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Impulsive reaction

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    (see attachment)


    2. Relevant equations



    3. The attempt at a solution
    I am clueless here. I can conserve angular momentum but how will i find the "impulsive reaction"?
     

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  2. jcsd
  3. Nov 13, 2012 #2

    haruspex

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    Let the impulsive reaction at A be I, a vector.
    How will the rod move just after the impact? Put in an unknown for that.
    You can now write down three equations: conservation of linear momentum (one vertical, one horizontal) and conservation of angular momentum.
     
  4. Nov 14, 2012 #3
    The rod should rotate about the hinged point A.
    Conserving linear momentum in horizontal direction:
    mv=mvx' (vx' is the horizontal velocity of rod after impact)
    vx'=v (is this correct?)

    I am not sure how conserving the linear momentum in vertical direction would help? Initially it is zero and after the impact, even if we assume that the rod has got some velocity in vertical direction, the velocity of rod in vertical direction turns out to be zero.
    0=mvy' (vy' is the velocity of rod in vertical direction)

    I am still not sure how to proceed further.
     
  5. Nov 14, 2012 #4

    haruspex

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    Yes, ultimately, but instantaneously what will be the motion of its c of g? Will it move horizontally, vertically, or somewhere between?
    No, you're still overlooking the reactive impulse at the hinge. You have to think of the rod as effectively struck by the hinge at the same instant in a way that leads to the expected subsequent motion. (If not, the top of the rod should move away from the hinge.) So in which direction will the reactive impulse act?
     
  6. Nov 14, 2012 #5
    I suppose it should move somewhere between.

    I still don't get it. :(
     
  7. Nov 14, 2012 #6
    Anyone?

    Are there any similar problems discussed here at PF?
     
  8. Nov 14, 2012 #7

    ehild

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    The angular momentum is conserved as there is no external torque acting on the system rod-particle. But the linear momentum is not conserved, as there is a force at the hinge during the impact.
    The change of linear momentum is m(rod) vcm(rod)-mv=ΔI where ΔI is the impulse of the force at the hinge. You can call it "impulsive reaction".
    So find the angular velocity of the rod after the impact,, you get the linear velocity of the CM from it.

    ehild
     
  9. Nov 14, 2012 #8
    Thanks ehild! That solved the problem and i got the right answer. :smile:
     
  10. Nov 14, 2012 #9

    ehild

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  11. Nov 14, 2012 #10
  12. Nov 14, 2012 #11

    ehild

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    The change of linear momentum is equal to the impulse of the external force. Here the force comes from the hinge, but that is opposite to the force the rod exerts on the hinge. ΔP=F(external)Δt can also be written as Pi=Pf+F(system)Δt, and that looks as conservation of impulse or momentum, just like when they speak about conservation of energy when friction does work. I would not say it conservation of energy or conservation of momentum, but other people do. Haruspex wanted you to include the impulse from the hinge when you wrote the change of linear momentum, he only called the law ΔP=F(external)Δt "conservation of momentum".

    ehild
     
  13. Nov 14, 2012 #12

    haruspex

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    There are basically two ways of solving it:
    Conservation of linear momentum (only need to consider horizontal since there is no vertical motion involved) gives you an equation involving the reactive impulse, I, and the speed of the rod directly after impact.
    The second equation is from conservation of angular momentum, but what equation you get will depend on where you take moments. If you take moments about the hinge it will not involve I; if you take them about the c of g of the rod it will.
    Either way, you end up with the same answer.
     
  14. Nov 14, 2012 #13

    ehild

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    Haruspex, why do you call the law "Change of momentum = impulse " "conservation of momentum"? There is an external force, so the momentum is not conserved.


    ehild
     
  15. Nov 14, 2012 #14

    haruspex

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    Does it contravene standard terminology? If so, I apologise for any confusion. To me it's the same. Initial quantity of x + input quantity of x = resulting quantity of x is a conservation law, whether x be momentum, energy, charge...
     
  16. Nov 14, 2012 #15
    Thanks you both for the replies, i understand it better after this discussion. :smile:
     
  17. Nov 14, 2012 #16

    ehild

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    Well, I saw that at several places already that people speak about conversation of energy when there is work of friction, or work of a man, that is work of non-conservative forces. I also saw using the word "impulse" for momentum mv. Work and energy are different concepts, impulse and momentum are also different in principle, just like internal energy is different either from heat or work, although the impulse is "input" for momentum and work is "input" for energy. Physicist do not use "input" so frequently as engineers.

    ehild
     
  18. Nov 15, 2012 #17
    Well , I suppose haruspex , meant impulse-momentum conservation and not just the "conservation of momentum." In fact "conservation of momentum" is a special case of impulse momentum conservation. Impulse momentum equation is given by :

    Δp = t1 t2∫Fexdt

    Where ∫Fexdt is just impulse.... If impulse =0 , we get Δp=0 , which is law of conservation of linear momentum...
     
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