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Angular momemtum conservation and collision

  1. Feb 3, 2010 #1
    i don't know what a homework problem is...i don't go to school..all i know is concepts are cleared when you think about all concept's possibilities and it's application..here is yet another application of COM ND COAM...
    suppose a sphere rolling with angular speed $ and center of mass speed v collides head on elastically with identical sphere initially at rest..how can we apply conservation of angular momentum??
  2. jcsd
  3. Feb 3, 2010 #2


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    You have to write the COE and COM and COAM equations (separately), and also use v = rω for the relation between velocity and angular velocity (assuming rolling without slipping). :smile:
  4. Feb 3, 2010 #3
    But centre's speed and angular speed are different..It's nt rolling without slipping..In case of elastic collision,the linear velocities of two will be exchanged as they r identical..Help me with applying cOAM...Does rotation and translation ke are seprately conserved?
  5. Feb 3, 2010 #4
    Rotational and translational kinetic energy are not separately conserved. Total energy is conserved. In the case of an elastic collision total kinetic energy is conserved: rotational + translational.

    Your problem, as stated, is underspecified. You have 4 variables in the final state: the velocity and angular velocity of both bodies. You have three constraints: 1 from each of conservation of energy, momentum, and angular momentum. Number of constraints < number of variables means that you don't have enough information to complete the problem. The conservation laws restrict you to a one parameter (1 parameter = 4 variables - 3 constraints) family of solutions. To figure out which particular solution actually happens you need to further specify the dynamics, for example, by stating what the contact torque between the two balls is and how long it is applied (or equivalently, the angular momentum transferred from one ball to the other).
  6. Feb 3, 2010 #5


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    Then you need to know what the coefficient of friction is. :wink:
    As gazebo_dude :smile: says, KE is interchangeable, you can't treat linear and rotational KE separately.
  7. Feb 3, 2010 #6
    if there is no friction between the two spheres, the conservation of angular momentum is easy, because no angular momentum is transfered. As a matter of fact, with no friction you cant even roll anything. it will just slide. I mean, lets take a look at the most cliche conservation of anything reference. Playing pool. We can assume that the pool balls have a very very small coefficient of friction between them. When you hit a pool ball, it spins and collides with an idle pool ball. When you watch, at the instance it collides, the pool ball that has just been struck doesnt spin at all. After travelling more and more, its kinetic energy gets split between its linear velocity, and its angular velocity, until it doesn't slide at all
  8. Feb 4, 2010 #7
    yea surface is frictionless
  9. Feb 4, 2010 #8
    dacruick.are you suggesting that sphere can't roll on a frictionless horizontal surface..??
  10. Feb 4, 2010 #9
    if the force is applied at center of mass ,only then it slides in case of horizontal surface...friction is necessary in case of rolling on incline, to provide torque as wieght passes thru center of mass ,so it's moment arm is zero..so its torque
  11. Feb 4, 2010 #10
    Why are u assuming that a sphere cannot have rotation nd translational simultaneously on a frictionless horizontal surface?
  12. Feb 4, 2010 #11
    So just to clarify for my own sake..We are saying that for a perfectly elastic collision, there is 0 transfer of angular momentum (since there is an infinitesmall amount of time for the two spheres to transfer angular momentum through friction). There is a 100% transfer of linear momentum however.

    So what you would see is the first ball spinning (by coincidence, for it is not due to the surface, which is frictionless) and approaching the second ball. It hits the second ball and stops flat linearly. The first ball stands in place, spinning as quickly as it was before, while the second ball moves with as much linear momentum as the first ball had. Is this correct?
    Last edited: Feb 4, 2010
  13. Feb 4, 2010 #12
    if you threw it like a bowling ball fine it can have spin. but the spin will be constant. the point i was getting at was if there is no friction, there is no change in angular momentum or anything like that.
  14. Feb 4, 2010 #13
    Yea..Dacruick..U r ryt that first ball will keep on rotating with same angular speed and ball B Which was at rest move with linear velocity V!!
    But why angular momentum is not transferred? Is it because surface was frictionless?At what value of friction ,100% angular momentum will be transferred?
  15. Feb 4, 2010 #14


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    Come on, Shahid0072, you know this …

    torque = rate of change of angular momentum …

    if there's no torque about the centre of mass, no angular momentum is transferred …

    and if there's no friction, where will the torque come from? :smile:
  16. Feb 4, 2010 #15
    I was trying to think of which lack of friction you were talking about. Did you mean the friction that could have come from the first ball in a non elastic collision? That would certainly introduce torque. Imagine the extreme inelastic case, where both balls stick together. The second ball certainly has acquired angular momentum from the interaction.

    Or did you mean the lack of friction from the surface the balls are on? The first ball hitting the second ball, even in an elastic collision, would transfer its linear momentum. If there was friction from the ground, the second ball would begin spinning by virtue of the fact it is moving linearly at all on a surface with friction. There would be torque there, from the ground.

    You could have meant both.

    But that led me to think about the billiard balls that dacruick talked about earlier in the thread. However, with a difference.

    Let's imagine a surface with extremely high friction. The first ball approaches the second ball. The first ball has both linear and angular momentum. Let's imagine the collision is perfectly elastic. Let me know if the following is true: The first ball hits the second and transfers its linear momentum to the second. It does not transfer any angular momentum to the second ball and so the first ball effectively ends it spinning because of the friction with the table (dispersed as heat).

    The second ball must begin spinning on this extremely high friction surface in order to linearly translate across it. It has taken 0% of the first ball's angular momentum, and 100% of the first ball's linear momentum and has turned that 100% linear momentum into part linear and part angular. Which means this second ball is not moving as quickly as the first ball was. Is this correct?
  17. Feb 4, 2010 #16
    Tiny tim..U r saying that torque=rate of change of angular momentum..Very well said..But u forgot one.. That angular momentum of system remains constant.And system includes both balls...Let me illustrate..Even force=rate of change on linear momentum..And there is no external force acting on sytem..Even then linear momentum is transferred...
  18. Feb 4, 2010 #17
    So,tim that reason that there is no external torque,thats why angular momentum is not transferred is not right..Its sth else
  19. Feb 4, 2010 #18


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    Hi DocZaius! :smile:
    Could and did :wink:
    I'm not keen on answering this without looking at the equations, but my immediate thought is that friction between each ball and the surface may well result in transferring some angular momentum between the balls. :confused:
    Hi Shahid0072! :smile:

    Yes, but with linear momentum, the linear force is along the line of impact, so linear momentum in that direction can be transferred.

    With angular momentum, the torque (angular force) is zero (because the linear force is through the centre of mass), so angular momentum can't be transferred. :wink:
  20. Feb 4, 2010 #19
    Thanks tim..M little new with this... Why isn't angular momentum not transferred ? Angular momentum can never be in line of action .What if the surface on which A waS moving is having friction? Will angular momentum be transffered?
  21. Feb 4, 2010 #20
    I have doubts..Please correct me if i am wrong:
    1.A sphere having linear speed V initially moving on a frictionless surface enters into a horizontal surface having friction! After some time ,the sphere will move with rolling without slipping..Assuming friction is sufficient enough to support rolling without slipping.
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