Thus far, physicists have not developed a concise definition of what constitutes mass and “ free space “. As a consequence, an overly simplistic definition of mass, albeit imperfect, may be used with convenience in laboratory settings. Mass, simply put, occupies free space.
Relatively simple analyses of forces acting upon massive objects can be made using free-body diagrams. Near the earth’s surface, Fw= mg quantifies the magnitude of the gravitational force of attraction acting upon an object. The “ w “ subscript clarifies that an object’s weight is being measured, m = mass in kilograms ( kg ), and g = the gravitational constant of acceleration ( 9.8 m/s2 ). When two masses ( m1 = m2 ) are positioned equal distances from a central pivot point ( x1 = x2 ), the mass of the system ( M = m1 + m2 ) can be regarded as being located at the central pivot point, which is the center of mass:
Recall that torque ( 𝛕 ) is the product of a force being applied to a lever ( r ) that is situated some distance away from an axis of rotation. Torque ( 𝛕 ) = Frsin θ, and the magnitude of a torque is maximized when the force is applied to a lever at a 900 angle. Since the sin 900 = 1, 𝛕 = Fwr = ( mg )( x ). The sum of the torques acting upon a system must be zero if the system’s components are in static equilibrium with one another. If we designate the clockwise direction of the system above as being negative, and we designate the counterclockwise direction as positive, F1 + ( – F2 ) = 0. Furthermore, F1 = F2, and ( m1g )( x1 ) = ( m2g )( x2 ). If we doubled the mass of m2 without doubling the mass of m1, this would necessitate doubling the length of the x1 lever to maintain balance. These summations will hold true as long as the vantage point of an observer within the system above is the pivot point. It is important to note, however, that the vantage point within a system can be located in any location of convenience. Let’s evaluate a circumstance in which an observer is located at some arbitrary distance to the left of m1:
In order for static equilibrium to exist in the system above, M = m1 + m2. The x3 lever, however, cannot be equal in length to x1 or x2. Since the center of mass on the right-hand side of the system lies between m1 and m2, the mass M on the left-hand side of the system must be located some distance x3 that is equal to the distance from the pivot point to the center of mass on the right-hand side of the system. Since the system is in static equilibrium, we may once again use the rules regarding torques to determine exactly where between m1 and m2 the center of mass is located. Notice, however, that F1 and F2 now exert negative torques on the system:
F3 + ( -F1 ) + ( -F2 ) = 0
F3 = F1 + F2
( Mg )( x3 ) = ( m1g )( x1 ) + ( m2g )( 2x1 )
Since m1 = m2, and x1 = x2,
( 2m )( g )( x3 ) = ( mg )( x + 2x )
The gravitational constant of acceleration may now be factored out of the equation:
( 2m )( x3 ) = ( m )( 3x )
x3 = ( 3/2 )( x ) = 1.5x
Thus, the center of mass, as predicted, is located between m1 and m2.
In the previous example, the equal mass values and lever lengths enabled us to conveniently simplify the system of torques via factorization. When a system’s masses and associated levers are more random in nature, a more general equation form suffices:
( m1 + m2 )( g )( x3 ) = ( m1g )( x1 ) + ( m2g )( x2 )
Note: Since x1 and x2 are no longer equal in value, it is important to recognize that x2 now represents the distance from the pivot to x2 ( as opposed to 2x1 ):
( m1 + m2 )( g )( x3 ) = ( g )( m1x1 + m2x2 )
( m1 + m2 )( x3 ) = ( m1x1 + m2x2 )
x3 = [ ( m1x1 + m2x2 ) / ( m1 + m2 ) ] = [ ( m1x1 + m2x2 ) / M ]
Although the gravitational constant of acceleration ( g ) was used to derive the center of mass equation, the absence of ( g ) in the final equation will allow us to evaluate elastic and inelastic collisions using the center of mass as a moving point of reference from which colliding particles are observed.
RENAISSANCE PHYSICS 