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Homework Help: Strange question in mechanics

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle of mass 1[kg] is moving with a velocity of: [tex]\vec{V}=2 \hat{x}+ \hat{y} -4 \hat{z} [m/s] [/tex]. At the instant t=0, the particle experiences a force which is given by:
    [tex] \vec{F}= \vec{A} \times \vec{V} - (\vec{A} \cdot \vec{V}) \hat {A} [/tex] where [tex] \vec{A}= \hat{x} -6 \hat{y} + 2 \hat{z} [/tex].
    What is the kinetic energy of the particle after a very long time?

    2. The attempt at a solution

    What I've done is to do these vectorial multiplications in order to obtain the force and from this expression I obtain the acceleration and by integration I obtain the time dependent velocity. However, if I let t approach infinity, then also the velocity will be infinity and the same thing for the kinetic energy, but this totally wrong.

    How can I solve this?
     
  2. jcsd
  3. Aug 27, 2010 #2

    D H

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    Please show your work. It's kind of hard to see where you went wrong if we can't see what you did.
     
  4. Aug 27, 2010 #3
    Ok...

    First, I found the force using the formula given:

    [tex] \vec{F} = 23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z} [/tex]

    So the acceleration is (m=1kg) the same:

    [tex] \vec{a} = 23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z} [/tex]

    Now, I integrate this last expression in order to obtain the time dependent velocity:

    [tex] \vec{v} = (23.65 \hat{x} -1.89 \hat{y} +16.29 \hat{z}) t+2 \hat{x}+ \hat{y} -4 \hat{z} [/tex]

    So, letting t approch infinity leads to an infinite velocity and hence an infinite kinetic energy, which is wrong because the answer is 8.74 [J]
     
  5. Aug 28, 2010 #4

    D H

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    The answer is correct.

    Your mistake is that you are assuming the force is constant. It isn't. It is a function of velocity.

    Hint: Think of the velocity as comprising components parallel to and normal to [tex]\vec A[/tex]. What does the force equation say will happen to those two components?
     
  6. Aug 28, 2010 #5
    Aha, I think I got you but i'm still not getting the right answer =/

    In the direction parallel to [tex]\vec{A}[/tex], a forces: [tex]-(\vec{A} \cdot \vec{V})[/tex] acts on the body, hence the equation of motion is:

    [tex] \frac{dV_{||}}{dt}=\frac{-\vec{A} \cdot \vec{V}}{m} [/tex]

    and the solution is a decaying exponential, so this velocity doesn't contribute to the kinetic energy at a very long time.

    Now, in the direction perpendicular to [tex]\vec{A}[/tex], no force acts on the body so it moves with a constant speed, which is:

    [tex] V=\sqrt{2^{2}+1^{2}+(-4)^{2}} [/tex]

    and the kinetic energy is:

    [tex] K=0.5mV^{2}=10.5 [J] [/tex]

    What's wrong in this solution?
     
  7. Aug 28, 2010 #6

    D H

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    You have the exponential decay part correct, but you do not have the big picture yet.

    The vector V has components parallel to and normal to A. The force has two components, one is AxV, the other is -(A·V)Â. The first is normal to A and only the normal component of V contributes to this term. The latter is parallel to A and only the parallel of V contributes to this term.

    What happens to each of those terms over time?
     
  8. Aug 28, 2010 #7
    I have no clue. I'm totally confused now :|
     
  9. Aug 28, 2010 #8

    D H

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    What are these two components of the velocity vector, initially?
     
  10. Aug 28, 2010 #9
    Well, at t<0 the velocity vector was: [tex] \vec{V}=2 \hat{x}+ \hat {y}-4 \hat{z} [/tex] and at t=0 , when the force acts on the body, the time dependent velocity which is parallel to [tex] \vec{A} [/tex] is given by (after solving the 1st order ODE):

    [tex] v=v_{0} e^{-At/m} [/tex] where [tex]v_{0} = \vec{V} \cdot \hat{A} = -1.8741 [/tex].

    Regarding the velocity which is perpendicular to [tex] \vec{A} [/tex], all I know is that it's a constant because no forces act perpendicular to [tex] \vec{A} [/tex].

    This is all I know ... I think.
     
    Last edited: Aug 28, 2010
  11. Aug 28, 2010 #10
    anyone?
     
  12. Aug 28, 2010 #11

    vela

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    In symbols, you have

    [tex]\vec{V} = \vec{v}_\perp + \vec{v}_\parallel[/tex]

    where [tex]\vec{A}[/tex] defines the perpendicular and parallel directions. You've figured out how [itex]\vec{v}_\parallel[/itex] evolves with time. What about [itex]\vec{v}_\perp[/itex]?
     
  13. Aug 28, 2010 #12
    Well, [tex]\vec{v}_\perp=constant [/tex] because no forces act in this direction, so it should be the given [tex] \vec{V} [/tex]. No?
     
  14. Aug 28, 2010 #13
    wait a minute. [tex] \vec{v}_\perp= \vec{V} \cdot \hat{A}_{\perp} [/tex]

    now, [tex] \vec{A}_{\perp}=a*(6,1,0)+b*(-2,0,1) [/tex] where a and b are constants.

    So there are infinite vectors which are perpendicular to [tex]\vec{A} [/tex] ! How can I figure out which is the right one?
     
  15. Aug 28, 2010 #14

    vela

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    Think of [tex]\vec{v}_\perp[/tex] as the part of [tex]\vec{V}[/tex] that's not parallel to [tex]\vec{A}[/tex]. In other words,

    [tex]\vec{v}_\perp = \vec{V}-\vec{v}_\parallel[/tex]
     
  16. Aug 28, 2010 #15

    vela

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    Just wanted to add that this approach would work, but you have to add the constraint that [tex]\hat{A}_\perp[/tex] lies in the plane defined by [tex]\vec{A}[/tex] and [tex]\vec{V}[/tex].
     
  17. Aug 28, 2010 #16

    vela

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    To be more precise, its magnitude is constant. The vector, however, isn't, because its direction changes with time.
     
  18. Aug 28, 2010 #17
    well, in that case after a very long time the parallel component drops downs and we're left with,

    [tex] \vec{v}_{\perp} = \vec{V} [/tex]

    so the perpendicular component equals the velocity given.

    Something stinks here for sure ... I can feel it ... but what it that? :D
     
  19. Aug 28, 2010 #18
    [tex] \hat{A}=\frac{1}{\sqrt{41}} (1,-6,2) [/tex]

    [tex] \vec{v}_{||}=\frac{v_{0}e^{-At/m}}{\sqrt{41}}(1,-6,2) [/tex]

    [tex] \vec{v}_{\perp} = (2,1,-4) - \frac{v_{0}e^{-At/m}}{\sqrt{41}}(1,-6,2) [/tex]

    so as [tex] t \to infinity [/tex]

    [tex] \vec{v}_{\perp} = (2,1,-4) [/tex]

    Is this, so far, right?
     
  20. Aug 28, 2010 #19

    D H

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    No! You do not know what the velocity vector is as t gets large. The reason is simple: It isn't constant.

    Fortunately, the velocity vector does not have to be constant to solve this problem.
     
  21. Aug 28, 2010 #20
    Then how am I supposed to solve it? :|
     
  22. Aug 28, 2010 #21

    vela

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    The particle has velocity [itex]\vec{v}(t) = \vec{v}_\perp(t)+\vec{v}_\parallel(t)[/itex]. The mistake you're making is assuming [tex]\vec{v}(t) = \vec{V}[/tex] for t>0. It's only true for t=0. After t=0, the applied force causes the velocity of the particle to change so it's no longer equal to [tex]\vec{V}[/tex].
     
  23. Aug 28, 2010 #22
    Let me arrange this stuff in my head:

    As you guys told me, the expression for [tex] \vec{v}_{||} [/tex] is correct. Now, we all agree that [tex] \vec{v}_{\perp} [/tex] is a constant ... right?

    Now from this point, what am I supposed to do? ... I'm not assuming anything, it's just that as far as I know, when no forces exist in some direction then the corresponding velocity is constant.
     
  24. Aug 28, 2010 #23

    D H

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    No!

    How can it possibly be constant? Look at the force equation.
     
  25. Aug 28, 2010 #24
    I'm looking and going nuts ...

    The second part of the force equation is already used for the parallel velocity component so I guess you're refering to the first one: [tex] \vec{A} \times \vec{V} [/tex].

    [tex]\vec{A} \times \vec{V} = \vec{A} \times (\vec{v}_{||} + \vec{v}_{\perp} ) \to \vec{A} \times \vec{V} = \vec{A} \times \vec{v}_{\perp} [/tex]

    all I can say is that the result is a vector which is perpendicular to the plane formed by [tex]\vec{A}[/tex] and [tex]\vec{v}_{\perp}[/tex] ... WHAT AM I MISSING HERE?!?!?!?!
     
  26. Aug 28, 2010 #25

    vela

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    What does a force that's perpendicular to velocity do?
     
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