CenterOfMass

# CenterOfMass - Center of Mass Shina Tan The center of mass...

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Center of Mass Shina Tan The center of mass is a vividly intuitive concept in everyday life. If you throw a pen into the air and watch how it rotates and falls, you will discover that its “center” apparently goes through a parabola trajectory, but because of the rotation, all other points on the pen wiggle through the air in some complicated trajectories. For any object thrown into the air, if the air’s resistance and all exter- nal forces other than gravity can be neglected, the center of mass goes through a parabola trajectory or vertical line, but the other parts may have complicated trajectories because of the object’s rotation or shape- changing movements. This fact can be proven mathematically. For simplicity let us first consider two point masses m 1 and m 2 linked by a rigid rod with negligible mass. Let the force exterted on m 1 by m 2 be f 12 ( t ), and the one exterted on m 2 by m 1 be f 21 ( t ). These forces depend on time t , but they satisfy Newton’s third law at all times: f 12 ( t ) + f 21 ( t ) 0 . Each point mass satisfies Newton’s second law: d 2 dt 2 m 1 r 1 ( t ) = m 1 g + f 12 ( t ) , d 2 dt 2 m 2 r 2 ( t ) = m 2 g + f 21 ( t ) . Adding these two equations, and using Newton’s third law, we get d 2 dt 2 m 1 r 1 ( t ) + m 2 r 2 ( t ) = ( m 1 + m 2 ) g , or d 2 dt 2 R ( t ) = g , (1) where R ( t ) = m 1 r 1 ( t ) + m 2 r 2 ( t ) m 1 + m 2 is the center of mass. Equation (1) has a simple solution: R ( t ) = R 0 + v 0 t + 1 2 g t 2 . If the initial velocity v 0 is not vertical, the trajectory of the center of mass is a parabola. If it is vertical, the trajectory of the C.O.M. is a vertical line. 1

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For a many-particle system, such as an athlete doing a high jump, or the pen that you throw into the air, the proof is essentially the same. Suppose that the system is composed of n point-like particles with masses m α (1 α n ). Each point mass obeys Newton’s second law: d 2 dt 2 m α r α ( t ) = m α g + X 1 β n ; β 6 = α f αβ ( t ) . In the above you have simultaneously written n equations (!) with α ranging from 1 through n . Note that n can be as large as you want. It may be of the order of the Avogadro’s constant! Adding all these equations together, you can check that all the internal forces f αβ ( t ) are cancelled with each other because of Newton’s third law. (If you are not sure, take n = 3 or 4 as an example.) So we are left with a simple equation: d 2 dt 2 n X α =1 m α r α ( t ) = M g , where M = n α =1 m α is the system’s total mass. Dividing both sides by M , and noting that M does not change with time, we get d 2 dt 2 R ( t ) = g . So we have indeed explained why the “center” of the pen does not wiggle through the air. And we have found a precise definition of this center: R ( t ) = 1 M X α m α r α ( t ) , (2) which is nothing but a weighted average of the positions of all the parts of the system.
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