Given a cylinder of mass M, radius R, length L and quadratically increasing density rho=ks^2, where s is the distance from the symmetry axis, we calculate the moment of inertia of a cylinder about its symmetry axis.
We begin with a reminder of the formula for the moment of inertia of a thin shell: I=MR^2, because we plan to split our cylinder into nested thin cylindrical shells.
Next, we get a reminder of the definition of density: rho=M/V, which can be turned around to say M=rho*V, but only if rho has a single well defined value over an entire volume.
This is part of the motivation for splitting into thin cylindrical shells: using a thin shell mass increment of dm, the entire volume of the shell wall is located at a single value of density ks^2, so we can express dm as rho*dV, where dV is the volume contained in the wall of the thin cylindrical shell.
Once we express dm, we can plug into the moment of inertia formula for a thin cylindrical shell and write dI=dm*s^2. We sub in our expression for dm and find an expression for the moment of inertia contribution dI in terms of s. Integrating this expression, we arrive at the moment of inertia of the cylinder.
We still want to express our moment of inertia as a multiple of MR^2, so we proceed to compute a mass integral - just the integral of dm. This gives us an expression for the total mass of the cylinder in terms of k, so we can solve for k in terms of M and sub into our moment of inertia formula.
Making this final substitution and simplifying, we arrive at a formula for the moment of inertia of a cylinder with non-uniform density function ks^2: I=2/3MR^2!
28 сен 2024
