### Motion of Center of Mass :

**Position vector of the centre of mass of a system of particle is given by :**

$ \displaystyle \vec{r_{cm}} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2} + ….+ m_n\vec{r_n}}{m_1 + m_2 + ….+ m_n } $

Differentiating with respect to time we get ,

$ \displaystyle \vec{v_{cm}} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2} + ….+ m_n\vec{v_n}}{m_1 + m_2 + ….+ m_n } $

Further Differentiating with respect to time we get ,

$ \displaystyle \vec{a_{cm}} = \frac{m_1 \vec{a_1} + m_2 \vec{a_2} + ….+ m_n\vec{a_n}}{m_1 + m_2 + ….+ m_n } $

$ \displaystyle \vec{a_{cm}} = \frac{\vec{F_1}+\vec{F_2}+ ….+ \vec{F_3}}{m_1 + m_2 + ….+ m_n } $

### Effect of External Forces on Center of Mass

$ \displaystyle \vec{a_{cm}} = \frac{1}{M}\Sigma_{i=1}^{n} m_1 \vec{a_i}$

$ \displaystyle M \vec{a_{cm}} = \Sigma \vec{F_{ext}} + \Sigma \vec{F_{int}}$

But the internal forces are in the form of action –reaction pairs. Hence they cancel each other. Thus

$ \displaystyle \Sigma \vec{F_{int}} = 0 $

$ \displaystyle M \vec{a_{cm}} = \Sigma \vec{F_{ext}} $

Thus centre of mass is effected by only external force acting on the system. Internal forces will have no effect on the motion of centre of mass.

**When no external force acts on the system:**

(a) acceleration of centre of mass is zero i.e.,

$ \displaystyle \Sigma \vec{F_{ext}} = M \vec{a_{cm}} $

$ \displaystyle M \vec{a_{cm}} = 0 $

$ \displaystyle \vec{a_{cm}} = 0 $

(b) Velocity of centre of mass is constant $ \displaystyle \vec{v_{cm}} = constant $

(c) Linear momentum of the system is constant $ \displaystyle \vec{p_{cm}} = constant $.

It is called the law of conservation of linear momentum.

**Solved Example : **Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speeds of 2 m/sec and 6 m/sec respectively on a smooth horizontal surface. Find the speed of the c.m. of the system

**Solution: **Velocity of c.m. of the system is given by

$ \displaystyle \vec{v_{cm}} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2} }{m_1 +m_2 } $

$ \displaystyle |m_1 \vec{v_1} + m_2 \vec{v_2}| = m_1v_1 + m_2 v_2 $

Since the particles m_{1} & m_{2} are moving in the same direction are parallel.

$ \displaystyle v_{cm} = \frac{m_1v_1 + m_2 v_2}{m_1 +m_2 } $

$ \displaystyle v_{cm} = \frac{1\times 2 + 0.5\times 6}{1 + 0.5 } $

= 3.33 m / sec

Exercise : In this illustration, if the second particle approaches the first with same speed, what will be the speed of centre of mass of the system?

#### Acceleration of the c.m. of a system of particles:

$ \displaystyle \vec{a_{cm}} = \frac{\vec{F_1}+\vec{F_2}+ ….+ \vec{F_3}}{m_1 +m_2 + ….+m_n } $

$ \displaystyle \vec{a_{cm}} = \frac{\sum \vec{F_{ext}}}{M} $

$ \displaystyle \sum \vec{F_{ext}} = 0 ; \: \vec{a_{cm}} = 0 $

Hence in absence of any external resultant force, centre of mass of a system of particles is either at rest or in uniform motion on straight line.

**Solved Example : **Two particles of mass 2 kg & 4 kg are approaching towards each other with accelerations 1 m/sec^{2} & 2 m/sec^{2} respectively on a smooth horizontal surface. Find the acceleration of centre of mass of the system.

**Solution: **The acceleration of c.m. of the system is given by

$ \displaystyle \vec{a_{cm}} = \frac{m_1 \vec{a_1} + m_2 \vec{a_2} }{m_1 + m_2 } $

$ \displaystyle a_{cm} = \frac{m_1 a_1 – m_2 a_2}{m_1 +m_2} $

$ \displaystyle a_{cm} = \frac{|2\times 1 – 4\times 2|}{2+4} $

a_{cm} = 1m/sec^{2}

Since m_{2}a_{2} > m_{1}a_{1}, the direction of acceleration of c.m. will be directed in the direction of a_{2}.

**Solved Example : **Find the acceleration of the centre of mass of two particles approaching towards each other under their own gravitational field.

**Solution: ** $ \displaystyle \vec{a_{cm}} = \frac{m_1 \vec{a_1} + m_2 \vec{a_2} }{m_1 + m_2 } $

$ \displaystyle \vec{a_{cm}} = \frac{ \vec{F_1} + \vec{F_2} }{m_1 + m_2 } $

Since , $\vec{F_1}$ & $\vec{F_2}$ are equal in magnitude and opposite in direction,

$\displaystyle \vec{a_{cm}} = \frac{ 0 }{m_1 + m_2 } = 0 $

### Mutual forces between two bodies:

(i) When two particles approach each other due to their mutual interaction, then they always meet at their centre of mass.

(ii) In a system of two particle of masses m_{1} and m_{2} , when m_{1} is pushed towards m_{2} through a distance d then shift in m_{2} towards m_{1} without altering CM positions is $\displaystyle -\frac{m_1}{m_2} d $

(iii) A boy of mass m is at one end of a flat boat of mass M and length l which floats stationary on water. If boy moves to the other end,

(a) The boat moves in opposite direction through a distance $\displaystyle d = \frac{m l}{M + m} d $

(b) The displacement of boy with respect to ground is $\displaystyle d’ = – \frac{M l}{M + m} d $

(iv) A boy of mass m is standing on a flat boat floating stationary on the surface of water. If the boy starts moving on the boat with velocity, V_{r} with respect to boat, then

(a) Velocity of the boat w.r.t. ground is $\displaystyle V = -\frac{m V_r}{M + m} $ , ‘-‘ indicates boat moves in opposite direction to the velocity of the boy.

(b) Velocity of boy w.r.t. ground is $\displaystyle V’ = \frac{M V_r}{M + m} $

**Solved Example:** Two particles A and B of mass 1 kg and 2 kg respectively are projected in the directions shown in figure with speeds u_{1} = 200 m/s and u_{2} = 50 m/s. Initially they were 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/s^{2})

**Solution:** Using m_{1} r_{1} = m_{2} r_{2}

or (1)(r_{1}) = (2) (r_{2})

or r_{1} = 2 r_{2}

and r_{1} + r_{2} = 90 m

Solving these two equations, we get

r_{1} = 60 m and r_{2} = 30 m

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

$ \displaystyle \vec{u_{com}} = \frac{m_1 \vec{u_1} + m_2 \vec{u_2} }{m_1 + m_2 } $

$ \displaystyle \vec{u_{com}} = \frac{1 \times 200 – 2 \times 50 }{1 + 2 } = \frac{100}{3} m/s$

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

$ \displaystyle v_{com}^2 = u_{com}^2 + 2 a_{com} h $

$ \displaystyle 0 = (\frac{100}{3})^2 – 2 \times 10 \times h $

$ \displaystyle h = \frac{100^2}{180} = 55.55 m$

Therefore, maximum height attained by the centre of mass is

H = 60 + 55.55 = 115.55 m

**Solved Example :** In the arrangement shown in figure, m_{A} = 2 kg and m_{B} = 1 kg. String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere.

**Solution: **Net pulling force on the system is (m_{A} – m_{B})g

or (2 – 1)g = g

Total mass being pulled is m_{A} + m_{B} or 3 kg

$ \displaystyle a = \frac{Net \;pulling \;force}{Total\; mass} = \frac{g}{3}$

$ \displaystyle \vec{a_{com}} = \frac{m_A \vec{a_A} + m_B \vec{a_B} }{m_A + m_B } $

$ \displaystyle = \frac{2 \times a – 1 \times a }{2 + 1 } $

$ \displaystyle = \frac{a}{3} = \frac{g}{9} m/s^2$

**Solved Example:** A projectile of mass 3m is projected from ground with velocity 20√2 m/s at 45°. At highest point it explodes into two pieces. One of mass 2m and the other of mass m. Both the pieces fly off horizontally in opposite directions. Mass 2m falls at a distance of 100 m from point of projection. Find the distance of second mass from point of projection where it strikes the ground. (g = 10 m/s²)

**Solution:** Range of the projectile in the absence of explosion

$ \displaystyle R = \frac{u^2 sin2\theta}{g} $

$ \displaystyle R = \frac{(20 \sqrt{2})^2 sin2\times 45^o}{10} $

R = 80 m

The path of centre of mass of projectile will not change, i.e., X_{com} is still 80 m. Now, from the definition of centre

of mass

$ \displaystyle X_{com} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} $

$ \displaystyle 80 = \frac{m \times x_1 + 2 m \times 100}{m + 2 m} $

Solving this equation, we get x_{1} = 40 m

Therefore, the mass m will fall at a distance x_{1} = 40 cm from point of projection

### Also Read :

→ Centre of mass of discrete & continuous mass distribution → Conservation of linear momentum & its Applications |