# Motion of Center of Mass , Velocity of Center of mass , Acceleration of Center of mass

### 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 m1 & m2 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/sec2 & 2 m/sec2 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

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

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

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

acm = 1m/sec2

Since m2a2 > m1a1, the direction of acceleration of c.m. will be directed in the direction of a2.

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 m1 and m2 , when m1 is pushed towards m2 through a distance d then shift in m2 towards m1 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, Vr 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 u1 = 200 m/s and u2 = 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/s2)

Solution: Using m1 r1 = m2 r2

or (1)(r1) = (2) (r2)

or r1 = 2 r2

and r1 + r2 = 90 m

Solving these two equations, we get

r1 = 60 m and r2 = 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, mA = 2 kg and mB = 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 (mA – mB)g

or (2 – 1)g = g

Total mass being pulled is mA + mB 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., Xcom 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 x1 = 40 m

Therefore, the mass m will fall at a distance x1 = 40 cm from point of projection