The center of mass (COM) is essentially the "balance point" of an object or a system of particles. It's the location where you can think of all the mass of the system being concentrated when analyzing motion. To put it simply, it's like the average position of all the mass in an object, but weighted by how much mass is at each point.This concept is super important for understanding how objects move, especially when you’re dealing with things like forces, torque, or the conservation of momentum. Whether you’re looking at individual particles, complex shapes, or even a continuous mass, knowing the center of mass helps you predict how the entire system will respond to external forces.
1.0Centre of Mass
The centre of mass is a point where we can concentrate the whole mass of the body and it behaves in a similar manner as a point object behaves under the same circumstances.
OR
The point in a system at which the whole mass of the system may be assumed to be concentrated for all translational effects of the system is called the Centre of Mass (CM).
2.0Properties of Centre of Mass
For symmetrical bodies having uniform distribution of mass, it coincides with the centre of symmetry or geometrical centre.
For a given shape it depends on the distribution of mass within the body and is closer to the massive part.
There may or may not be any mass present physically at the centre of mass and it may be within or outside the body.
3.0Centre of Mass of Discrete Mass System
Position vector of centre of mass of all n-particles
Volumetric mass density (For 3-D Objects): ∫dm=∫ρdV
6.0Uniform Symmetric Bodies
For symmetrical bodies having homogeneous distribution of mass, CM coincides with the centre of symmetry or the geometrical centre.
7.0Centre of Mass of Composite Bodies
Composite bodies are made up of two or more individual bodies. To find the center of mass, each component is treated as a particle with mass equal to the body’s mass, located at its center of mass. The center of mass of the entire body is then calculated using this approach.
md and ms are masses of disc and square xd and xs are x-coordinates of centre of mass of disc and square respectively yd and ys are y-coordinates of centre of mass of disc and square respectively.
8.0Centre of Mass of Truncated Bodies
To find the centre of mass of truncated bodies or bodies with cavities we can make use of superposition principle
Restore the removed portion in its original position to get the complete body.
Add the removed portion back to the truncated body, keeping its position fixed relative to the coordinate frame.
Treat the remaining portion as if it's a new, intact body after the removal.
[Original mass (M) – mass of the removed part (m)] = {original mass (M)} + {– mass of the removed part (m)}
Linear momentum of a system of particles is equal to the product of mass of the system with velocity of its centre of mass. From Newton's second law,Fnet=dtd(MvCM)
10.0Effect of External Force on Centre of Mass
From Newton's second law, Fext=dtd(MvCM)
MaCM=F1+F2+F3+⋯⇒Fnet=MaCM
11.0Conservation of Linear Momentum of System
Fext=dtd(MvCM)=dtd(psystem)
IfFext=0,thenpsystem=constant
This means that the total linear momentum of a system of particles remains constant over time if no external forces act on the system (i.e., the impulse of external forces is zero). The momentum of the system is unaffected by internal forces, and if the net external impulse during a given time interval is zero, the total momentum of the system is conserved during that period. pinitial=pfinal
This is the principle of conservation of momentum. Since force, impulse, and momentum are vectors, the momentum component of a system in a specific direction is conserved if the net impulse of external forces in that direction is zero.
Example :
Block-Bullet System
Conserving momentum of bullet and block, mv+0=(M+m)V
Velocity of block V=M+mmv
By conservation of mechanical energy
21(M+m)V2=(M+m)gh⇒V=2gh
Maximum height gained by block h=2gV2=2g(M+m)m2v2
If bullet emerges out of the block
Conserving momentum
mv+0=mv1+Mv2
m(v−v1)=Mv2
Conserving energy
21Mv22=Mgh⇒v2=2gh
From above equations we get
m(v−v1)=M2gh⇒h=2gM2m2(v−v1)2
12.0Sample Questions on Centre of Mass
Q-1. A body of mass 3m at rest explodes into three identical pieces. Two of the pieces move with speed v in perpendicular directions. What is the total kinetic energy released in the explosion?
Solution:
mv′=2mv⇒v′=2v
K.E.=(21mv2)×2+21m×(2v)2=2mv2
Q-2. A sphere of mass m is rolling down on a wedge of mass M. When sphere reaches at lowest point then find distance moved by wedge in horizontal direction.
Solution: Let distance moved by wedge is 'x'
m(R−x)−Mx=0
⇒mR−mx−Mx=0⇒mR−x(M+m)=0⇒x=M+mmR
Q-3. For a given system of particles find out velocity of COM.
Q-4. A semi-circular disc of radius r is removed from a rectangular plate of side 'r' and as '2r' as shown in figure. Then find out the position of COM of the remaining part.