ENERGY AND MOMENTUM: Parallel Axis Theorem/Moment of Inertia of a Rod

The moment of inertia ( I ) is the rotational equivalent of mass possessed by an object. This proclamation, however, comes with a caveat: massive objects are not created equal. Different objects of equal mass have differing abilities to resist changes in rotational motion. Additionally, the location of an object’s axis of rotation influences its inertial characteristics. Consider a system where two equivalent masses are connected by a relatively massless rod:

This system is most easily placed into rotation when its axis of rotation is at the center of mass ( cm ). Each radius ( x ), along with their associated masses ( m ), contribute to the inertia of the system; thus, the moment of inertia that each mass contributes to the system is summed up as follows:

I = mr²

I_total =  ∑mr² = ( m₁r₁ + m₂r₂ )

Let’s assume that m₁ = m₂ = ( 5 kg ), and x₁ = x₂ = ( 4 m ):

I_total = [ ( 5 kg )( 4 m )² + ( 5 kg )( 4 m )² ]

I_total = [ ( 5 kg )( 16 m² ) + ( 5 kg )( 16 m² ) ]

I_total = [ ( 80 kg*m² ) + ( 80 kg*m² ) ]

I_total = 160 kg*m²

Let’s now use the center of ( m₁ ) as our axis of rotation:

I_total = [ ( 5 kg )( 0 m )² + ( 5 kg )( 8 m )² ]

I_total = [ 0 + ( 5 kg )( 64 m² ) ]

I_total = 320 kg*m²

By shifting the axis of rotation away from the center of mass, we increased the system’s ability to resist changes in rotational motion. The parallel axis theorem provides a simpler approach to solving such problems. Fortunately, moment of inertia values are available for a wide variety of objects. The ( I ) value of a uniform-density rod is given below. We are now ready to use the information provided to determine the moment of inertia when the axis of rotation is shifted to the rod’s left end:

In the equation above, ( d ) represents the distance that separates the axis of rotation from the rod’s center of mass, and ( M ) represents the mass of the rod. It is important to note that this formula only works in systems where mass is equally distributed about ( cm ):

I_total = ¹/₁₂ ML² + Md²

I_total = ¹/₁₂ ML² + ( M )( ½ L )²

I_total = ¹/₁₂ ML² + ( M )( ¼  L² )

I_total = ¹/₁₂ ML² + ¼ ML²

I_total = ¹/₁₂ ML² + ³/₁₂ ML²

I = ⁴/₁₂ ML²

I = ¹/₃ ML²