# Homework Help: Moment of Inertia Help

1. Jun 7, 2006

### cukitas2001

Hey guys more help need on my behalf....the topic im having trouble with now is moment of inertia. I reallyd ont understand how it is calculated and when i look at my notes, book, and tutorials online it looks even more confusing. Can soemone give me a dumb mans walk through of moment of inertia. I'm currently working on a problem and im stuck as you might have guessed.

1) 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.

part a asks me to find the moment of inertia about an axis perpendicular to the bar thorugh its center. It made sense to me to use (1/12)ML^2 and just add the individual balls moments of inertia of 2(0.5)*1^2 and i got it right :tongue:

part b now asks: Find the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls.

Now since its going through one of the balls i decided to try to use:
(1/12)ML^2+(0.5)*1^2 and add M(1^2) according to the parallel axis theorem. Obviously this came out wrong so what did i do wrong? was it the application of the theorem?

2. Jun 7, 2006

### vsage

where did you pull the two 1^2 from? The new axis is at one of the balls. The center of gravity of the rod is 1 meter from this, yes, but what about the other ball's center of gravity?

3. Jun 7, 2006

### cukitas2001

the 1^2 would be from the parallel axis theorem that says the Icm+Md^2 the distance that was moved for the new axis was jsut one meter so i put 1^2.....what all this center of greavity talk im still on center of mass/moment of inertia

4. Jun 7, 2006

### src2206

You have used the FORMULA not the THEOREM. reread the theorem. Igive youa hint: In this case the distance of one of the balls from the axis is 0 and the other is 4m. You did not take this into account. Do not go by the formula but go by basic knowladge :).

5. Jun 7, 2006

### vsage

The center of gravity and center of mass are two phrases for the same thing. Yes you moved the axis 1 meter, but you already had to apply the parallel axis theorem to the "far" ball (the ball that is on the opposite end of where the axis sits in the 2nd part of the problem) for the first part, now you need to apply it again with its new distance from the axis.

6. Jun 7, 2006

### cukitas2001

oh....k i tried Ip=Icm+Md^2 where Icm=(1/12)ML^2 and Md^2=0.5(4)^2
which equates to 9.33 but this is still wrong? what am i unbelievably leaving out now?

7. Jun 8, 2006

### Hootenanny

Staff Emeritus
Your close now. You have taken into account the moment of inertia of the rod about its centre of mass (highlighted in red) where L is the length of the bar. However, the rod is not rotating about its centre of mass, therefore you need to add in the moment of inertia of the contre of mass about the end, thus;

$$I_{rod} = \frac{1}{12}ML^{2} + M\frac{L}{2}^{2}$$

Then add to this the moment of inertia of the particle (ML2) as you did above.

Do you follow?

8. Jun 8, 2006

### cukitas2001

Why is it (1/2) ML^2 ? Is that how point masses are treated when dealin with moment of inertia?

Last edited: Jun 8, 2006
9. Jun 8, 2006

### Hootenanny

Staff Emeritus
Sorry, it would have been better written as;

$$I_{rod} = \frac{1}{12}ML^{2} + M\left( \frac{L}{2} \right)^{2}$$

Sorry for the confusion, it is 1/2L because that is the distance of the centre of mass from the axis of rotation (end of the rod).

10. Jun 8, 2006

### cukitas2001

oh ok i get it now. Now I'm asked: Find the moment of inertia of this combination about an axis parallel to the bar through both balls. Could i just add the other balls moment of inertia in the for of:
(1/2)ML^2+M(L/2)^2+mL^2+mL^2 ?
or would it be something like:
(1/2)ML^2+M(L/2)^2+M(L/2)^2+mL^2+mL^2 since theres two axis?

11. Jun 8, 2006

### Hootenanny

Staff Emeritus
If the axis is parallel to the rod and goes through both balls (point masses), then the axis of rotation must be about the centre of the rod, down its length. Now, if the balls lie on the axis of rotation (and have no volume becuase they are point masses) then they do not contribute to the moment of inertia. Thus, the moment of inertia is due solely to the rod rotating about its central axis. So the moment of inertia is...

Last edited: Jun 8, 2006
12. Jun 8, 2006

### cukitas2001

(1/12)ML^2 god please let me be right

13. Jun 8, 2006

### Hootenanny

Staff Emeritus
Well, not quite. Think of the rod as a cylinder. Remember that it is rotating about its central axis. Imagine a rolling coke can, thats what you've got here.

HINT: The moment of inertia will be independent of length

14. Jun 8, 2006

### cukitas2001

(1/2)MR^2 ? but then what will R be? the ball radius?

15. Jun 8, 2006

### Hootenanny

Staff Emeritus
Looks good to me.
Nope, in your original post you said;
This means the balls have no volume and hence no raidus, they are points. R in this case would be the radius of the bar.

16. Jun 8, 2006

### cukitas2001

the radius of the bar isn't givent hough but would it be safe to assume that since the problem states small balls attached to the end of the bar that the radii of the ball equal the radius of the bar?

17. Jun 8, 2006

### Hootenanny

Staff Emeritus
I'm not sure of this question, in my opinion it is very poorly written; it states that the balls are to be considered as particles then gives a radius for them. I am not sure what they are asking here.

My best guess is this. If no radius for the bar is given you have to assume it is thin (no radius). Therefore, we must now consider the balls as rigid bodies.

Last edited: Jun 8, 2006
18. Jun 8, 2006

### cukitas2001

the answer was 0.....i guess since R wasnt given it was used as 0 in the equation.....i am not having as much fun in this chapter as i'd like...

the last part asks: Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.

okay hopefully i can get this....so were using parallel axis again and the distance this time is .5m right? so my equation should look like:
(1/12)MR^2+M(.5)^2+(soemthing with the balls)?

Last edited: Jun 8, 2006
19. Jun 8, 2006

### Hootenanny

Staff Emeritus
I thought it may be a trick question. However, I disagree with how they present it. They say the balls are to be considered as point masses and yet they define their radius?! :grumpy: I have replied to your other thread btw.