Center of Mass and Moment of Inertia of Extended Objects

Question

Consider a uniform rod with length L and mass M placed on the x-axis between the points

(-1/3*L, 0) and (2/3*L, 0).

Determine the moment of inertia of the rod for rotation about the y-axis by integration.

Answer 1

Hi Megan!

 

Following the comment below about integration, we can go straight to the definition of moment of inertia as an integral:

 

I = ∫r²dm

 

With this thin rod, we effectively have one dimension to worry about, so we can define a differential unit of mass in terms of a differential unit of length along the x axis, as:

 

dm = λdx

 

where λ = linear mass density; the mass per unit length

 

In our case, it is nice because the rod is uniform, which means λ is constant and equal to M/L.

 

Also, since we are confined to one axis, our r² position vector becomes just x².

 

Thus, the moment of inertia integral becomes:

 

I = ∫x²(λdx) = λ∫x²dx = (M/L)∫x²dx

 

This has to be integrated from x = -(1/3)L to x = (2/3)L.

 

See if that helps.  If you have any more questions about this, or where to go from here, or you would like to check an answer, just let me know.