What happens to the moment of inertia when a rod is folded?

[OrlandoLewis]

Homework Statement

Consider one rod of length L which is spun at its center perpendicular to its length; it will have a certain moment of inertia. Now, if the same rod is folded in the middle creating a certain angle and still spun at its center, what happens to its moment of inertia?

Homework Equations

Thin rod about axis through center perpendicular to length
$$I = \frac{1}{12}ML^2$$
Thin rod about one edge
$$I = \frac{1}{3}ML^2$$

The Attempt at a Solution

For a rod that is not folded its moment of inertia will be:
$$\frac{1}{12}ML^2$$
For a folded one, I tried using L to be the length divided by two so:
$$I = 2(\frac{1}{3}\frac{M}{2}(\frac{L}{2})^2) = \frac{1}{12}ML^2$$

So the moment of inertia will be the same.

 

[kuruman]

o its moment of inertia will increase by half making the folded rod harder to spin. Is that correct?

You mean increase to twice of its former value according to your equation. Think about your answer again.

On edit: Also think about TomHart's suggestion.

 

[TomHart]

For a folded one, I tried using L to be the length divided by two so

What if you calculated the inertia of the one long rod. Then you cut it in half but didn't change the angle between the halves. What would you expect to happen to the moment of inertia?

 

[OrlandoLewis]

What if you calculated the inertia of the one long rod. Then you cut it in half but didn't change the angle between the halves. What would you expect to happen to the moment of inertia?

Then it should remain the same right?

 

[OrlandoLewis]

You mean increase to twice of its former value according to your equation. Think about your answer again.

On edit: Also think about TomHart's suggestion.

Yes, I meant decrease by the former value.

 

[kuruman]

Then it should remain the same right?

Yes, because in your solution you doubled the moment of inertia of two half rods of mass M instead of doubling the moment of inertia of two half rods of mass M/2.

 

[OrlandoLewis]

Yes, because in your solution you doubled the moment of inertia of two half rods of mass M instead of doubling the moment of inertia of two half rods of mass M/2.

Oh yeah, missed that. So theoretically, whatever angle I fold, the rod's moment of inertia will just be the same?

 

[kuruman]

Yes, because you have the same amount of mass at the same distance from the reference point regardless of angle.

 

[OrlandoLewis]

Yes, because you have the same amount of mass at the same distance from the reference point regardless of angle.

Oh, that's an interesting concept

 

[kuruman]

Here is a related concept you might also find interesting. A thin ring of mass $m$ and radius $R$ has moment of inertia $mR^2$. If you squeeze that mass into a semicircle of the same radius, the moment of inertia of the semicircle will still be $mR^2$. Keep squeezing into a quarter-circle down to a point and the moment of inertia will not change while you do so.