# Is angular momentum always conserved?

• alba
In summary: Angular momentum is conserved, period. The point you choose to evaluate it about has nothing to do with that.The kinetic energy is conserved only if the impact is elastic, and I would not assume that unless explicitly stated in the problem statement. In the case of an elastic impact, all you need is conservation of kinetic energy and conservation of linear momentum, regardless of where you choose to evaluate the angular momentum.In the case of an inelastic impact, the situation is more complex. In the absence of external forces and torques, the center of mass will continue to move at constant velocity, but the velocities of the individual objects will not necessarily be conserved.
alba

A ball (m = 1 Kg , v = p =+22 m/s, Lm = +11, Ke = 242 J) hits the tip of a rod (M = 10Kg , length = 1m, i = 5/6 ).

the ball bounces back with v, p = -11.846 m/s , L = -5.923, (Ke = 70.16) the rod translates with v = 3.3846m/s , p = 33.846 , (Ke = 57.2J) and rotates about its CoM with ##\omega## = 16.58 (Ke = 114.556).

Ke is conserved : 70.16 + 57.2 + 114.6 = 242
linear momentum is conserved: 33.846 - 11.846 = 22

L
was +11
after the impact we have
Lm = -5.923
Lr = 16.58 (##\omega## * i ) 5/6 = 13.82
13.82 - 5.92 = 7.9

It seems that angular momentum is not conserved. Did I go wrong somewhere?

Yes, angular momentum is always conserved. You have not shown how the velocities and rotation rates resulting from the collision were derived. With three conserved quantities (linear momentum, angular momentum and energy) you should be able to solve for three results (ball velocity, rod velocity and rod rotation) in such a way as to conserve all three.

This is a homework-like problem and should probably have been posted in a more appropriate sub-forum.

jbriggs444 said:
This is a homework-like problem and should probably have been posted in a more appropriate sub-forum.
You are wrong, as I said these figures were given in a thread in this same forum and not in the HW forum. The figures were given by mentor DaleSpam, and you yourself posted in that same thread. I did not add anything, was just noting something odd, and asking for clarification, that is all

alba said:
these figures were given in a thread

DocZaius said:

In that thread, multiple sets of numbers were bandied about, often without a careful explanation of how they were derived or what assumptions went into the associated scenario.

Of the numbers that Alba apparently picked for the OP in this thread, it seems that the velocity of the rebounding ball came from DaleSpam in a scenario where the rod was anchored to a pivot and the angular velocity of the rod came from Bobie in a scenario where the rod was not anchored.

I haven't followed the details discussed here, but it's worth remembering that angular momentum referenced to the same point in space (i.e. angular momentum "about" a point) is conserved. However, computations of angular momentum about different points need not have any relation to each other. Whether there are "moving points" (such as the center of mass) about which angular momentum is conserved would be interesting to discuss.

I worked out the result of an impact in which L is conserved:

the bouncing speed should be -9.428 m/s , the rod should translate at +3.1428 (p = +31.4) and ##\omega## = 20.3 (L =15.7)

Can someone explain why, in order to conserve angular momentum, the bouncing speed must be different with or without a pivot?

Is there a formula to calculate that difference?

Physically, the pivot pushes on the rod, changing its velocity. The rod pushes on the ball, changing its velocity.

Mathematically, you are solving two different systems of simultaneous equations and obtaining two different results. Obtain expressions for the two solutions and subtract them. There's your formula for the difference.

Angular momentum about a fixed (inertial) pivot point is conserved. If you move the pivot point around, the angular momentum might change.

Thanks,

Can you tell me, now, how do I get the result if in the same example the ball hits the rod not on the tip (at .5 m from CM) but at .4m (that is 10 cm from the tip)? What is the moment of inertia, does it change?

Why not model the collision as an impulse of unknown magnitude at the point of collision? Write down formulas for how this impulse will modify the momentum (and therefore the velocity) of the ball and rod. Write down a formula for how this impulse will alter the angular momentum (and therefore the rotation rate) of the rod about its center.

Use those formulas to write down a formula for the resulting kinetic energy of rod and ball combined.

If you want to assume a perfectly elastic collision, solve for final kinetic energy = initial kinetic energy.

Angular momentum is only conserved if there is no external torque. In the referenced thread it was unclear if there was an external torque or not due to the poor description of the problem.

jbriggs444 said:
Why not model the collision as an impulse of unknown magnitude at the point of collision? Write down formulas.
I was referring to the example in my OP using those data. I am asking if I must assume the same moment of inertia = 5/6 or not. mass and length are the same (10Kg,1m) but the point of impact is not at .5m from thr CM but at .4.
I am asking if I must change the formula to i = 10 *.8^2/ 12

## 1. What is angular momentum and how is it defined?

Angular momentum is a measure of the amount of rotation an object has around a fixed point, or axis. It is defined as the product of an object's moment of inertia and its angular velocity.

## 2. Is angular momentum always conserved?

Yes, angular momentum is always conserved in a closed system where there are no external torques acting on the system. This means that the total angular momentum of the system will remain constant, regardless of any internal changes or motions.

## 3. How is angular momentum conserved?

Angular momentum is conserved due to the law of conservation of angular momentum, which states that in the absence of external torques, angular momentum will remain constant. This is similar to the law of conservation of linear momentum, which states that in the absence of external forces, linear momentum will remain constant.

## 4. Can angular momentum change in a system?

In a closed system with no external torques, angular momentum cannot change. However, if external torques are present, angular momentum can change. This is because external torques can cause changes in an object's moment of inertia or angular velocity, which will result in a change in its angular momentum.

## 5. What are some real-life examples of angular momentum conservation?

Some real-life examples of angular momentum conservation include the rotation of planets and moons in the solar system, the spinning of a gyroscope, and the motion of ice skaters performing spins. In all of these cases, the angular momentum remains constant due to the absence of external torques acting on the system.

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