Moment of inertia of two rods (T shape)

[IanAlpace]

Homework Statement

Hey, just wanted to ask a quick question about how to find the moment of inertia of two rods (of equal mass and length) attached together in a T shape (the axis is at the bottom of the vertical rod). My calculations got the correct result but I just wanted to check whether my equation is correct.

Homework Equations

m = 1,00kg
l = 1,00m
I = (1/3)*m*l²+(1/12)*m*l²+m*l²
I = 1,42kgm²

The Attempt at a Solution

This is a part of my homework and the answer given by my equation matches the answer given by my teacher, but I'm a little confused about the equation. I'm mostly curious about why the equation contains three masses, when the problem only contains two objects with mass. The initial equation that I came up with (I = (1/3)*m*l²+(1/12)*m*l²) got the wrong answer which left me a bit confused.

Thanks in advance.

 

[kuruman]

Hi IanAlpace and welcome to PF.

The moment of inertia of the vertical part of the "T" is $\frac{1}{3}ml^2$. The other two terms are the moment of inertia of the horizontal part according to the parallel axis theorem.

 

[IanAlpace]

To anyone who may stumble upon this thread..
I figured out my equation is indeed correct for this specific problem. Because there is two rods with the same mass and length, in order to find the correct equation and answer I used Steiner's parallel axis theorem, according to which:

Moment of inertia of the T shaped contraption is:
inertia of vertical rod + inertia of horizontal rod + mass*distance²
(distance is the distance between the two axis of the two rods, in this case distance is equal to l)

I = (1/3)ml²+(1/12)ml²+ml²