CENTRE OF MASS

In a system of extended bodies there is one special point that has some interesting and simple properties no matter how complicated the system is. This point is called the center of mass.

For a system of n particles whose position vectors are r⃗ 1,r⃗ 2,..........,r⃗ n as shown in the figure, the position vector for the center of mass r⃗ cm is  defined as

r⃗ cm=m1r⃗ 1+m2r⃗ 2+........+mnr⃗ nm1+m2+...........+mn

or r⃗ cm=∑mir⃗ iM

where M=∑mi is the total mass of the system.
The components of the above equation may be written as

xcm=∑mixiM, ycm=∑miyiM, zcm=∑miziM

The location of the centre of mass is independent of the reference frame used to locate it. The centre of mass of the system of particles depends only on the masses of the particles and the positions of the particles relative to one another.
A rigid body, such as a meter stick, can be thought of as a system of closely packed particles. Hence it also has the centre of mass. The number of particles in the body is so large and their spacing so small, however, that we can treat the body as though it has a continuous distribution of mass. For continuous distribution of mass, the centre of mass is defined as

r⃗ cm=1M∫r⃗ dm

where r⃗  is the position vector of the centre of mass of a small mass element dm.
The components of this equation are

xcm=1M∫xdm, ycm=1M∫ydm and zcm=1M∫zdm

Often we deal with homogeneous objects having a point, a line, or a plane of symmetry. Then the centre of mass will lie at the point, on the line, or in the plane of symmetry. For example, the centre of mass of a homogeneous rod will lie at the centre of the rod, the centre of mass of a homogenous sphere will be at the centre of the sphere, the centre of mass of a cone will be at the axis of the cone, etc.

Application 1

Find the center of mass of a uniform semicircular rod of radius R.

Solution

From the symmetry of the body we see at once that the centre of mass must lie along the y axis, so x_cm = 0.

In this case it is convenient to express the mass element in terms of the angle θ, measured in radians. The element, which subtends an angle dθ at the origin, has a length Rdθ and a mass dm=λRdθ. (l = mass per unit length)
Its y coordinate is y=Rsinθ.

Therefore, ycm=1M∫ydm takes the form
ycm=1M∫0πλR2sinθdθ=λR2M[−cosθ]π0=2λR2M
The total mass of the ring is M=πRλ; therefore, ycm=2Rπ

Motion of the Centre Of Mass

Now we can discuss the physical importance of the centre of mass concept. Consider the motion of a group of particles m₁, m₂, ……., m_n and whose total mass is M which is a constant. From the definition of the centre of mass, we have

Mr⃗ cm  = m_1r⃗ 1 + m_{2r⃗ 2 +………………+} m_nr⃗ n

Differentiating this equation w.r.t. time, we obtain

Mv⃗ cm=m1v⃗ 1+m2v⃗ 2+.........+mnv⃗ n

where v⃗ 1(=dr⃗ 1dt) is the velocity of the first particle, etc., and v⃗ cm(=dr⃗ cmdt) is the velocity of the centre of mass.

Differentiating the above equation w.r.t. time, we obtain

Ma⃗ cm=m1a⃗ 1+m2a⃗ 2+.........+mna⃗ n

where a⃗ 1 is the acceleration of the first particle, etc., and a⃗ cm is the acceleration of the centre of mass.  Now, from Newton’s second law the force F⃗ 1 acting on the first particle is given by F⃗ 1=m1a⃗ 1. Likewise, F⃗ 2=m2a⃗ 2, etc. We can then write the above equation as

Ma⃗ cm=F⃗ 1+F⃗ 2+.........+F⃗ n

Among all these forces the internal forces are exerted by the particles on each other. However, from Newton’s third law, these internal forces will occur in equal and opposite pairs, so that they contribute nothing to the sum. The right hand sum in the above equation then represents the sum of only the external forces acting on all the particles (system). We can then rewrite the above equation as

Ma⃗ cm=∑F→external

This states that the centre of mass of a system of particles moves as though all the mass of the system is concentrated at the centre of mass and all the external forces were applied at that point.
One important situation is that in which ∑F⃗ external=0→ then v⃗ cm = constant.

Application 2

A dog of mass 10 kg stands on a stationary boat of mass 40 kg so that he is 20 m from the shore. He then walks 8 m on the boat towards the shore and halts. How far is he from the shore now? Assume that there is no friction between the boat and water.

Solution

In the horizontal direction, there is no force on the system (dog + boat).  Therefore, the centre of mass of the system does not move in the horizontal direction. Since the dog moves towards the shore, the boat moves away from the shore to keep  centre of mass stationary. Let d be the distance by which the boat moves backwards and let x be the initial distance of the boat from the shore.

The initial x-coordinate of the centre of mass = 10×20+40.x10+40

The final  x-coordinate of the centre of mass
= 10(20−8+d)+40(x+d)10+40

Equating the two, we get
d=85m
The  dog is 20 – 8 + 85 = 13.6 m from the shore.

Application 3

Two balls with masses m₁ = 3 kg and m₂ = 5 kg have initial velocities v₁ = v₂ = 5 m/s in the directions as shown in the figure. They collide at the origin.
(a) Find the velocity of the center of mass 3 s before the collision
(b) Find the position of the center of mass 2 s after the collision

Solution

(a) The given time is of no consequence since v_cm is constant for all times.
From equation,  in component form

vcmx=m1v1x+m2v2xM=(3kg)(−5cos37om/s)+(5kg)(0m/s)8kg=−1.5m/s

vcmy=m1v1y+m2v2yM=(3kg)(−5sin37om/s)+(5kg)(5m/s)8kg=+2m/s

Thus, v⃗ cm= -1.5i^+ 2j^m/s

(b) Since the collision occurs at the origin, the position of the center of mass  2s later is

r⃗ cm=v⃗ cmt = -3i^+ 4j^m

Application 4

A 75 kg man stands at the rear end of a platform of mass 25 kg and length 4m, which moves initially at 4 m/s over a frictionless surface. At t = 0, he walks at 2 m/s relative to the platform and then stops at the front end. During the period of walking, find the displacement of
(a) the platform,
(b) the man,
(c) the center of mass,

Solution

Initially the man, the platform, and the cm have the same velocity, 4m/s. When he begins to walk forward, his increase in momentum must be compensated by a decrease in the platform’s momentum. Let us say that the velocity of the platform relative to the ground while he is walking i.e.

v⃗ PG=vpi^

The velocity of man relative to the ground is then

v⃗ MG=v⃗ MP+v⃗ PG=(2+vp)i^

From the conservation of momentum, we have
(75 + 25) × 4 = 75( 2 + v_p) + 25 v_P
Thus, the velocity of the platform is v_P = 2.5 m/s, and the velocity of the man
is v_m = 4.5 m/s.

Since the velocity of man relative to platform is 2m/s, therefore, it takes  t=4m2m/s=2s for him to walk from the rear to the front.

Displacement of platform is               x_p = v_pt = (2.5)(2) = 5m
Displacement of the man is               x_m = v_mt = (4.5)(2) = 9m
Displacement of the center of mass   x_cm = v_cmt = (4)(2) = 8m