# Kishan K Sinha

Labor omnia vincit improbus.

## Friday, November 10, 2017

## Tuesday, June 20, 2017

### Concept of Center of Mass : Discrete Distribution of Mass

Let us consider a rigid body in motion.
By definition, a rigid body consists of a number of particles separated by
constant distances and cannot be distorted by application of any amount of
force. In order to completely determine the state of motion of such a system
(object), we need first to specify the motion of all the constituting particles
of the system. Let this rigid body be composed of

*N*number of particles of masses*m*_{1},m_{2}*, .....,m*with position vectors_{N}**r**,_{1}**r**, .....,_{2}**r**, respectively with respect to the origin_{N}*O*of an arbitrary co-ordinate system. If we want to find out the total momentum of this object, we will have to add the momenta of all its constituent particles. Mathematically this can be done in the following way :
Similarly if we want to find out the
total kinetic energy of the object, we will have to add the individual kinetic
energies of all the constituent particles :

Taking into consideration the large
number of particles involved into making a classical object, specifying motion
of each and every particle is obviously a hopeless task. Thus before attempting
to analyze the motion of such a multi-particle object, we first need to devise
a technique to simplify the problem. This is where the importance of choice of
co-ordinate system (and reference frame) in physics comes into picture.

Let us consider a co-ordinate system in
which the following condition holds :

..... (a)

where

**r’**,_{1}**r’**, .....,_{2}**r’**are the position vectors of masses_{N}*m*respectively with respect to the origin_{1},m_{2}, .....,m_{N}*O’*of the new co-ordinate system. Thus if we add the products of masses and corresponding position vectors of all the particles of the object in this co-ordinate system, we get zero. Choice of such a co-ordinate system tremendously simplifies the problem. Let us see how.
Let

**R**be the position vector of*O’*with respect to the origin*O*of the old co-ordinate system as shown in the figure. Then by triangle law of addition of vectors, position vector of mass*m*in old co-ordinate system_{i}**r**, position vector of mass_{i}*m*in the new co-ordinate system_{i}**r’**and_{i}**R**are related by the following equation :If we differentiate the above equation with respect to time we get relation for velocities of the particle in the two co-ordinate systems :

The total momentum of the object in the
two co-ordinate systems can now be written in the following manner :

If we differentiate (a) with respect to
time we get the following :

Using this in (b) we get an astonishing
result :

In words, we now don’t have to worry
about the momentum of every constituent particle of the object, we just need to
know the total mass of object and the velocity of the origin of our new
co-ordinate system with respect to the old one. Thus the new co-ordinate system
has considerably simplified our problem of finding the total momentum of an
object consisting of a large number of particles. It is as if we are observing an point particle of mass

*M*and velocity**V**. Such a co-ordinate system (actually the reference frame) is called as*center of mass co-ordinate system*(reference frame) and the origin is known as*the center of mass (or the barycenter) of the object*and it is unique for an object.
Just
as we did with the velocities in equation (b) we can do the same for position
vectors of the particles :

Or

This gives the position of the center of
mass (origin of the new co-ordinate system) in terms of the position vectors of
particles in the old co-ordinated system.

Now let us see how the expression for
the total kinetic energy of object turns out in the new co-ordinate system.
Substituting for

**v**in the expression for the total kinetic energy we get :_{i}
Here the first term contains

*v’*(velocities of particles w.r.t. the new coordinate system). Since we are considering a rigid body case, the distances between all the particles in the body remains constant. As a result of this the distances of particles from_{i}^{2}*O’*must also remain fixed. This implies that the only motion these particles can perform with respect to*O’*is rotation about*O’*. Thus, if*ω*is the rotational angular velocity of the body then,
Substituting this in the expression we
obtained for the total kinetic energy we get :

The term in the bracket is known as the

*moment of inertia*of the rigid body and is denoted by the letter*I*. Thus finally we have :
A
very simplified equation for the total
kinetic energy of the body taking into account both the rotational as
well as
the translational motion. Thus observing the body with reference to the
center of mass greatly simplifies the problem mathematically.

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