# What is the moment of rotational inertia about the axis indicated?

• vertex78
So in this case, the moment of inertia of the rectangle about the axis indicated is the sum of the moments of inertia of the thin wire sides about that axis. In summary, the moment of inertia of a rectangle with uniform thin wire sides about the axis indicated is the sum of the moments of inertia of the side with the largest length (side b) and the side with the smallest length (side a).

## Homework Statement

The rectangle shown has a mass, m. The sides are made of uniform thin wire. The short side is of length a, and the long side is of length b. If m = 5.50 g, a = 0.05 cm and b = 0.1 cm, what is the moment of rotational inertia about the axis indicated?

(ignore the x's)
axis
|
|-----------------|
|xxxxxxxxxxxxxxxxx| side a = 0.05cm
|xxxxxxxxxxxxxxxxx|
|-----------------|
| side b = .1 cm

## Homework Equations

$$I = \frac{1}{3}mL^2$$

## The Attempt at a Solution

$$I = \frac{1}{3}(0.0055kg*.001^2) = 1.833x10^{-9}$$

Can't I treat this as a high density rod along the bottom of the rectangle since the height of the particles do not matter?

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You are ignoring the contribution by side a.

D H said:
You are ignoring the contribution by side a.

it seems like the height of the rectangle is irrelevant since each particle at a particular length will have the same rotational inertia regardless of the height. Am I missing something here?

vertex78 said:
it seems like the height of the rectangle is irrelevant since each particle at a particular length will have the same rotational inertia regardless of the height. Am I missing something here?

yeah, precisely, each mass element on this "a"-rod will contribute the same amount... BUT this contribution should not be zero... as you have assumed so far.
also, judging by the wording of the question, the rectangle is of mass m, not each individual sides! Hence, uniform distribution of mass means that each side only get a portion of this m... and obvious the side which lie on the axis of rotation does not contribute.. and hence effectively you are rotating something that is of total mass less than m...

mjsd said:
yeah, precisely, each mass element on this "a"-rod will contribute the same amount... BUT this contribution should not be zero... as you have assumed so far.

I don't really understand how I have assumed the contribution is zero. I know this must be from my lack of understanding this correctly but I keep thinking that since I am using the full weight of the object and the length of side b then I would taking every particle in account. Can you elaborate on this a little more for me?

also, judging by the wording of the question, the rectangle is of mass m, not each individual sides! Hence, uniform distribution of mass means that each side only get a portion of this m... and obvious the side which lie on the axis of rotation does not contribute.. and hence effectively you are rotating something that is of total mass less than m...

Again my thinking was to put all the mass on one side so I would be thinking of it as a really dense rod of length side b, instead of a rectangle. Of course I could not get the correct answer thinking of it in this way

Could you give a little more explanation that can help me a little more towards understanding this so I can get a correct answer?

Look up the parallel-axis theorem. See if that helps.

vertex78 said:
it seems like the height of the rectangle is irrelevant since each particle at a particular length will have the same rotational inertia regardless of the height. Am I missing something here?

Yes, you are missing something here. Side "a" has some mass.

The moment of inertia is an additive quantity. In other words, if you can split some object into components, the moment of inertia of the object as a whole about some axis is the some of the moments of inertia of each of the components about that same axis.