- #1

Eggue

- 13

- 2

- Homework Statement
- A thin, uniform rod is bent into a square of side length a. If the total mass is M, ﬁnd the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem

- Relevant Equations
- Parallel axis theorem, Moment of inertia of a rod about it's end

I tried to find the moment of inertia of 2 rods connected at the corners by adding up their moments of inertia:

[tex] \frac{1}{3}(\frac{M}{4})a^2 + \frac{1}{3}(\frac{M}{4})a^2 = \frac{1}{6}Ma^2 [/tex]

I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel axis theorem

[tex] \frac{1}{6}Ma^2 = I_\text{cm} + \frac{M}{2}(\frac{\sqrt{2}a}{2})^2 [/tex]

However, when i try to solve for I_cm i get a negative value. I was going to multiply I_cm by 2 to get the final answer

I know the correct solution is finding the moment of inertia about the center of mass of each rod and using the parallel axis theorem but I'm wondering why my approach is wrong?

[tex] \frac{1}{3}(\frac{M}{4})a^2 + \frac{1}{3}(\frac{M}{4})a^2 = \frac{1}{6}Ma^2 [/tex]

I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel axis theorem

[tex] \frac{1}{6}Ma^2 = I_\text{cm} + \frac{M}{2}(\frac{\sqrt{2}a}{2})^2 [/tex]

However, when i try to solve for I_cm i get a negative value. I was going to multiply I_cm by 2 to get the final answer

I know the correct solution is finding the moment of inertia about the center of mass of each rod and using the parallel axis theorem but I'm wondering why my approach is wrong?