In this article, we will approach the concepts of Center of Mass and collisions in a new perspective. These concepts are very important for Board exams as well as the competitive exams.
Center of Mass
The concepts covered in Motion in 1-D, Motion in 2-D, Laws of motions dealt with point bodies, i.e bodies whose dimensions had little to do as far as the distances asssociated with their motion are concerned. But in real world things are a little different. Mass is distributed over space rather than concentrating on a single point. Also they can have complicated geometries. This is where the concept of Center of mass comes into picture.
Now, if the entire mass can be represented as a point mass at a single location, we can conveniently solve such problems. And this single point where the entire mass is located is called as the center of mass.
If we can represent a system of particles/ an object whose mass is distributed over space, by a single point where the whole mass is concentrated, that point can be called as the Center of Mass. This point can Center of Masspletely represent the object and how it will behave to different forces.
Center of Mass is more like finding the average, and the expression is rather simple.
Assume a system of n particles, each is located in space with co-ordinates (xi, yi, zi) and mass mi where i∈[0,n-1].
Then the position of Center of Mass, (xm, ym,zm) can be found using:
xm=∑(ximi) / ∑(mi),
ym=∑(yimi) / ∑(mi),
zm=∑(zimi) / ∑(mi),
where the limits of summation is i= 0 to n-1.
A continuous body can be imagined as a system of infinitesimally small particles. So the number of particles tends to infinity, and the summation is better expressed as an integral.
Find the Center of Mass of a semicircular lamina of negligible thickness, radius R and mass M distributed uniformly over the area.
We know that Center of Mass is more equivalent to average, and this shape is symmetrical about the x axis. So by simple logic we can see that the x co-ordinate of Center of Mass is 0. Also, this is a 2-D shape and zm is of no significance.
To find ym , consider a narrow element of thickness dr at a distance r as shown. The mass of the portion would be
2*M*r dr/R2. (area of the small element*total mass/total area).
So the lamina is made up of infinite number of elements like this, and thus, the integral would be:
ym = (∫r * 2*M*r dr/R2 )/M , where the limit of integration is from 0 to R.
which on solving gives ym= 4R/3π.
If the Center of Mass of an irregular shape is to be found, like a portion cut off from a circle, just imagine that the cutoff portion and the remaining portion together made a proper shape, whose Center of Mass is known to us. Then substitute back to get the answer, like the following question. Again, no direct formula is helpful for such questions.
Concept of Internal forces and Motion of Center of Mass
Concept of Center of mass helps us to neglect all the internal forces too. Consider the case where a missile is launched as a projectile and it explodes mid air. This explosion can be considered as an internal force. Thus, the system doesn’t experience any external forces as such. So we can safely assume that the path of the Center of Mass is not changed- it still follows the same parabolic path. This is helpful because we can now avoid a lot of internal forces while calculation.
Conservation of Linear Momentum
It states that the net momentum in any direction in any isolated system (net external force=0) remains constant.
- Recoil of a gun
- Man jumping out of a boat
- Collision of two or more objects(to be discussed in detail)
- Rocket Propulsion
For clarity, collision between two objects is discussed for the following contexts:
- Elastic collision in single dimension
- Elastic collision in 2D
- Inelastic collision and coefficient of restitution
Elastic collision in single dimension
Principle assumption: The materials are elastic.
Elastic materials are those materials which doesn’t deform during a collision, ie, they have a tendency to regain their shape. This is more of an ideal situation and we can safely assume that momentum and energy are conserved.
The underlying principles to be kept in mind while approaching any collision problem are conservation of energy and conservation of momentum. Let’s consider the following scenario:
2 spherical bodies travelling along x axis of masses m1 and m2, and initial velocities u1,u2 are collided and after collision their velocities are v1 and v2.
So according to principle of conservation of momentum, initial momentum = final momentum.
∴ m1u1 + m2u2 = m1v1 + m2v2.
and total Kinetic energy in conserved.
m1u21 + m2u22 = m1v21 + m2v22. (eliminating the common term, 1/2).
Solving these two equations and substituting we get two interesting results.
- u2 – u1 = v1 – v2 : This means the velocity of approach and velocity of separation are the same in case of an elastic collision.
- The final expression of v1 in terms of u1, u2, m1 and m2:
v1 = ((m1-m2)u1 + 2m2u2)/(m1+m2)
v2 can be found by interchaging 1 and 2 in the above equation.
Here, a lot of different cases could be derived, like a light object with high velocity hitting a heavy body at rest etc., where values of v1,v2,m1,m2 could be assumed 0 or infinity and expression could be simplified.
Elastic collision in 2D
Principle of conservation of momentum states that Net Momentum along a line is conserved in case of an isolated system. So the approach is to resolve the initial velocities into x and y axes and solve like two separate collisions in 1D problems. Due to large possibilities in setting the question and bigger number of variables involved, it is suggested to understand the method of approaching this kind of problems rather than mugging up the result.
Inelastic collision and Coefficient of restitution
We have seen that velocity of approach = velocity of separation in case of an elastic collision. In case of an inelastic collision, it is:
velocity of separation = e*(velocity of approach),
where e = coefficient of restitution, and it is a value between 0 and 1.
e = 1 implies the collision is perfectly elastic and no energy is lost.
e = 0 implies perfectly inelastic, and the two bodies stick together after collision.
Here, some deformation happens to the objects and the principle of conservation of energy does not hold good. Still, principle of conservation of momentum is valid here.