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Do Collisions With Objects With Angular and Linear Momentum Produce Same Acceleration

  1. Oct 29, 2012 #1

    morrobay

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    If 3 spheres have equal masses and the stationary object is in a collision
    with a sphere with linear momentum . p= mv and linear speed 3m/sec
    And in the second case the collision with the stationary object is with a sphere with angular momentum L = rxp
    With angular speed 1 radian/ sec , radius = 3m.
    Does the stationary object in these two collisions have the same acceleration ?
     
    Last edited: Oct 30, 2012
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  3. Oct 29, 2012 #2

    Simon Bridge

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    Re: Do Collisions With Objects With Angular and Linear Momentum Produce Same Accelera

    You'd want to try each case using conservation of momentum, and angular momentum.
    In general, if angular, as well as linear, momentum is involved, the effects of the collision will not be the same as for linear momentum alone.
     
  4. Oct 30, 2012 #3

    morrobay

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    Re: Do Collisions With Objects With Angular and Linear Momentum Produce Same Accelera

    Case #1 is an object at rest in a collision with an object with linear momentum p= mv
    Case #2 is the same object at rest in a collision with an object with angular momentum L = rxp
    My question is whether the object at rest in both of these separate collisions is
    accelerated equally . All three objects have same mass. Case#1, object with linear speed 3m/sec
    Case #2, object with angular speed 1 radian/sec. radius = 3m.
    note: the underlying question is whether the stationary object is accelerated more in
    the collision with object with angular momentum ?
     
  5. Oct 30, 2012 #4
    Re: Do Collisions With Objects With Angular and Linear Momentum Produce Same Accelera

    How do you define your angular momentum, in respect to what point?
    In case A there is also angular momentum, in respect to any point which is not on the ball's trajectory. And in case 2 there is linear momentum as well.
    What is different in case 2? Is the ball spinning?
     
  6. Oct 30, 2012 #5

    mfb

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    Re: Do Collisions With Objects With Angular and Linear Momentum Produce Same Accelera

    It depends on friction and, in the case of nonzero friction, on the collision point.
    Without friction, the rotation does not change anything.
     
  7. Oct 31, 2012 #6

    morrobay

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    Re: Do Collisions With Objects With Angular and Linear Momentum Produce Same Accelera

    Consider 3 non-spinning billiard balls of equal masses:
    #1 is moving in translational motion on the x axis , v = 3m/sec , p=mv
    #2 is at rest on x axis at x=3m
    #3 is in uniform circular motion in xy plane on a frictionless track at 3m from the origin, x0y0 L = rxp
    In collision #1 there is one dimensional elastic collision at x =3m
    So after collision v2 = (2m1/m1+m2)v1 , v2 for billiard ball #2
    In collision # 2 the orbiting billiard ball is in collision with stationary billiard at x=3m y = 0
    The linear speed of billiard ball in circular motion , v = ωr, (1 radian/sec ) (3m) = 3m/sec
    So Im asking if the acceleration of the stationary billiard ball after collision in these two separate cases is equal ?
     
    Last edited: Oct 31, 2012
  8. Oct 31, 2012 #7
    Re: Do Collisions With Objects With Angular and Linear Momentum Produce Same Accelera

    I don't see the relevance of the historic trajectory of ball #3. In the case of non-spinning billiard balls and inelastic collisions, all that ever matters are the initial/final masses/velocities. The two collisions will produce the same effect since these conditions are identical (apart from the orientation). Acceleration is not usually considered because it gets tricky to define a time interval of the interaction, and most of the time, all we need to use is conservation of momentum.

    I'd also add that when you discuss angular momentum in the context of collisions, it is commonly assumed that you're referring to an axis of rotation which passes through the centre of mass of the body, ie. it's spinning. From the replies so far it seems that this is what everyone has understood of your question.
     
    Last edited: Oct 31, 2012
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