Question

In lecture we found that the center of mass of a rod with a non-uniform density profile (where the density changes linearly from one end to the other, doubling in density from one end to the next) was 5/9L when measured from the less-dense end. We also found that the rotational inertia around the less-dense end came out to be 7/18ML2. Use these facts and the parallel-axis theorem to find the rotational inertia of this rod around its center of mass and around its other (more-dense) end.

Answer 1

Rotational inertia at the less-dense end is

I 7 -ML 18

Center of mass of rod is at distance $R_{l}=\frac{5}{9}L$ from the less-dense end.

Center of mass of rod is at distance from more-dense end

4 Rm = L - L ==L 9 9

Moment of inertia at the less-dense end can be written as

I1 = Icm + MR

$I_{cm}=I_l-MR_l^2$

$I_{cm}=\frac{7}{18}ML^2-M\left ( \frac{5}{9} L\right )^2=0.08\,ML^2$

Moment of inertia at the more-dense end can be written as $I_m=I_{cm}+MR_m^2$

$I_m=I_l-MR_l^2+MR_m^2$

$I_m=\frac{7}{18}ML^2-M\left ( \frac{5}{9}L \right )^2+M\left ( \frac{4}{9}L \right )^2=\frac{5}{18}ML^2$