# Homework Help: Moment of Inertia of a particle

1. Jun 8, 2005

### verd

Hey,

I know that this Moment of Inertia stuff is really easy, but I don't understand why I'm getting this one wrong.

[Image]
Find the moment of inertia $$I_{x}$$ of particle a with respect to the x axis (that is, if the x axis is the axis of rotation).
Particle a has mass m, and particle b has mass 2m.

...So wouldn't the Moment of Inertia just be:
(Moment of Inertia of a particle)
I = mr^2
$$I_{x} = m3r^2 = 3mr^2$$?

Because they only asked me for the moment of inertia of particle a, and not the total moment of inertia, that should be right, shouldn't it?

Last edited: Jun 8, 2005
2. Jun 8, 2005

### OlderDan

Where does that 3 belong?

3. Jun 8, 2005

### verd

That image link... if you click it, you can see the 'wonderous and beautiful' diagram. If the x-axis is the axis that's rotating, then the vertical distance to particle a in the positive y-direction is 3r.

So I get:

$$I_{x} = m3r^2$$

4. Jun 8, 2005

### Pyrrhus

Yes indeed the distance is actually r. but $I_{x} = mr^2$

I agree with OlderDan, confusing diagram... :uhh:

Last edited: Jun 8, 2005
5. Jun 8, 2005

### OlderDan

Actually, you are twice wrong. The diagram is confusing, but the distance is just r, not 3r. If it were 3r then it would be

$$I_{x} = m(3r)^2 = 9mr^2$$

6. Jun 8, 2005

### verd

I tried exactly that... The first time around. And I was told that I was off by a "multiplicative factor"

7. Jun 8, 2005

### verd

wait, so it's just:

$$I_{x} = mr^2$$?

EDIT: Ah, I see... The 3r is the x-value... Kind of confusing placement. Upon first glance, a lot of people would think they were referring to the y.

Egh. Either way, I don't get any credit for it. The computer fouled up between windows, and apparently it submitted a blank answer. Now I get no credit. ...Oppose homework via computer. I can't tell you how many times the computer has erred and caused me grade points.

Last edited: Jun 8, 2005
8. Jun 8, 2005

### OlderDan

Yes, it got me too at first, but since the 3 was not being squared properly it got me to look again.

9. Jun 8, 2005

### verd

...If the axis was parallel to the z axis, what in the world would $$I_[z]$$ be?

How do I go about finding this?

10. Jun 8, 2005

### OlderDan

For the z axis, you need the distance from the particle to the origin, which you can find by Pythagoras. Since you really need distance squared for I, you don't even have to take a square toot.

11. Jun 8, 2005

### verd

Ah. Okay. I understand that part... You'd have $$3r+r$$ for the most part.
but to find the Moment of Inertia, don't you need to multiply all of that by m?

So $$I_{z} = \sqrt{m(3r)^2+mr^2}$$?

12. Jun 8, 2005

### verd

eh? Is that what you meant?

13. Jun 8, 2005

### OlderDan

There is no square root. Look at the definition of I. The distance is not $$3r+r$$, but I think you already know that.

Last edited: Jun 8, 2005
14. Jun 8, 2005

### verd

Ah. Understood. Got confused with the radical, and whether or not I was to take the square root of the mass.

Eh, and again, I'm a bit confused... I'm asked to find this rod's moment of inertia:

[Image]

To find the Moment of Inertia of the rod in the diagram, wouldn't this be the correct integral?

$$\int_{0}^{l} x^2\displaystyle{\frac{m}{l}}dx$$?

For just dI, I get $$\displaystyle{\frac{1}{3}}(\displaystyle{\frac{m}{l}}dx)lsin(\theta)$$
am I going in the right direction here? Or am I still confused?

15. Jun 8, 2005

### verd

The above problem has me clueless... My textbook doesn't give me much for information regarding this subject... And what I got for dI was apparently incorrect because i didn't make use of x. Even though I thought I had computed it using l and theta.

Anyone have any ideas?

...I have another problem similar to that one. Granted, it's not as difficult, but given the information the text book gave me, I don't see how to solve it.

The moment of inertia of a particle or a system of particles is given by:
$$I = m_{1}r_{1}^2+m_{2}r_{2}^2...$$ So I know this.

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and with mass 4.00 kg, while the balls each have mass 0.500 kg and can be treated as point masses.
c.) Find the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls.
d.) Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.

a & b just asked for the moment of inertia with an axis perpendicular to the center of the bar, and with a moment perpendicular but at the end of the bar. ...Which I was easily able to figure out.

But because I don't have a radius for that bar, I don't see how I can compute the moment for either of the two above. I know that for d, I have to use the parallel-axis theorem... But where in relation to the center of the bar is this new axis? Would everything be 5m?

EDIT: I've never had so many problems with such an easy chapter. (The concepts and what the book presents is extremely easy and understandable. What was gone over in the lecture was also very easy and understandable) However, I seem to be encountering problem after problem that is either explained in a confusing manner, or is confusing and is lacking details about the format of the answer. I'm stuck on two more problems. And now I don't know where to turn

Last edited: Jun 8, 2005
16. Jun 9, 2005

### verd

Any help? Please. I'm so stuck... I'm sorry I keep bothering you... But I'm lost.

17. Jun 9, 2005

### OlderDan

The moment of inertia of this rod depends on where the rotation axis is. If it is the vertical axis, it is completely different from an axis out of the page. What axis are you using?

Last edited: Jun 9, 2005
18. Jun 9, 2005

### OlderDan

I don't see the difference between c and "with a moment perpendicular but at the end of the bar" although I'm not clear on the use of the word "moment" in this context. For d) you have to ignore the radius of the rod since it is not given. If you knew the radius, it would be the same as for any disk or cylinder about the long axis, plus the added term for the .5m by the parallel axis theorem.

19. Jun 9, 2005

### verd

I'm using the vertical axis... So it'd be swinging around almost like it would if it were attached at the end, (if there was no angle, I know I could use 1/3ML^2) However, the angle complicates things.

I'm not looking for you to answer my question-- I don't want you to get the wrong idea. I'm just somewhat lost, and was wondering if you could point me in the right direction... (Thanks by the way)

And for the next question,
oh geez, I'm sorry. I wrote the wrong one in c.
This is the correct one:
c.) Find the moment of inertia of this combination about an axis parallel to the bar through both balls.

so in ignoring the radius, how would I find the moment of inertia? I know I can't ignore the mass of the bar. But I can't just treat it like a particle... Could I?

20. Jun 9, 2005

### OlderDan

For the rod at an angle to the vertical axis you need

$$I = \int_{0}^{l} r^2\frac{m}{l}}dx$$

where r is the distance from the axis to the element of mass $$\frac{m}{l}}dx$$ and $$l$$ is the length of the rod. You are on the right track with the trig relation, but did not get it quite right. What is r in terms of x? All of r has to be squared, not just the x part.

My guess is that for c) they want you to find the general expression for a parallel axis at some distance (say D) parallel to the rod's long axis. Then d) is a specific case of the general result in c). In these cases all the mass is at the same distance from the rotation axis, so it is just like a point mass.

21. Jun 9, 2005

### verd

if r is the distance from the axis to the element of mass, then r would have something to do with lcos(theta)... wouldn't it? And as y increased, lcos(theta) would decrease, right? would that be r?

Last edited: Jun 9, 2005
22. Jun 9, 2005

### Staff: Mentor

If you are finding the moment of inertia about a vertical axis, then this should be:
$$\int_{x = 0}^{x = l} r^2\frac{m}{l}dx$$
where x is the distance along the rod, and r is the distance from the axis. Of course, $r = x \sin \theta$.

(Note that the moment of inertia of an object depends only on the distance each mass element is from the axis. So you could imagine sliding the mass elements around to form a stick of length $l \sin \theta$ that is being rotated perpendicular to one end.)

23. Jun 9, 2005

### verd

Okay. I got it. Thank you so much...