How Does Angular Impulse Affect a Rotating Cylinder's Momentum?

AI Thread Summary
A rotating cylinder experiences a change in angular momentum when an angular impulse is applied perpendicularly to its existing momentum. The resulting angular momentum, L-result, maintains the magnitude of the original momentum while shifting direction counter-clockwise by a specific angle derived from trigonometry. The cylinder does not simply point in the new direction but undergoes precession due to the addition of the angular momentum vectors. When the angular momentum vectors are perpendicular, they alter the direction of the final momentum without affecting its magnitude. The conservation of angular momentum is expressed as L = L1 + ΔL, with the magnitude calculated using the Pythagorean theorem.
24forChromium
Messages
155
Reaction score
7
True or false.jpg

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 amount 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?

Additional question:
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?
 
Physics news on Phys.org
The vectors add, so the resulting angular momentum will have a larger magnitude.

24forChromium said:
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?
Neither. It will precess.
 
mfb said:
The vectors add, so the resulting angular momentum will have a larger magnitude.

Neither. It will precess.
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.
 
  • Like
Likes 24forChromium
24forChromium said:
Also, it would be the most helpful of all if you can give the final angular momentum expressed in terms of L1 and ΔL.
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.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top