Assume a system consist of a collection of particles, for example the atoms that make up a solid object. Consider a special point, called the center of mass (CM) of the system, whose position is given by
where M is the total mass of the system.
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The sum is over all the particles that make up the system.
The position vector of the center of mass is
.
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Why do we care about this special point called the center of mass?
Let us find its velocity and its acceleration.
The velocity of the center of mass multiplied by the total mass of the system is equal to the total momentum of the system.
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The acceleration of the center of mass multiplied by the total mass of the system is
equal to the total force acting on the system.
Newton's 2nd law, F = ma, when
applied to an extended object, predicts the motion of a particular reference
point for this object. This reference point is the center of mass.
The total force on a particle, Fi, is the vector sum of all internal and external forces acting on the particle. If we sum the forces acting on all the particles of the system, then in this vector sum every internal force that particle 1 exerts on particle 2 is cancelled by the internal force that particle 2 exerts on particle 1. This is a consequence of Newton's third law. We therefore have
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The total force on the system is the vector sum of all the external forces.
But
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The total momentum of the system only changes, if external forces are acting on the system. The center of mass of the system only accelerates, if external forces are acting on the system.
If only internal forces are acting on the particles that make up the system, then the center of mass does not accelerate, its velocity remains constant. The total linear momentum of the system is conserved.
For the total linear momentum of a system to be constant, the particles making up the system can move with respect to each other, and can accelerate with respect to each other, but they must move and accelerate in such a way that
,
.
The center of mass of a system moves as if the total mass of the system were concentrated at this special point. It responds to external forces as if the total mass of the system were concentrated at this point.
How do we experimentally find the center of mass (CM) of of an object?
| Every extended object has
a center of mass. If
near the surface of Earth an object, in any orientation, is
supported at a location directly below its center of mass or
suspended from a location directly above its center of mass, it will
be balanced and it will not start to rotate. Usually, but not always, the center of mass of an object lies within the object itself. For example, the center of mass of a ball is the very middle of the ball, and the center of mass of a book is the middle of the book. If an object has parts that can move with respect to each other, the location of the center of mass depends on the positions of the parts. For example, when a cheerleader lifts her arms, her center of mass moves to a higher position in her body than when her arms are at her side.
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For a homogeneous, symmetrical system, the center of mass always lies on the symmetry
axis.
Consider a square with sides of length a. We can position our coordinate system as shown.
The x and y axes are symmetry axes. We can easily verify that xCM = 0, yCM = 0 from symmetry. The center of mass lies on the intersection of the symmetry axes, at the origin.
 | Four objects are situated along the y-axis as follows: a 2kg object is at
+3m, a 3kg object is at +2.5m, a 2.5kg object is at the origin, and a 4kg
object is at -0.5m. Where is the center of mass of these objects?
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| A uniform carpenter's square has the shape of an L, as shown in the
figure. Locate the center of mass relative to the origin of the
coordinate system.
|
Assume you are 100m away from your car and you want to walk to your car.
You push with
your foot on the ground in a direction opposite to the direction you want to go.
The
ground pushes back and pushes you in the direction you want to go. You accelerate in the
desired direction. (Strictly speaking, some parts of
your body are pushing one of your legs backward, while your leg pushes the rest
of your body into the forward direction. These internal forces are equal
in magnitude and opposite in direction, and they cannot, by themselves,
accelerate your center of mass. But the reaction force from the ground due
to static friction cancels the internal force that is directed into the backward
direction, and the internal force directed into the forward direction now
accelerates you and does work.)
Once you have reached your car, you get in and drive off.
Again it is
the interaction with the ground via frictional forces that makes your motion possible.
If,
however, you are an astronaut in a spacesuit floating untethered 100m from your spaceship,
how can you move over to your ship?
To have a force acting on you, you push on another object. The object then pushes back. But here in space there is nothing to push against. You better carry something with you that you can throw away. As you throw away an object, you exert a force F on it for a time Dt. The momentum of the object changes by an amount Dp = FDt. The object pushes back with a force equal in magnitude and opposite in direction. Your momentum changes by an amount -Dp = -FDt. If you and the object were initially at rest, you now have a velocity v = -FDt/m, where m is your mass. The total momentum of you and the object together has not changed. The total momentum of interacting objects, which are not acted on by outside forces, is always conserved. You and the object have both gained kinetic energy, because the forces you have exerted on each other have done work. They have converted energy that was stored in the system in some other form into kinetic energy.
| An astronaut's manned maneuvering unit (MMU) is a one-man, nitrogen-propelled
backpack. Using rotational and translational hand controllers, the crewmember
can fly with precision in or around the orbiter cargo bay. The MMU propellant
is gaseous nitrogen stored under high pressure. The gas has potential energy.
When an astronaut fires the thrusters, nitrogen gas expands through a nozzle
into low-pressure space. Potential energy is converted into ordered kinetic
energy as the net force on the escaping gas due to the pressure gradient does
work. The gas, which had no net momentum in its compressed state gains momentum
in the direction the nozzle is pointing. The astronaut is gaining the same
amount of momentum in the opposite direction. The gas gains no net momentum in
the direction perpendicular to the direction the nozzle is pointing, even so it
also expands in this direction. There are gas molecules moving in all possible
perpendicular direction, and as we add their momenta vectorially, the
perpendicular components cancel out. The nozzle is designed to minimize the
perpendicular components and maximize the components in the direction the nozzle
is pointing.
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When the astronaut reaches the ship, the astronaut and the ship can accelerate together by firing a rocket. The rocket works on the basis of the same physical principle as the astronaut's jetpack. Gas is ejected and gains momentum in a given direction. The ship gains momentum in the opposite direction. High-pressure gas is obtained by burning fuel. After the rocket has been fired, chemical energy, which was stored in the rocket fuel, has been converted into kinetic energy of the gas and the ship. The total momentum of the gas and the ship stays constant (zero if the ship is initially at rest). The center of mass of the system does not accelerate.
If a rocket takes off from a launch pad on earth, the force of gravity is acting on it. The center of mass of the system (rocket ship + exhaust gases) accelerates towards the ground. The force with which the exhaust gases push against the rocket must be greater than the remaining rocket’s weight if the rocket is to lift off.
For more information on rockets see click the following links:
| Brief History of Rockets |
| Rocket Principles |
| Rocket Principles II |
For a single stage rocket in free space, the change in speed is given by
vf - vi = veln(Mi/Mf),
where ve is the speed of the exhaust gases. The increase in speed is proportional to the exhaust speed and the natural logarithm of the ration Mi/Mf. To reach a high speed, the exhaust speed must be high, and the initial mass of the rocket must include the mass of a large quantity of fuel.
The ultimate speed a rocket can reach is governed by the amount of fuel it can carry and by the speed of its exhaust gases. Because both of these quantities are limited, multistage rockets are used in the exploration of space. A smaller rocket is mounted on the front of a larger one. When the fuel in the first stage (the large rocket) has burned, the empty fuel tank and the motor are ejected. The second stage (the smaller rocket) is already moving rapidly. Since it does not have to carry the motor and the fuel tank of the first stage, it can now reach a much higher final velocity by ejecting its exhaust gases.
Link:
| The motion of a two-stage rocket |
| The first stage of a Saturn V space vehicle consumes fuel at the rate of
1.5´104kg/s, with an exhaust speed of
2.6´103m/s.
(a) Calculate the thrust produced by these engines.
|
When a satellite is launched, big rockets are used to give it its initial kinetic energy. Once it is in space, smaller rockets are used to increase its kinetic energy and change its direction of travel. Assuming that the satellite's initial kinetic energy must equal the sum of its final kinetic energy and the change in its potential energy, we can calculate the initial speed needed to lift it into any desired orbit.
For a satellite to orbit the earth in a circular orbit of radius rf its velocity vf is determined by mvf2/rf = GMm/rf2, since the force of gravity provides the centripetal acceleration. (Here m is the mass of the satellite and M is the mass of the earth.) The satellites kinetic energy therefore is K = (1/2)mvf2 = GMm/2rf. Its potential energy is U = -GMm/rf. The gravitational force is a conservative force, K + U = constant. We therefore have (1/2)mvi2 - GMm/ri = GMm/2rf - GMm/rf = -GMm/2rf. (Here ri is the radius of the earth.) vi2 = 2GM/ri - GM/rf.
The bigger the radius of the desired orbit, the bigger is the required initial kinetic energy. When the radius of the desired orbit goes to infinity, the satellite escapes the gravitational pull of the earth. If rf goes to infinity then (1/2)mvi2 - GMm/ri = 0, vi2 = 2GM/ri. The escape velocity is 11.2 km/s.
