PhysicsLAB Resource Lesson
Thin Rods: Center of Mass

Let's derive the equation needed to calculate the center of mass of a uniform, continuous rod. Remember that the equations used to calculate the center of mass of a collection of discreet masses were
 
            and            
 
The center of mass in the vertical dimension, ycm, would be somewhere along the rod's central axis and will be included at the end of the derivation. For now, we will focus on calculating the position of the rod's horizontal center of mass, xcm.
 
Since the rod is a continuous mass, we must add up all of the small mass segments, Δmi, located at a distance xi from the left hand side to calculate its center of mass.
 
When we take the limit of this summation as Δm approaches zero we obtain a general expression for the center of mass.
 
 
But we can not integrate x with respect to dm, so we must first develop a relationship between dm and x. This is done with a quantity called λ, or the mass per unit length .  For a uniform rod, λ would equal a constant value. For example, a rod might have a mass per unit length of λ = 0.4 kg/m.
 
 
Substituting λdx for dm now allows us to integrate to calculate the rod's center of mass.
 
 
Note that the expression λL equals the mass of the rod, M; that is, when we multiple the mass per unit length by the length of the rod we obtain the rod's entire mass. 
 
This result, that the center of mass which was painstakingly obtained would have probably been your first logical guess. The center of mass of a uniform rod is at the center of the rod. But what if the mass per unit length, λ, was not uniform, but also a function of x? Then we would need the process outlined above to obtain our answer. Let's look at two more complicated examples.
 
 Where would the center of mass be located for a continuous body that has a mass per unit length distribution of λ = Ax (this rod would look like a rough "baseball bat")?

 Where would the center of mass be located for a continuous body that has a mass per unit length distribution of λ = Ax2 (this rod would look like a rough "trumpet bell")?





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