# Angular momentum and impulse

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1. Aug 21, 2015

### 24forChromium

Hopefully the image is self-explanatory, if not:

A cylinder is rotating around its central axis with angular momentum L1; an angular impulse, ΔL is then added to the cylinder, perpendicularly with respect to L1.

The hypothetical result is: the cylinder has one angular mometum at the end, L-result. L-result is equivalent in magnitude to L1, and its direction is shifted counter-clockwise by a certain ammount so that a line parallel to L1 draw from the tip of L-result will meet the tip of ΔL, mathemetically, the angle has increased in the counter-clockwise direction by (90-arccos(ΔL / L1)) degrees. (<-- that stuff is got from basic trignometry)

Is this true? If it is false, what is the right way to do it?

will the cylinder "follow" the angular momentum arrow and also point in the new direction, or will it remain in place and then rotate around the new arrow in a weird fashion?

2. Aug 22, 2015

### Staff: Mentor

The vectors add, so the resulting angular momentum will have a larger magnitude.

Neither. It will precess.

3. Aug 22, 2015

### 24forChromium

I heard from various sources that if the two angular momentum vectors are perpendicular to one another, they will only change the direction of the final angular mometum. I understand, albeit to a shallow extent, precession, and I think precession only occurs when a torque is continuously exerted on a wheel. Also, it would be the most helpful of all if you can give the final angular momentum expressed in terms of L1 and ΔL.

4. Aug 22, 2015

### A.T.

5. Aug 23, 2015

### Staff: Mentor

As vector: L = L1 + ΔL
This is conservation of angular momentum.
As magnitude: $|L|=\sqrt{L1^2 + ΔL^2}$ - using the right angle between the two components.

See the section "Torque-free" in the wikipedia article.