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Rigid body (rectangular) collision with wall

  1. Jul 1, 2012 #1
    Hello,
    I have a rectangular object with the following properties:
    Mass (m) = 1 units
    Moment Of Inertia (I) = 4.41 units
    initial velocity ([itex]v_{i}[/itex]) = (5, 0)
    initial angular velocity ([itex]w_{i}[/itex]) = 0

    If the block collides with a solid rigid wall with distance vector r = (1.1, 1.4) from the center of mass of the body, what will be the final angular velocity and final velocity of the body (velocity of center of mass) after the collision?

    This is how i have tried the problem:
    - find the velocity of the point on the body, where collision occurs using the formula:
    [itex]v_{pi}[/itex] = [itex]v_{i}[/itex] + ([itex]w_{i}[/itex] x r)
    [where, [itex]v_{pi}[/itex] = initial velocity of point, [itex]v_{i}[/itex] = initial velocity of center of mass]


    - find the final velocity of point using the formula:

    [itex]v_{pf}[/itex] = -[itex]v_{pi}[/itex]
    [where, [itex]v_{pf}[/itex] = final velocity of point]


    - find the angular velocity of the body using formula:

    [itex]w_{f}[/itex] = (r x [itex]v_{pf}[/itex]) / I
    [where, [itex]w_{f}[/itex] is the final angular velocity]


    - find the final velocity of the body using the formula:

    [itex]v_{f}[/itex] = [itex]v_{pf}[/itex] - ([itex]w_{f}[/itex] x r)
    [since [itex]v_{pf}[/itex] = [itex]v_{f}[/itex] + ([itex]w_{f}[/itex] x r)]

    This doesn't seem to work out well. I am not well versed in application of cross products. I would like to know if this approach is valid and what further calculations need to be performed in my approach.

    Any help is very much appreciated.

    Thanks in advance.
     
    Last edited: Jul 1, 2012
  2. jcsd
  3. Jul 1, 2012 #2

    mfb

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    In general, I would not expect this.


    This formula has wrong units. I has a mass, the other three quantities have not. You hide this error by using dimensionless quantities, but it is still there.


    Assuming a perfect elastic collision with a point-like "wall", the direction of momentum transfer is orthogonal to the surface of the body, with an unknown magnitude. You can now use energy and angular momentum conservation (use the collision point as center) to solve the equations.
     
  4. Jul 1, 2012 #3
    Thanks for the reply.

    I thought in a perfectly elastic collision, the point would travel with an equal and opposite momentum. Am i this wrong?


    Sorry, this is the actual formula i am using:
    [itex]w_{f}[/itex] = m * (r x [itex]v_{pf}[/itex]) / I
    I did not include m since m is 1.

    I'm at loss what how to proceed with your method. Should energy conservation be taken into account to solve this problem or is it possible without factoring it in?
     
  5. Jul 1, 2012 #4

    mfb

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    It might be possible to derive effective formulas which do not look like energy conservation. But these formulas would still contain it.

    I am not sure, but I highly doubt this for any specific point in a solid object.
     
  6. Jul 1, 2012 #5
    I have some confusion with conservation of angular momentum. Since the rectangular body has zero angular momentum initially and gains angular momentum after collision, how do i relate the pre-collision and post-collision angular momentums in this case?
     
  7. Jul 1, 2012 #6

    mfb

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    It has no angular momentum with respect to its center of mass. However, it has an angular momentum with respect to the collision point, unless its velocity and "center of mass <-> collision point" are aligned.
     
  8. Jul 1, 2012 #7
    how do i calculate angular momentum around point of collision?
    using L = r x p ?
     
  9. Jul 1, 2012 #8

    mfb

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    This is the angular momentum from the translational movement, if you use the momentum of the whole body. In this case, you have to add the rotation around the center of mass.
     
  10. Jul 1, 2012 #9
    But rotation around center of mass is zero. As i have stated, [itex]w_{i}[/itex] = 0.
    Do i have to use another equation?
     
  11. Jul 2, 2012 #10

    mfb

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    No, you just have to add 0 for the initial state (=>you can ignore it). However, you have to consider it for the final state after the collision.
     
  12. Jul 2, 2012 #11
    So you mean to say that L[itex]w_{i}[/itex] + L[itex]w_{f}[/itex] = 0 ? I'm not able to understand/visualize this. Can you please give it in an equation form?
     
  13. Jul 2, 2012 #12

    mfb

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    [itex]m r \times v_i + I \omega_i = m r \times v_f + I \omega_f[/itex]
    With [itex]\omega_i=0[/itex] in your case. r is the vector between center of mass and collision point, if you use the position at the time of the collision it is the same for initial and final state.

    In a similar way, energy conservation:
    [itex]m v_i^2 + I \omega_i^2 = m v_f^2 + I \omega_f^2[/itex] (multiplied by 2 to remove the common 1/2)
     
  14. Jul 2, 2012 #13
    I'd say you can't solve it if you want to approximate the wall as immovable. Laws of conservation don't work if you don't compute the energy/momentum/angular momentum transferred to the wall. Actually conservation of energy can work (think of a frontal elastic collision) but the other two don't. And I think you need all three of them to solve it.
    It's been a long time since I last tackled this kind of problems though, so I may be wrong. Is there a quicker way than finding the limit for mass of the wall [itex]\rightarrow\infty[/itex]?
     
  15. Jul 3, 2012 #14
    I didn't think of considering [itex]m r \times v_i [/itex] as part of initial angular momentum before. I just tried using this equation, but it does not yield the correct post collision angular velocity [itex]w_{f}[/itex]. Using distance vector r = (1.1, 1.4), m = 1, I = 4.41, [itex]v_{i}[/itex] = (5,0) and [itex]v_{f} = (-5, 0)[/itex], i get a negative angular velocity where a positive final angular velocity is expected.
    Here's the type of collision i'm talking about: http://img513.imageshack.us/img513/4419/expq.png [Broken]

    The red object is the rigid body and the black object is the wall.

    Am i applying something wrong?

    I don't think so there is, or atleast i am not aware of any other way. In my case, the wall needs to immovable, so i've taken it to have infinite mass compared to the object. Thats how i have seen in some books solving for the momentum of such system.
    Do you have a suggestion on how i should tackle this?
     
    Last edited by a moderator: May 6, 2017
  16. Jul 3, 2012 #15

    mfb

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    The momentum transfer occurs at the collision point only, and therefore transfers no angular momentum with respect to this point.
    => You can use the conservation of angular momentum.

    @alterecho: Your final velocity looks wrong. Energy conservation would force a zero rotational motion afterwards.
     
  17. Jul 4, 2012 #16
    You're right. The thing is, i've been modeling the translational and rotational motion based on assumptions. I really don't know how a rectangular body will react on collision with a wall (in absence of gravity). So how do you think this rectangular body will move after colliding with the wall?
    I'll link the visualization of the case again: http://img513.imageshack.us/img513/4419/expq.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  18. Jul 4, 2012 #17

    mfb

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    With conserved rotational momentum around the collision point and conserved energy in each collision.
    In the case of the screenshot, I would expect two independent collisions, one at the upper edge and probably one at the lower edge later (depends on the geometry).
     
  19. Jul 4, 2012 #18
    That seems to be a good expectation but that image is just rough representation. I'm at loss at how to proceed further. I'll have to think of another way taking all that you said into consideration. Please do let me know if you have a suggestion or a method on how to proceed.
     
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