Inertia is a measure of a system’s ability to resist a change in motion, and it is directly proportional to a system’s massiveness. Such a system or object could be stationary with respect to an observer, or it could move with a constant velocity. When a system moves with constant velocity with respect to an observer, its inertia is determined by its momentum ( p ). The momentum possessed by an object is the product of its mass ( m ) and velocity ( v ):
p = mv
It is important to realize that inertia and momentum, although very similar, are not always synonymous. A system’s rotational inertia depends on its geometric mass distribution about an axis of rotation. Neglecting wind resistance, a cylinder and sphere of equal mass will roll downhill at different rates. A figure skater can increase her angular momentum by pulling her arms inward by some radial distance ( r ). Within rotating systems, the moment of inertia ( I ) is the rotational equivalent of mass:
I = mr2
Although mass and geometry influence the inertia of rotating systems, the momentum of any object or system, regardless of shape, can only be changed with an applied force. Newton’s First Law of Motion states that an object ( or system ) will move with a constant velocity ( or remain at rest ) indefinitely until it is acted upon by an unbalanced force. Under the influence of an unbalanced force, an object’s change in velocity ( Δv/Δt ) is referred to as acceleration ( a ):
Fnet = Δp/Δt = ( m )( Δv/Δt ) = ma
With an unambiguous definition of inertia, we are now ready to examine how a system’s ability to resist changes in rotational motion depends upon the axis around which the system’s mass revolves. Let’s first determine the moment of inertia of a system where two equal masses extend equal distances outward from the system’s axis of rotation:
Within a two-mass system, the moment of inertia of the system is the sum of each mass’ individual moment of inertia:
Itotal = ∑mr2 = ( m1r21 + m2r22 )
Let’s assume that m1 and m2 are 5 kg each, and each mass is located 4 meters from the axis of rotation:
Itotal = [ ( 5 kg )( 4 m )2 + ( 5 kg )( 4 m )2 ]
Itotal = [ ( 5 kg )( 16 m2 ) + ( 5 kg )( 16 m2 ) ]
Itotal = 80 kg*m2 + 80 kg*m2 = 160 kg*m2
Let’s now determine the system’s moment of inertia when we choose the center of m1 as our axis of rotation:
Itotal = [ ( 5 kg )( 0 m )2 + ( 5 kg )( 8 m )2 ]
Itotal = 0 + ( 5 kg )( 64 m2 ) = 320 kg*m2
Within this two-mass system, rotation is more difficult when the axis of rotation differs from the geometric center of mass. When systems have irregular shapes and/or different masses situated at different distances from the axis of rotation, calculating a moment of inertia is considerably more difficult.
Notice how the second ( Itotal ) value is exactly twice that of the first. When the chosen axis of rotation was the center of mass, the individual ( I ) values associated with m1r12 to m2r22 were added together; however, a more simplistic mathematical approach is more practical when evaluating complicated systems. If the moment of inertia about a system’s center of mass is known ( Icm ), we can determine the moment of inertia of any location relative to the center of mass. Let’s once again determine ( Itotal ) when the axis of rotation is at the center of m1:
Itotal = Icm + Md2
In the equation above, ( d ) represents the distance separating the axis of rotation from the system’s center of mass. It is important to note that this formula only works in systems where mass is distributed symmetrically about ( cm ):
Itotal = 160 kg*m2 + ( 10 kg )( 4 m )2
Itotal = 160 kg*m2 + ( 10 kg )( 16 m2 )
Itotal = 160 kg*m2 + 160 kg*m2
Itotal = 320 kg*m2
Let’s now use the parallel axis theorem to determine the moment of inertia of a system that is more complicated:
Icm = ∑mr2 = ( 4 )( 5 kg )( 16 m2 ) = 320 kg*m2
Let’s now move the axis of rotation 10 meters to the left:
The leftmost masses are 6 meters from the new axis of rotation:
( 2 )( 5 kg )( 36 m2 ) = 360 kg*m2
The masses to the right are 14 meters from the new axis of rotation:
( 2 )( 5 kg )( 196 m2 ) = 1,960 kg*m2
Itotal = ∑mr2 = 360 kg*m2 + 1,960 kg*m2 = 2,320 kg*m2
Let’s now use Itotal = Icm + Md2 to determine the system’s moment of inertia:
Itotal = 320 kg*m2 + ( 20 kg )( 10 m)2
Itotal = 320 kg*m2 + ( 20 kg )( 100 m2 )
Itotal = 320 kg*m2 + 2,000 kg*m2
Itotal = 2,320 kg*m2
The parallel axis theorem has wide applications involving the analysis of collisions between rigid bodies of uniform mass. If, for example, a baseball bat is modeled as a rod of uniform mass, we can use the parallel axis theorem to determine what happens to the system after the collision has ensued.
RENAISSANCE PHYSICS 