A General Question about Inertia

[student34]

Homework Statement

My textbook says that I = (1/3)ML^2 for a slender uniform rod, and I = (1/2)MR^2 for a uniform solid cylinder.

So let's say that there is cylinder that is a thin disk with M = 5kg, and R = 7m. And let's say that a thin rod also is 5kg and has a length of 7m.

Why don't both objects have the same inertia?

Homework Equations

Inertia for a thin uniform rod is I = (1/3)ML^2.

Inertia for a thin uniform cylinder is I = (1/2)MR^2.

The Attempt at a Solution

My textbook says that some point mass circling an axis has the same inertia as a thin ring of the same mass (if this isn't true, then my issue in this thread is irrelevant).

So I will divide the mass of the rod into infinitesimally small masses dm. I should be able to do the same for the cylinder with each infinitely thin ring equaling dm. Apparently, a ring can be thought of as a point mass. Then the disk can be thought of as a thin rod with the same mass and the same length as the other. But rods and cylinders have different formulas.

 

[lightgrav]

a wedge is a portion of a cylinder, so they'll have the same rotational inertia "fraction".
the uniform rod has each piece having the same mass (dm; half its mass on the outside half).
But the cylinder has most of its mass on the outside, where the circumference is longer,
so the pieces have dm proportional to (r/R).

 

[student34]

Oh wow, thanks a lot!