in the inverse ratio of the masses.

Concept. For a continuous body, $y_{cm}=\dfrac{1}{M}\int y\,dm$. For a wire, take small arc elements and add their contributions.

Step 1. Choose an element. Put the centre of the semicircle at the origin, ends on the $x$-axis. An element at angle $\theta$ from the $x$-axis has length $R\,d\theta$, mass $dm=\lambda R\,d\theta$ with $\lambda=\dfrac{M}{\pi R}$, and height $y=R\sin\theta$.

Step 2. Integrate from $\theta=0$ to $\pi$. $$y_{cm}=\frac1M\int_0^{\pi}R\sin\theta\cdot\lambda R\,d\theta=\frac{\lambda R^2}{M}\Big[-\cos\theta\Big]_0^{\pi}=\frac{\lambda R^2}{M}\cdot2=\frac{M}{\pi R}\cdot\frac{2R^2}{M}=\frac{2R}{\pi}$$

Step 3. Symmetry. Every element at $+x$ has a partner at $-x$, so $x_{cm}=0$.

Answer. $\left(0,\dfrac{2R}{\pi}\right)$, so $|x|=2$. Remember it: semicircular wire $\dfrac{2R}{\pi}$; semicircular plate $\dfrac{4R}{3\pi}$.

Concept. A body with a hole = full body minus the removed piece. Treat the hole as a "negative mass" at its own centre: $x_{cm}=\dfrac{M x_1-m x_2}{M-m}$. For a uniform plate, mass is proportional to area.

Step 1. Locate the hole. Disc radius $R=20$ cm at the origin. The hole (radius $5$ cm) touches the rim, so its centre is at $x_2=20-5=15$ cm.

Step 2. Masses by area. Take the mass per unit area as $\sigma$. $$M=\sigma\pi(20)^2=400\pi\sigma,\qquad m=\sigma\pi(5)^2=25\pi\sigma$$

Step 3. Apply the negative-mass formula. The full disc's centre is at $x_1=0$: $$x_{cm}=\frac{400\pi\sigma(0)-25\pi\sigma(15)}{400\pi\sigma-25\pi\sigma}=\frac{-375}{375}=-1\ \text{cm}$$

Answer. $1.0$ cm from the origin, on the side opposite the hole. The centre of mass always shifts away from a hole.