# Rod - Moment of inertia

1. Mar 12, 2016

### reminiscent

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

2. Relevant equations
I = (1/3)ML2
Parallel axis theorem: Inew = Iold + mr2

3. The attempt at a solution
I already solved part a) and I got (1/12)ML2. For b), I get that I have to use the parallel axis theorem but I don't know how to go from there. We have not gone over this concept yet in class (not even moment of inertia) so bear with me...

2. Mar 12, 2016

### Simon Bridge

... start from the equation for the parallel axis theorem. You can look it up: the meanings of the symbols I mean.
I think you were supposed to use it for part (a) so how did you do that?

Last edited: Mar 12, 2016
3. Mar 12, 2016

### drvrm

how did you do the part (a)-- if you have doneit correctly then (b) should not be a problem?

the above is not relevant -did you use it?

4. Mar 12, 2016

### Simon Bridge

5. Mar 13, 2016

### reminiscent

All I did for part a) was that since the rod is bent at its center, that means both segments of the rod have masses of M/2 and lengths of L/2. So for one rod, I = (1/24)ML2. For 2 rods, then it's just 1/2(ML2)

6. Mar 13, 2016

### SammyS

Staff Emeritus
Is that a typo ?

I thought that you got 1/12 ML2 , (which is correct.)

7. Mar 14, 2016

### reminiscent

Yes, sorry, that was a typo. That is what I got. But how am I able to use the parallel axis theorem for b)? We did not go over it in class completely yet...

8. Mar 14, 2016

### SammyS

Staff Emeritus
What is the location of the center of mass ?

9. Mar 16, 2016

### reminiscent

The location of the center of mass for each segment of the rod is the midpoint of each segment, correct?

10. Mar 16, 2016

### reminiscent

I don't understand the axis that it is being rotated. Is it the midpoint between the 2 ENDS of each segment of the rod? So does it boil down to a trig problem?

11. Mar 16, 2016

### SammyS

Staff Emeritus
So then where is the center of mass of the whole bent rod?

12. Mar 16, 2016

### reminiscent

The midpoint of the rod, at the bent center?

13. Mar 16, 2016

### SammyS

Staff Emeritus
No, not at the bend.

Make a sketch.

14. Mar 16, 2016

### reminiscent

Is it the midpoint of the ends of the two segments if they were connected?

15. Mar 16, 2016

### SammyS

Staff Emeritus
No. That's the pivot point for part (b).

It should be mid-way between CM of each segment, Right?

16. Mar 16, 2016

### reminiscent

So like 1/4 of the way of each segment?

17. Mar 16, 2016

### SammyS

Staff Emeritus
No.

That's a correct statement.

The C.M. for the bent rod is mid-way between the center of mass for each of the two segments

Last edited: Mar 16, 2016
18. Mar 16, 2016

### Arkun

I have the same problem as OP has, and while I know where the C.M. of the bent rod is, how can you use the center of mass of the two segments of the rod to algebraically obtain the center of mass of the whole rod?

What I did was determine the distance of the center of mass from the rotational axis and use that as the L for the moment of inertia equation.

My attempt at finding $I_{CM}$:

$I_{CM} = m*(\sqrt{((L / 2) * cos30)^{2} + ((L / 2) * sin30)^{2}})^{2}$

This is for part (a) of the problem.

19. Mar 16, 2016

### SammyS

Staff Emeritus
If you don't understand that, then it's doubtful that do know the correct location of the center of mass of the whole rod .

Simplifying what you have for $\ I_\text{CM}\$ gives:
$\displaystyle \ m\cdot\left(\sqrt{((L / 2)^2 \cdot \cos^2(30^\circ) + ((L / 2)^2 \cdot \sin^2(30^\circ)}\right)^{2} \$

$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 \cdot( \cos^2(30^\circ) + \sin^2(30^\circ))}\right)^{2} \$

$\displaystyle \ = m\cdot\left(\sqrt{(L / 2)^2 }\right)^{2} =mL^2/4\$​

It's not at all clear, how that is the moment of inertia about any point.

20. Mar 16, 2016

### Arkun

Isn't the center of mass of each segment of the rod at their midpoint? Which means that the distance between the center of mass for each segment of the rod and the axis of rotation is $L / 4$?