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When a collision is elastic and no external torque acts on a system, angular momentum is conserved

I found this example and checked the results:

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, $I = 10*1^2/12$ = 5/6 ) in an elastic collision.

If the rod is pivoted, the ball bounces back with v, p = -11.846 m/s , L = -5.923, (Ke = 70.16) and the rod rotates with $\omega$ = 20.3 , L is conserved : Lr = (20.3 *5 /6) = 16.923 and Ke = 70.166 + 171.834 = 242 J

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).

If the rod is not fixed to a pivot, in order to conserve linear momentum the rod must translate with p = (11.846 + 22) = 33.846 (v = 3.3846, Ke = 57.28) and the energy of the rotating rod becomes Ke = 114.5 and $\omega = \sqrt(2E/I)$ = 16.58

angular momentum 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. Is there a case in which also L is conserved?

In that case the bouncing speed must necessarily be different from 11.8 m/s, if that case exists, can you explain why the bouncing speed is different, whereas the masses are the same?

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  • $\begingroup$ Hi GreenRay. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Mar 24 '15 at 19:55
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Yes, angular momentum is conserved if you do the problem correctly.

If you assume the ball bounces back along exactly the same path it followed before the collision, there are three degrees of freedom: the velocity of the ball, the velocity of the rod, and the rotational rate of the rod. There are three constraints: conservation of linear momentum in the direction of motion of the ball, conservation of angular momentum, and conservation of kinetic energy. You seem to understand this.

From there, it's unclear what your approach was. How did you get the particular numbers you cite? There are infinitely-many ways to conserve linear momentum and kinetic energy. Conserving just those two applies two constraints to a problem with three degrees of freedom, so there is an entire one-dimensional manifold of solutions. IE you could give the ball any velocity you like (up to a maximum), then choose the rod's velocity and rotation rate to fit the two constraints. If you simply pick one of these solutions at random, then it is very unlikely to conserve angular momentum. You must use all three constraints to solve the problem.

You should do this an confirm that the correct figures are $$v_{ball} = -\frac{66}{7} m/s$$

$$v_{rod} = \frac{22}{7} m/s$$

$$\omega = \frac{132}{7}s^{-1}$$

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  • $\begingroup$ @GreenRay In order to use any conservation law, the conditions for its validity have to be met. It would be an instructive exercise to justify the validity of conservation of energy, momentum, and linear momentum in this problem. $\endgroup$ – garyp Jan 15 '15 at 14:59
  • $\begingroup$ Mark, you sound a bit condescending. If you give me incorrect numbers (not just with a typo) how can I focus on the idea? Even now, if the numbers are exact, you do not give me a clue on how you got to those numbers. If you care to explain, can you also say if the speed of rebound (66/7) is the same if the rod is pivoted on the CoM and cannot translate? $\endgroup$ – user59485 Jan 15 '15 at 17:25
  • $\begingroup$ GreenRay - @Mark gives correct numbers and he did give you "how" he got those numbers: conserve linear momentum assuming the ball rebounds along the initial path, conserve kinetic energy, and conserve angular momentum about the center of the rod. The algebra is left to you. $\endgroup$ – Bill N Jan 15 '15 at 20:08
  • $\begingroup$ @BillN My first response had a typo in the figures which led to some back-and-forth which I deleted as unproductive. $\endgroup$ – Mark Eichenlaub Jan 15 '15 at 20:09

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