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mukulss (493)

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Just as we derived the kinematics equations, we will do the same with those of other fields in mechanics. The reason is: a familiarity with calculus and its use in physics is important and necessary.

Center of Mass

Center of mass (also called center of gravity, or just CM) is a concept that is very commonly used in physics. Before we can go into that we need to explain the total momentum of a system because the two are interrelated. The total momentum (P) of a collection of particles is the vector sum of all the individual momenta of the individual particles in the collection:
P = p1 + p2 + p3 + ...
or
MV = m1v1 + m2v2 + m3v3 + ...

Now the next step is that the total mass is simply the sum of all the masses, right?
M = m1 + m2 + m3 + ...

Thus:
V = (m1v1 + m2v2 + m3v3 + ...)/(m1 + m2 + m3 + ...)


Now, CM is defined as the point which is representative of the whole system. As a result, the velocity at which the CM of a system moves is also representative of the momentum of the entire system. In other words, the momentum of the CM of a system is also the momentum of the whole system.

If you take the derivative with respect to time of the equation 4-1 above, you will get the position (displacement) of the center of mass:
S[CM] = (m1s1 + m2s2 + m3s3 + ...)/(m1 + m2 + m3 + ...)


If you break it into component form:
X[CM] = (m1x1 + m2x2 + m3x3 + ...)/(m1 + m2 + m3 + ...)


Y[CM] = (m1y1 + m2y2 + m3y3 + ...)/(m1 + m2 + m3 + ...)



If the total momentum of the system is constant, the velocity of the CM has to be constant as well (the only other variable is mass and we know that it will not spontaneously gain or lose it). What also has to be true in this situation is that there are no net external forces. This is the first case.

The second case is when there is an external force, the CM has an acceleration. If we take the original total momentum equation and integrate it with time, we get:
MA = m1a1 + m2a2 + m3a3 + ...

Wait a minute, aren't mnan forces? So, we proved that the sum of all external forces acts on the system of all the masses just as if all the masses were compressed at one single point, the CM:
summation(F[ext]) = MA



Gravitational Potential

By now, as a physics student, you should know all about Newton's Law of Gravitation and it will be of great help in discussing gravitation potential. We mean potential energy when we speak of potential here. Now we know that gravity is a force and that work is a what a force does over a distance. Now we also know that work is related to potential energy.

Let's say we have a mass m that is at a height of y from the reference plane, which we is at a height 0. The downward force of gravity on the mass is called its weight, which is denoted as FW. The direction of the gravitational force is downward opposite to upward motion, and the work of this force is:
Wgravity = FWy = -mgy

Why is it negative? Negative work? It is actually a way of showing which direction the work is working in. Since the reference plane is 0 and the distance it is away from that plane is a positive y, negative work means the work is pushing it down in the negative direction.

What if the object is not falling all the way to the reference plane? What if you just want to get it from one point (y2) to another point (y1), falling but not reaching the place designated as zero? Then you must find the difference in the work of moving to the two places. In other words:
Wgravity = -FW(y2 - y1) = -(mgy2 - mgy1)

Now you can move it from y2 to y1 anywhere, even below the reference plane, and you will still get a correct answer. This quantity, the product of the weight and the height, is called the gravitational potential energy (Ugravity):
Ugravity = mgy


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