1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Angular momentum in a collision

  1. Nov 16, 2014 #1

    bobie

    User Avatar
    Gold Member

    Suppose a point mass B (m = 2 Kg) is rotating on a massless string (r = 2m) at v = 3m/s. Then KE = 9 J, p = 6 Kg m/s and L = p * r = 12 Nm (left in the sketch)


    8ZTe8.jpg
    Suppose B collides with A (m=2) the bob of a pendulum on a massless rod r = 1m. Is L conserved? It seems that there is no external torque.

    If L is conserved then the speed of A must be 12 /( 1 * 2) = 6 m/s but in this case KE is 36 J, that is 4 times greater. If E is to stay the same, as it seems obvious, v must remain 3 m/s but angular momentum will then be L = 6 * 1 = 6 Nm, that is half the original 12 Nm.

    - 1) Is L conserved or there has been an external torque?
    - 2) If we want to use the general formula L = I / ω , L = 12, ω = v / C = 3/ 4π → I = 12*4π/3 = 48 π/ 3 is this correct?

    Thanks for your help
     
    Last edited: Nov 16, 2014
  2. jcsd
  3. Nov 16, 2014 #2

    jbriggs444

    User Avatar
    Science Advisor

    Where is the reference point about which the angular momentum of B is reckoned?
    Where is the reference point about which the angular momentum of A is reckoned?
    What is the sign of the angular momentum of B before the collision?
    What is the sign of the angular momentum of A after the collision?

    In the absence of an external torque, angular momentum with respect to a particular reference point is conserved. But if you change reference points, you will (if total linear momentum is non-zero) change angular momentum.
     
  4. Nov 16, 2014 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Around the origin of the massless string: If the two circles touch at the interaction point so there is no momentum transfer along the 1m rod, yes.

    You have to be consistent with the point where you calculate your angular momentum.
     
  5. Nov 16, 2014 #4

    bobie

    User Avatar
    Gold Member

    So L remains 12 wrt to F and becomes 6 wrt to the new axis of rotation .
    Can you reply also to the second question, please
     
  6. Nov 16, 2014 #5

    jbriggs444

    User Avatar
    Science Advisor

    Yes. The angular momentum wrt the new axis of rotation was 6 immediately prior to the collision and remains so afterward.

    ##\omega = \frac{v}{r}## The circumference, C, does not enter in.
     
  7. Nov 17, 2014 #6

    bobie

    User Avatar
    Gold Member

    But what is the angular momentum of A with reference to F when it is rotating? L will not stay 12, it will always change from 12 to 24 Nm, all the time.

    Secondly, doesn't A exert an external torque on B when they collide?
     
    Last edited: Nov 17, 2014
  8. Nov 17, 2014 #7

    jbriggs444

    User Avatar
    Science Advisor

    Indeed, it will not remain 12. It will vary due to the external torque corresponding to the external centripetal force applied by the pendulum's axis on the bob, A.
    That depends on where you draw the boundaries of the system of interest. If you consider the system to consist of B alone then yes, it is an external torque.

    If you consider the system to consist of A and B together then it is part of an internal torque pair.
     
  9. Nov 17, 2014 #8

    Dale

    Staff: Mentor

    I am assuming that two of your unspoken assumptions are an elastic collision and frictionless passive bearings. Can you please confirm?

    As was mentioned before, you have to choose where you measure angular momentum relative to. However, anywhere you choose you will have an external torque on the system. Angular momentum is not conserved.
     
  10. Nov 18, 2014 #9

    bobie

    User Avatar
    Gold Member

    Yes , elastic, frictionless. If we consider the system A,B, how do you describe conservation ov L? does the centripetal force of the rod exert a torque? when a skater pulls in her arms in the normal direction there is an (internal) torque? does L (A) vary from 12 to 24 because of the rod?
     
  11. Nov 18, 2014 #10

    jbriggs444

    User Avatar
    Science Advisor

    Taking the system of interest to be A and B together then at the moment of the collision there are three interactions. There is the collision itself. There is the pull of the string on B and there is the pull or push of the pendulum rod on A.

    The pull of the string on B is in the same direction as the displacement of A from F. It can exert no torque. The pull or push of the rod on B is in the same direction as the displacement of B from F. Further, the rod is massless and is on frictionless bearings. It can exert no torque. The collision itself is an internal interaction. It cannot change the total angular momentum of the system. Conservation of angular momentum applies. The total angular momentum of {A,B} about F will be the same before and after the collision.

    The angular momentum is 12 just prior to the collision. It remains 12 just after the collision.

    If ball B comes to rest after the collision (as it must to conserve energy in an elastic collision), it will have zero angular momentum. We can simplify the problem by shrinking the system of interest to consider A alone.

    There is one interaction while A moves in a circular arc at the end of the pendulum rod. That is the external force of the pendulum rod on A. This force will be centripetal (i.e. toward the center of A's circular arc) and non-zero. It will have a varying component perpendicular to the displacement of A from F. At all but two points in the arc, this perpendicular component will be non-zero. That means that at almost all times it exerts a varying, non-zero torque on A about F. The angular momentum of A about F will vary over time due to this external torque.

    When a skater pulls in her arms the third-law partners in the force pair (the arms on the body and the body on the arms) add to a net of zero. The change in angular momentum due to the pair is zero. In the absence of any other external torques, angular momentum will be conserved. Any change in her moment of inertia must be accompanied by a change in her angular velocity.
     
  12. Nov 18, 2014 #11

    Nugatory

    User Avatar

    Staff: Mentor

    Bobie, you are starting to go off in every different direction again. There are a bunch of good answers already in this thread, and you need to understand them before you raise any new issues.

    So I've closed this thread. I will reopen it after you have sent me by PM a satisfactory answer to the following questions:
    1) what is the total angular momentum of the system around the point F, before the collision?
    2) what is the total angular momentum of the system around the pivot of the pendulum, before the collision?
    3) what is the total angular momentum of the system around the point at which the collision happens, before the collision?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Angular momentum in a collision
  1. Angular Momentum (Replies: 2)

  2. Angular momentum (Replies: 0)

  3. Angular momentum (Replies: 3)

  4. Momentum and Collision (Replies: 3)

Loading...