# Question involving L and ω where L does not equal Iω

1. Oct 24, 2009

### azure kitsune

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

Suppose in the given figure that mB=0; that is, only one mass, mA, is actually present. If the bearings are each a distance d from O, determine the forces FA and FB at the upper and lower bearings respectively.

(This is from Physics for Scientists and Engineers, by Giancoli)

2. Relevant equations

Equations for angular momentum and torque? τ=dL/dt

3. The attempt at a solution

I'm having trouble understanding the example where both masses are present. If L does not always equal Iω, then when exactly is this true?

I think I should understand the example where both masses are present first. If I want to find both forces in the figure, what equation should I start with?

#### Attached Files:

• ###### img_4520_resize.jpg
File size:
23.4 KB
Views:
57
2. Oct 26, 2009

### tiny-tim

Hi azure kitsune!
The angular momentum vector of a rigid body is only aligned with the axis of rotation if that axis is a principal axis of the body.

You're probably thinking that LA on its own obviously is vertical (and so is LB), so why isn't the total L also vertical?

The diagram is misleading … it suggests that each mass has an "intrinsic" L.

L depends on the point about which you measure it …

looking only at mass A, and measuring the moment of inertia about a point P a distance r from mA, there's an extra mAr2 term only in directions perpendicular to PA, so only those directions (and the single direction PA itself) are principal directions …

so if P is the point PA, on the axis at the same height as A, then the angular velocity is perpendicular to PAA, and so is along a principal axis, and therefore so is the angular momentum, LA

but if P is the point O (or PB), then the angular velocity is not along a principal axis, so LA is at a slight angle to the vertical.

From the PF Library on moment of inertia …

Moment of inertia tensor:

Surprisingly, angular momentum is not generally aligned with rotation.

Since the angular momentum vector of an unforced rigid body must be constant (in space), the axis of rotation (if not already aligned along it) must move around it: this is precession.

The moment of inertia tensor converts the angular velocity vector of a rigid body into the angular momentum vector: $\tilde{I}\,\bold{\omega}\ =\ \bold{l}$

A tensor converts one vector to a different vector.

The eigenvectors of the moment of inertia tensor of a rigid body are its principal axes, and the eigenvalue of each principal axis is the (ordinary) moment of inertia about that axis.

Every rigid body has either:
i] three perpendicular principal axes
ii] principal axes in every direction in a particular plane (all with the same moment of inertia), and a perpendicular principal axis
iii] principal axes in every direction (all with the same moment of inertia)

In particular, any axis of rotational symmetry of a rigid body is a principal axis.

3. Nov 5, 2009

### azure kitsune

Thanks for replying. I'm still a little confused, but I'm not sure what to ask right now.

How would you measure moment of inertia with respect to a point? Isn't it done with respect to an axis? What extra mAr2 term are you referring to?

Okay... so L and ω are go in the same direction if and only if ω lies on a principle axis? And when this is true, we must have L = Iω?

I'm still a little confused about the principle axis part but I'll try to think about it a little more.

(And I was able to solve the given problem in the textbook, but I still want to fully understand the situation.)

4. Nov 6, 2009

### tiny-tim

Hi azure kitsune!
ah, I started saying "a point" when I was talking about the angular momentum, L
and then I decided to carry on the theme

yes, strictly speaking I should have said "measuring the moment of inertia about an axis through a point P a distance r from mA, there's an extra mAr2 term only in directions perpendicular to PA"
It's in the parallel axis theorem … see http://en.wikipedia.org/wiki/Parallel_axis_theorem

IP = Ic.o.m. + mAr2

meaning that the moment of inertia about an axis through P equals the moment of inertia about the parallel axis through the centre of mass, A, plus mass times the square of the perpendicular distance, r, from A to the original axis.

(so for the axis parallel to PA, r = 0, and so IP = Ic.o.m. as expected, but for the perpendicular axes, r = |PA|)

5. Nov 6, 2009

### azure kitsune

Thanks for clearing that part up. It makes more sense now. So because angular velocity and angular momentum are measured with respect to different things, they don't always point in the same direction.

I'm still wondering something though. When you were talking about the mass A and the point P, you talked about a lot of different axes of rotation. Isn't there only one axis of rotation for the object though? So are these other axes ever important?

And another question: If L and ω do not point in the same direction, is there still a relation between them?

Thanks for your help tiny-tim! :)

6. Nov 6, 2009

### tiny-tim

No, angular momentum is measured about a point, while angular velocity is not measured about any point or axis, it just "is".

To put it another way:

Angular velocity is different for different inertial frames (frame means the velocity of the observer), but in each frame it is independent of position (go further away, and the angular velocity stays the same).

Angular momentum is also different for different inertial frames, and in each frame it is also different for different reference positions (go further away, and the angular momentum gets bigger).

And moment of inertia is measured about an axis (go further away from the parallel axis through the centre of mass, and the moment of inertia gets bigger).
Moment of inertia is measured about an axis. Those are the only axes I was talking about.

(And yes, there is only one axis of rotation.)
Yes, L = Iω, where I is the moment of inertia tensor.