# Which rod is harder to spin?

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1. Feb 17, 2017

### OrlandoLewis

1. The problem statement, all variables and given/known data
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?

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

3. 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.

Last edited: Feb 17, 2017
2. Feb 17, 2017

### kuruman

On edit: Also think about TomHart's suggestion.

3. Feb 17, 2017

### TomHart

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?

4. Feb 17, 2017

### OrlandoLewis

Then it should remain the same right?

5. Feb 17, 2017

### OrlandoLewis

Yes, I meant decrease by the former value.

6. Feb 17, 2017

### kuruman

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.

7. Feb 17, 2017

### OrlandoLewis

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

8. Feb 17, 2017

### kuruman

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

9. Feb 17, 2017

### OrlandoLewis

Oh, that's an interesting concept

10. Feb 17, 2017

### 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.