# Moment of Inertia in rod

1. Nov 26, 2013

### Panphobia

1. The problem statement, all variables and given/known data
A rod of mass M and length l rotates in a vertical plane about its centre which is on a frictionless, horizontal pivot. On the ends of the rod are point-like masses m1 and m2, where m1 != m2.

a)moment of inerta about the center of the rod
b)Determine the angular momentum when the angular velocity is ω
c) Determine the angular acceleration when the rod makes an angle of θ with respect to the horizontal z-axis. For m1 > m2 what is the direction of the torque?
d) At what angle θ is the angular velocity ω the greatest?

3. The attempt at a solution
So I would know how to do this question without the point masses m1 and m2, what they do is change the centre of mass of the rod, so does this question have something to do with the change in the centre of mass? Also what does it mean when the rod makes an angle of θ with respect to the horizontal, does this mean it is slanted?

2. Nov 26, 2013

### Simon Bridge

I'm reading this problem as implying that the rod is initially placed horizontal.
Ignoring the problem for a moment, just picture this situation - what will happen when the rod is let go?

3. Nov 26, 2013

### Panphobia

What do you mean? Its already in motion.

4. Nov 26, 2013

### nil1996

Just you need to find the moment of inertia first.The moment of inertia of a rod around its centre is ML^2/12.
Now find the moment of inertia when the two masses are attached.

5. Nov 26, 2013

### Panphobia

What is the difference? I know there is one, but in class we have only gone so far as to calculate moments of inertia, not any of this. Would I add the moments of inertia of the points to the moment of inertia of the rod?

Last edited: Nov 26, 2013
6. Nov 26, 2013

### nil1996

Yes as they are point masses you can add them!

7. Nov 26, 2013

### Panphobia

so (1/12)ML^2 + (1/2)m1L^2 + (1/2)m2L^2 = I ?
But since the moment of inertia for a point mass is mR^2 and R = L/2 would it actually be (1/4)mL^2?

Last edited: Nov 26, 2013
8. Nov 26, 2013

### nil1996

moment of inertia is given by MR2 where M==mass of the object;R==distance of the object from the axis around which we want moment of inertia.

9. Nov 26, 2013

### Panphobia

ahhh so it is (1/4)m(1+2)L^2, thanks! I can figure out a) and b) with that...now for the others. What does c) mean by angle θ with respect to the horizontal? Is it slanted?

10. Nov 26, 2013

### nil1996

wait what is that (1/4)m(1+2)L^2 ?
why you have done m(1+2)

11. Nov 26, 2013

### Panphobia

Because 1/4m1L^2 + 1/4m2L^2 = (1/4)m(1+2)L^2

12. Nov 26, 2013

### nil1996

It should like this-
1/4m1L2+1/4m2L2

those are only different masses indicated by m1 and m2

I=ML2/12 + 1/4m1L2 + 1/4m2L2

I=ML2/12+1/4L2(m1+m2)

13. Nov 26, 2013

### Panphobia

So about my question in post #9?

14. Nov 26, 2013

### nil1996

when it is at an angle (theta ) then it is slanting to the z axis.

see the attachment

#### Attached Files:

• ###### moment.png
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15. Nov 26, 2013

### Panphobia

what significance does m1 > m2 have?

16. Nov 26, 2013

### nil1996

it is necessary for rotating the rod, for giving it angular acceleration.

17. Nov 26, 2013

### Panphobia

This has something to do with T = I$\alpha$ correct?

18. Nov 26, 2013

### nil1996

right:)

19. Nov 26, 2013

### Panphobia

Hmm this may be a dumb question but it says it is making an angle with the horizontal axis, but I$\alpha$ does not incorporate an angle, but T = r x F does, but then I am still stumped.

20. Nov 26, 2013

### nil1996

T = r x F
yes it is a cross product and needs the angle

τ=r*F*sinθ

τ is the magnitude of the torque,
r is the displacement vector (a vector from the point from which torque is measured to the point where force is applied),
F is the force vector,
θ is the angle between the force vector and the lever arm vector.