Question on finding moment of inertia about the center of mass

[melizabeth]

So I have a question. I (moment of inertia) is basically mr^2 right? And r is supposed to be the distance from the axis of rotation. When the axis of rotation is directly through the center of mass, how is there Icm (moment of inertia about the center of mass). It's confusing to me, because so many problems demand using the Icm when the axis is through the center of mass. Wouldn't Icm be zero? But then they expect you to use the radius of the cylinder. The I of a solid cylinder is given as 1/2(Mr^2). Someone please help me. I think it might just be my lack of understanding.

 

[SteamKing]

It is your lack of understanding. Just because the axis of rotation runs thru the c.o.m., it does not necessarily follow that the moment of inertia is automatically zero. Remember, I = m*r^2, and that r^2 is what is key to understanding why I is not zero.

As an example, let's say that I have four masses m arrayed about the origin:
Mass 1 is located at (1,1)
Mass 2 is located at (-1,1)
Mass 3 is located at (-1,-1)
Mass 4 is located at (1,-1)

Clearly, the c.o.m. is located at (0,0). Each mass is Sqrt(2) units from the origin.

MOI = SUM (mi * ri^2), for 1 = 1 to 4

The MOI = m1 * 2 + m2 * 2 + m3 * 2 + m4 * 2 = 4*m*2 = 8m

 

[haruspex]

The moment of inertia is only mr² when the whole mass is at distance r from the axis, e.g. as would be the case for a small mass element of a larger body. To get the MI of an arbitrary object you need to integrate over these mass elements: ∫r².dm. Since dm and r² are both non-negative, the result never cancels out.