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Interactions between parts of a system transfer momentum between the parts, but do not change the total momentum of the system. We can define apoint called the center of mass that serves as an average location of a system of parts.
The center of mass need not necessarily be at a location that is either in or on one of the parts. For example, the center of mass of a pair of heavy rods connected at oneend so as to form a "V" shape is somewhere in space between the two rods.
Having determined the center of mass for a system, we can treat the mass of the system as if it were all concentrated at the center of mass.
For a system composed of two masses, the center of mass lies somewhere on a line between the two masses. The center of mass is a weighted average of the positions of the twomasses.
Facts worth remembering -- Center of mass for two objects
For a pair of masses located at two points along the x-axis, we can write
xcm = (m1*x1/M) + (m2*x2/M)
where
Multiple masses in three dimensions
When we have multiple masses in three dimensions, the definition of the center of mass is somewhat more complicated.
Facts worth remembering -- Center of mass for many objects
Vector form:
rcm = sum over all i(mi*ri / M)
Component form:
xcm = sum over all i(mi*xi / M)
ycm = sum over all i(mi*yi / M)
zcm = sum over all i(mi*zi / M)
where
It can be shown that in an isolated system, the center of mass must move with constant velocity regardless of the motions of the individual particles.
It can be shown that in a non-isolated system, if a net external force acts on a system, the center of mass does not movewith constant velocity. Instead, it moves as if all the mass were concentrated there into a fictitious point particle with all the external forces acting on that point.
This section contains explanations and computations involving momentum, impulse, action and reaction, andthe conservation of momentum.
This section contains several examples involving momentum
Use the Google calculator to compute the momentum of a 70-kg sprinter running 30 m/s at 0 degrees.
Answer: 2100 kg*m/s at 0 degrees
Use the Google calculator to compute the momentum in kg*m/s of a 2205-lb truck traveling 33.6 miles per hour at 0 degrees when the changes listed belowoccur:
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