DEFINITION OF CENTRE OF MASS

Centre of mass: Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system. When the system moves under some external forces than this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at his point of translational motion. This point is called the centre of mass of the system.

Centre of mass of system of n point masses or n particles is that point about which moment of mass of the system is zero, it means that if about a particular origin the moment of mass of system of n point masses is zero, then that particular origin is the centre of mass of the system.

The concept of centre of mass is a pure mathematical concept. If there are n particles having mass m₁, m₂ ….m _n and are placed in space (x₁, y₁, z₁), (x₂, y₂, z₂) ……….(x _n, y _n, z _n) then centre of mass of system is defined as (X, Y, Z) where

= $\frac{1}{M}$ $\sum _i{m}_i{x}_i$

Y = $\frac{1}{M}$ $\sum _i{m}_i{y}_i$

and Z = $\frac{1}{M}$$\sum _i{m}_i{z}_i$

where M = $\sum _i{m}_i$ is the total mass of X the system.

Locate the point with coordinates (X, Y, Z). This point is called the centre of mass of the given collection of the particles. If the position vector of the i ^th particle is r_i, the centre of mass is defined to have the position vector.

$\overrightarrow{R_{CM}}=\frac{1}{M}\sum _i{m}_i\overrightarrow{r_i}$

Where M = m₁ + m₂ + ……….. + m_n = $\sum _{i=1}^n{m}_i$

where ${\overrightarrow{r}}_i=x_i\stackrel{\wedge }{i}+y_i\stackrel{\wedge }{j}+z_i\stackrel{\wedge }{k}$ \,\overset{\to }{\mathop{{{R}_{CM}}}}\,=\dfrac{1}{M}\,\left[ \begin{align} {{m}_{1}}({{x}_{1}}i\,+\,{{y}_{1}}j+{{z}_{1}}k)+{{m}_{2}}({{x}_{2}}i\,+\,{{y}_{2}}j+{{z}_{2}}k)+.. \\ {{m}_{n}}({{x}_{n}}i\,+\,{{y}_{n}}j+{{z}_{n}}k) \\ \end{align} \right]

=\dfrac{1}{M}\,\left[ \begin{align} i({{m}_{1}}{{x}_{1}}\,+\,{{m}_{2}}{{x}_{2}}+..+m{{x}_{n}})+j({{m}_{1}}{{y}_{1}}\,+\,{{m}_{2}}{{y}_{2}}+..+m{{y}_{n}})+.. \\ k({{m}_{1}}{{z}_{1}}\,+\,{{m}_{2}}{{z}_{2}}+..+m{{z}_{n}}) \\\end{align} \right]

= i x _cm + j y _cm + k (z _cm)

$\overrightarrow{R_{CM}}=\frac{\sum _{i=1}^n{m}_i\overrightarrow{r_i}}{M}$This is equation for centre of mass of n point masses.

Illustration 1: Three equal masses are situates at vertices of equilateral triangle as shown in figure. Find centre of mass of the system.

Solution:Let m be the mass of three masses and X_CM, Y_CM and Z_CM be the centre of masses of along the X axis, Y axis and Z axis, then ${\text{X}}_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

${\text{X}}_{\text{CM}}=\frac{m\left(0\right)+m\left(a\right)+m\left(a/2\right)}{m+m+m}=\frac{a}{2}$

${\text{Y}}_{\text{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$

${\text{Y}}_{\text{CM}}=\frac{m\left(0\right)+m\left(0\right)+m\left(\sqrt{3}a/2\right)}{m+m+m}=\frac{a}{2\sqrt{3}}$

${\text{Z}}_{\text{CM}}=\frac{m_1{z}_1+m_2{z}_2+m_3{z}_3}{m_1+m_2+m_3}$

${\text{Z}}_{\text{CM}}=\frac{m\left(0\right)+m\left(0\right)+m\left(0\right)}{m+m+m}=0$

Hence coordinate of centre of mass is $\left(\frac{a}{2},\frac{a}{2\sqrt{3}},0\right)$

CENTRE OF MASS OF CONTINUOUS BODIES

If we consider the body to have continuous distribution of matter, the summation in the formula of centre of mass should be replaced by integration. If x, y, z are the coordinates of this small mass dm, we write the coordinates of the centre of mass as

$X=\frac{1}{M}\int xdm,Y=\frac{1}{M}\int ydm,Z=\frac{1}{M}\int zdm$

The integration is to be performed under proper limits so that as the integration variable goes through the limits, the elements cover the entire body. We illustrate the method with three examples.

Illustration 2: Find centre of mass of a uniform straight rod of mass m and length l.

Solution: Let M and L be the mass and the length of the rod respectively. Take the left end of the rod as the origin and the X–axis along the rod (figure). Consider an element of the rod between the positions A and B of the rod. Let the element be at a distance x from the centre O and its width be dx. So as x varies from 0 through L, the elements cover the entire rod.

As the rod is uniform, the mass per unit length is M/L and hence the mass of the element is dm = (M/L) dx.

The x–coordinate of the centre of mass of the rod is

$X=\frac{1}{M}\int xdm=\frac{1}{M}\underset{0}{\overset{L}{\int }}x\left(\frac{M}{L}dx\right)=\frac{1}{L}{{\left[\frac{x^2}{2}\right]}^L}_0=\frac{L}{2}$

The y–coordinate is Y = $Y=\frac{1}{M}\int ydm=0$ and similarly Z = 0. The centre of mass is at $\left(\frac{L}{2},0,0\right)$, i.e. at the middle point of the rod.

MOTION OF CENTRE OF MASS

Motion of the Center of Mass: Let us consider the motion of a system of n particles of individual masses m₁, m₂, ……., m_n and total mass M. It is assumed that no mass enters or leaves the system during its motion, so that M remains constant. Then, as we have seen, we have the relation

${\overrightarrow{r}}_{CM}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+......+m_n\overrightarrow{r_n}}{m_1+m_2+.......+m_n}$

$=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+......+m_n\overrightarrow{r_n}}{M}$

Or $M{\overrightarrow{r}}_{CM}=m_1{\overrightarrow{r}}_1+m_2{\overrightarrow{r}}_2+......+m_n$

Differentiating this expression with respect to time t, we have$M\frac{{\overrightarrow{dr}}_{CM}}{dt}=m_1\frac{{\overrightarrow{dr}}_1}{dt}+m_2\frac{{\overrightarrow{dr}}_2}{dt}+.....m_n\frac{{\overrightarrow{dr}}_n}{dt}$

Since,$\frac{\overrightarrow{dr}}{dt}$ = velocity

Therefore, $M{\overrightarrow{v}}_{CM}=m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+.....+m_n{\overrightarrow{v}}_n$… (i)

Or velocity of the Center of Mass is ${\overrightarrow{v}}_{CM}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2+.....+m_n{\overrightarrow{v}}_n}{M}$ Or ${\overrightarrow{v}}_{CM}=\frac{\sum _{i=1}^n{m}_i{\overrightarrow{v}}_i}{M}$

Further, $m\overrightarrow{v}$ = momentum of a particle $\overrightarrow{P}$.

Therefore, Eq. (i) can be written as${\overrightarrow{P}}_{CM}={\overrightarrow{P}}_1+{\overrightarrow{P}}_2+.....+{\overrightarrow{P}}_n$

Or ${\overrightarrow{P}}_{CM}=\sum _{i=1}^n{\overrightarrow{P}}_i$

Differentiating Eq. (i) with respect to time t, we get$M\frac{{\overrightarrow{dv}}_{CM}}{dt}=m_1\frac{{\overrightarrow{dv}}_1}{dt}+m_2\frac{{\overrightarrow{dv}}_2}{dt}+......+m_n\frac{{\overrightarrow{dv}}_n}{dt}$

Or $M{\overrightarrow{a}}_{CM}=m_1{\overrightarrow{a}}_1+m_2{\overrightarrow{a}}_2+......+m_n{\overrightarrow{a}}_n$ … (ii)

Or ${\overrightarrow{a}}_{CM}=\frac{m_1{\overrightarrow{a}}_1+m_2{\overrightarrow{a}}_2+.....+m_n{\overrightarrow{a}}_n}{M}$

Or ${\overrightarrow{a}}_{CM}=\frac{\sum _{i=1}^n{m}_1{\overrightarrow{a}}_1}{M}$

Further, in accordance with Newton's second law of motion $\overrightarrow{F}=m\overrightarrow{a}$. Hence, Eq. (ii) can be written as${\overrightarrow{F}}_{CM}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+.....+{\overrightarrow{F}}_n$ Or ${\overrightarrow{F}}_{CM}=\sum _{i=1}^n{\overrightarrow{F}}_i$

Thus, as pointed out earlier also, the centre of mass of a system of particles moves as though it were a particle of mass equal to that of the whole system with all the external forces acting directly on it.

Illustration 3: Two particles A and B of mass 1 kg and 2 kg respectively are projected in the directions shown in figure with speeds u_A = 200 m/s and u_B = 50 m/s. Initially they wee 90 m apart. Find the maximum height attained by the centre of mass of the particles. Assume acceleration due to gravity to be constant. (g = 10 m/s²).

Solution: Using m_Ar_A = m_Br_B

Or (1)(r_A) = (2)(r_B) Or r_A = 2r_B… (i)

And r_A + r_B = 90 m… (ii)

Solving these two equations, we get r_A = 60m and r_B = 30m;

i.e., CM is at height 60m from the ground at time t = 0

Further, ${\overrightarrow{a}}_{CM}=\frac{m_A{\overrightarrow{a}}_A+m_B{\overrightarrow{a}}_B}{m_A+m_B}$= g = 10 m/s² (downwards)

As ${\overrightarrow{a}}_A={\overrightarrow{a}}_B=g$ (downwards)

${\overrightarrow{u}}_{CM}=\frac{m_A{\overrightarrow{u}}_A+m_B{\overrightarrow{u}}_B}{m_A+m_B}$

$=\frac{\left(1\right)\left(200\right)-\left(2\right)\left(50\right)}{1+2}=\frac{100}{3}m/s$ (upwards)

Let, h be the height attained by CM beyond 60 m. Using

$v_{CM}^2=u_{CM}^2+2a_{CM}h$

Or $0={\left(\frac{100}{3}\right)}^2-\left(2\right)\left(10\right)h$

Or $h=\frac{{\left(100\right)}^2}{180}=55.55m$

Therefore, maximum height attained by the centre of mass is , H = 60 + 55.55 = 115.55 m