Is angular momentum always conserved?

  • Thread starter alba
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  • #1
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I read the following example in a thread in this forum:

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?
 

Answers and Replies

  • #2
jbriggs444
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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.
 
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  • #3
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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
 
  • #4
A.T.
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these figures were given in a thread
Link?
 
  • #6
jbriggs444
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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.
 
  • #7
Stephen Tashi
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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.
 
  • #8
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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?

Thanks for your help
 
  • #9
jbriggs444
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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.
 
  • #10
Khashishi
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Angular momentum about a fixed (inertial) pivot point is conserved. If you move the pivot point around, the angular momentum might change.
 
  • #11
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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?
 
  • #12
jbriggs444
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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.
 
  • #13
Dale
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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.
 
  • #14
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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
 

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