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Conservation of momentum in combination of angular and linear momentum

  1. Feb 16, 2011 #1
    angular momentum and linear momentum is conserved, but what happen when combination angular momentum and linear momentum occurs?

    for example a ball hits a horizontal paddle wheel on a base(which is free to move in any directions). then what happen to linear momentum of ball and paddle wheel? will it be equal to linear momentum before collision? or does the linear momentum after collision will be reduced by angular momentum gained by of paddle wheel (paddle wheel was initially at rest and due to tangential collision of ball it starts rotating after collision)
  2. jcsd
  3. Feb 16, 2011 #2

    Doc Al

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    Both linear and angular momentum will be separately conserved during the collision.
  4. Feb 16, 2011 #3
    Yes, momentum before = momentum after; likewise angular momentum.

    It is not that linear momentum gets "converted to" angular momentum. It is that your system of ball and paddle wheel already has angular momentum before the collision occurs. The ball's trajectory before collision is a straight line with tangential separation [itex]r[/itex] from the centre of mass of the paddle wheel. [itex]\vec L = \vec r \times \vec p[/itex]. Since [itex]r[/itex] and [itex]p[/itex] are non-zero, the ball-paddle system possesses angular momentum.
  5. Feb 16, 2011 #4
    you mean the ball and paddle system possesses angular momentum before collision due to and [itex] \vec p [/itex] of ball ( which travels in straight line before collision) with respect to paddle wheel's CG.

    But what about Kinetic Energy(KE) of ball and paddle system before and after collision, will it be split into rotational KE and Linear KE and their sum equals initial KE of Ball
    Last edited: Feb 16, 2011
  6. Feb 16, 2011 #5

    Doc Al

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    The angular momentum of the system depends on where you measure it from. Regardless, as long as there's no external torque on the system, angular momentum will be conserved.

    If the ball travels in a straight line towards the paddle wheel's COG, why do you think it will start rotating? Originally you said the collision was 'tangential', which I interpreted to mean not along the COG.

    In general, kinetic energy is not conserved--unless the collision is perfectly elastic (which is just a way of saying that the kinetic energy is conserved).
  7. Feb 16, 2011 #6
    no ball travels parallel to center of mass of paddle wheel and hit paddle wheel tangentially.

    assume collisions are perfectly elastic. then what about rotational kinetic energy and linear kinetic energy, before and after collision. Will it conserved independently just as in case of linear and angular momentum. Or will it be sums of rotational and linear kinetic energy
  8. Feb 17, 2011 #7

    Doc Al

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    The sum of linear and rotational KE will be conserved, not each separately.
  9. Feb 17, 2011 #8
    that means linear kinetic energy will be less after collision due to the rotational kinetic energy gained by paddle wheel, which was at rest before collision. right?
  10. Feb 17, 2011 #9

    Doc Al

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  11. Feb 17, 2011 #10

    what happens when ball travels through center of mass of paddle wheel and base and hits paddle wheel tangentially . now how is going to be the conservation of angular momentum and linear momentum.

    Also tell me about the conservation of Rotational and Linear Kinetic Energy
  12. Feb 17, 2011 #11
    sr241, in response to your diagram: my best guess of the situation after collision is as follows. (Let the point CM be the centre of mass of the paddle-base system. The ball has initial momentum [itex]{p_{{\rm{b,i}}}}[/itex] and final momentum [itex]{p_{{\rm{b,f}}}}[/itex].)

    • The paddle wheel rotates anticlockwise about its axle with angular momentum [itex]L[/itex] w.r.t. CM.
    • The paddle-base system rotates clockwise about CM with angular momentum [itex]-L[/itex] w.r.t. CM.
    • The paddle-base system has final linear momentum [itex]{p_{{\rm{pb}}}}[/itex] (rightwards).
    • The ball has final linear momentum [itex]{p_{{\rm{b,f}}}} = {p_{{\rm{b,i}}}} - {p_{{\rm{pb}}}}[/itex].

    Momentum before = momentum after = [itex]{p_{{\rm{b,i}}}}[/itex].
    Angular momentum before = angular momentum after = 0 (all w.r.t. CM).

    (This is ignoring any complicating factors, such as the possibility of the ball's trajectory being deflected off-axis upon collision with the paddle. I'm assuming that the ball continues rightwards along its original linear trajectory, or else bounces normally off the paddle.)

    This is just an intuitive guess, so please take my answer with a pinch of salt until somebody either confirms or refutes it.
  13. Feb 17, 2011 #12
    from your earlier response it seems that ball and paddle wheel possess some angular momentum. I have made a vector drawing of it.

    If ball possess angular momentum before collision then that would be P2 and corresponding linear momentum would be P3. you mean to say P2 and P3 will separately conserved, then in a way it is the sum of P2 + P3 = P1 (actual linear momentum before collision) that is conserved. or in other words sum of angular momentum and linear momentum will be conserved. or say some part of initial linear momentumP1 will go into angular momentum
  14. Feb 18, 2011 #13

    in the video linear momentum is conserved. and angular momentum is conserved by ball starts rotating in opposite direction of bar.
    Last edited by a moderator: Sep 25, 2014
  15. Feb 18, 2011 #14

    Doc Al

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    It took me a while to understand what you were doing with the vectors P1, P2, and P3. Vector P1 is the initial linear momentum of the ball. P2 and P3 are the components of P1 in the tangential and radial directions with respect to the center of mass of the paddle.

    To calculate the angular momentum of the ball with respect to the original position of the paddle's center of mass, you would evaluate r X P1, where r is the position vector of the ball measured from the paddle's center of mass. This a vector cross product. The magnitude of that angular momentum will equal rP2. (Note: P2 is the tangential component of P1, not the angular momentum.)

    In this collision, two things are conserved: The angular momentum and the linear momentum. The angular momentum I discussed above; the linear momentum is just P1 = P2 + P3.
  16. Feb 18, 2011 #15
    if two balls of same mass (a,b - a moving initially and b at rest) collides along center of mass; after collision a will be at rest and b will be moving with same velocity as "a". but if a's path is parallel to b's center of mass, a will not be at rest after collision; a will be moving along same path with velocity = a's initial velocity - b's final velocity, ( thus linear momentum is conserved) . but a and b starts rotating with respect to their center of mass in opposite direction. thus angular momentum is conserved, right ?

    How to calculate a's and b's final velocity and angular velocity from the path and initial velocity of "a" ? does moment of inertia has anything to do with this?
  17. Feb 18, 2011 #16
    In the video, after collision, the ball and paddle on the left rotate in opposite directions; but the ball and paddle on the right rotate in the same direction. I don't know what software you (or somebody) used to create this simulation, but it does not appear to be physically accurate. For a start, upon collision, the balls start rotating anticlockwise instantaneously of their own accord. There is no physical reason why they should do this. For the purposes of understanding this discussion, please ignore what you see in that video.

    I might have confused things when I wrote, in my first post, "...the ball-paddle system possesses angular momentum." I ought to have added, "about its (the ball-paddle system's) own centre of mass".

    Please note that the above applies to your diagram in post #12, i.e. in the case where the ball-paddle (or ball-paddle-base) system possesses non-zero angular momentum about its own centre of mass.

    For your final question, regarding the two balls of equal mass, we need only consider the conservation laws which Doc Al has already stated: conservation of linear momentum, and conservation of angular momentum.

    (For simplicity, let all trajectories lie in the x-y plane.)

    Let's say ball A travels with velocity [itex]\vec u[/itex] along the positive [itex]x[/itex] axis. Ball A strikes ball B off-centre. Let the final velocities of A and B be [itex]\vec v[/itex] and [itex]\vec w[/itex] respectively. Then, due to conservation of linear momentum:

    [tex]\left| {\vec u} \right| = {u_x} = {v_x} + {w_x}[/tex]
    [tex]{v_y} + {w_y} = {u_y} = 0[/tex]
    [tex] \Rightarrow {v_y} = - {w_y}[/tex]

    For a perfectly elastic collision, balls A and B each rebound with zero spin, and their velocities are at 90º to each other. This conserves linear momentum and linear kinetic energy.

    If the balls impart some rotation/spin to one another upon collision (due to friction, deformation etc.) then a portion of the initial kinetic energy of A is transferred to rotational/spin kinetic energy of A and B. The remainder of A's initial kinetic energy is transferred to linear kinetic energy of A and B. The balls' rebound velocities will then necessarily make an angle 0º < [itex]\theta [/itex] < 90º with each other. (Say if you need help proving this.)

    My feeling is that if the balls do impart some spin angular momentum to one another, then the angular momentum of A will be equal and opposite to that of B. (I haven't proved this. Perhaps someone could comment...?)

    The angular momentum of the system (w.r.t. some arbitrary point P) is constant. The linear momentum of the system is constant. The linear kinetic energy of the system can change. The rotational kinetic energy of the system can change.
    Last edited by a moderator: Sep 25, 2014
  18. Feb 19, 2011 #17

    the video shows two equal mass bars collide their linear momentum is separately conserved. but they are spinning in same direction. could it be because of direction of n in the cross product for angular momentum = p*r*sin theta * n

    in the first bar n is towards us and in second bar n away from viewer. thus angular momentum can be conserved separately even if bars are rotating in same direction.

    then what is the significance of n in angular momentum?
    Last edited by a moderator: Sep 25, 2014
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