# Signs associated with angular momentum

1. Dec 11, 2007

### cardinalboy

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

A light but stiff rod of length R is attached at an angle theta to a shaft along the z-axis; the rod is used to rotate (a single) mass M about the shaft. The mass moves with speed v in a CCW direction. Describe the angular momentum, L, of the mass with respect to the attachment point of the rod.

2. Relevant equations

L = r x p
L = rmv

3. The attempt at a solution

Since L has both horizontal and vertical components, there will be two equations. By the right hand rule, L should be perpendicular to the plane containing the rod (R) and the linear momentum vector (p) and so it points at an angle theta relative to the xy plane in a direction toward the z axis.

The vertical component of L points upward, parallel to the z axis, and so it is positive: Lsin(theta), where L = Rmv.

The horizontal (radial) component is perpendicular to the z axis and pointing inward toward the z axis, so it is positive or negative? I am not sure how to find the sign associated with this horizontal component. I see that it should have a magnitude of Lcos(theta), but I am not sure how to determine the sign.

Last edited: Dec 11, 2007
2. Dec 12, 2007

### Shooting Star

A vector has a magnitude and a direction, and is completely determined by these two. What exactly do you mean by asking whether it's positive or negative? If a fixed direction is given, then the component of a vector along that direction may be said to be positive or negative, as you have done in the case of the component of L along the z-axis.

For example, if I have the vector 2i+j, it's meaningless to say whether it's positive or negative.

3. Dec 12, 2007

### cardinalboy

Second Attempt At Angular Momentum Post

I'm sorry about my poor description previously. I tried as best I could to give a diagram of the problem I've described above (see attachment).

The triangle that I have drawn on the mass (and I've given a bigger version of this triangle off to the right) represents the angular momentum vector L, as well as its components Lz, which is vertical component and parallel to the z-axis, and LR, which is pointing radially inward toward the z axis.

I see that, by the triangle I've drawn, the vertical component Lz should be Lsin(theta) and should be positive since upwards is defined as the positive direction. I also recognize that the radial (or horizontal) component of L, or LR, should have a magnitude of Lcos(theta).

The authors of a book I am reading claim that the magnitude of LR is negative. Is this negative simply because the authors have decided that radially inward is the negative direction and radially outward is the positive direction, or is there some deeper meaning here that I am missing. Could I, if I wanted to, define inward as the positive direction and still get the proper solutions to a problem involving angular momentum. Any assistance would be greatly appreciated. Thanks!

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Last edited: Dec 12, 2007
4. Dec 13, 2007

### Staff: Mentor

Magnitudes are always postitive by definition, so I assume you (or the book) meant that the radial component is negative.
Sure. What really counts is the actual direction, which is inward (assuming the mass moves in a CCW direction when viewed from the +z axis looking down). Whether you call a component negative or positive depends on your sign convention/coordinate system.

5. Dec 13, 2007

### Shooting Star

(The picture took an awful lot of time to be approved!)

All right, so they mean negative in the r direction.