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Classical Mechanics

  1. Jan 4, 2006 #1
    given a particle of mass m where r is the position vector for its trajectory
    F is the force it experiences
    p is its momentum
    r0 represents the initial position vector
    and r0 F, p are all constant vectors
    show that
    [tex] \vec{r}' = \frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t [/tex] where
    [tex] \vec{r}' = \vec{r} - \vec{r_0} [/tex]
    represents teh parametric equations of a parabola if F is not parallel to p
    ok so the angular momentum of the particle is
    [tex] \vec{L} = \vec{r}' \times \vec{p} = (\frac{1}{2} \frac{\vec{F}}{m} t^2 + \frac{\vec{p}}{m} t) \times \vec{p} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) + \frac{t}{m} (\vec{p} \times \vec{p}) [/tex]

    now certainly p cross p is zero
    and we end up with
    [tex] \vec{L} = \frac{t^2}{2m} (\vec{F} \times \vec{p}) [/tex]
    and this is not zero because F is not parallel to p
    so this is a parabola because it is second order with responect to t?
    this is the part that im not too sure about...
    Is this right at all first of all?

    Thank you for your help!
     
    Last edited: Jan 4, 2006
  2. jcsd
  3. Jan 4, 2006 #2

    George Jones

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    I don't understand why you're looking at angular momentum. Why don't you just integrate both sides of Newton's second law twice with respect to time?

    How can force and monentum both be contant? Did you mean to say that p is the initial momentum?

    Regards,
    George
     
    Last edited: Jan 4, 2006
  4. Jan 4, 2006 #3
    are you saying that I should integrate [tex] \frac{dp}{dt} = F [/tex] with respect to time twice?

    I dont understand how that would relate the question at hand

    r0, F and p are the initial values
    Im trying to prove that r' is the parametric equation of a parabola...
     
  5. Jan 4, 2006 #4

    George Jones

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    Sorry, I miunderstood your original question. One way to do this is to choose an xyz coordinate system such that: F is along the y-axis; FxP is along the z-axis; z'_0 = 0. Then it is fairly easy to show that r' = (x', y', z') is such that z' = 0 and y' = Ax' + Bx'^2, where A and B are constants.

    Regards,
    George
     
  6. Jan 4, 2006 #5
    how or why are we making the assumption that z0 = 0?? Isn't what you're suggesting making hte problem lose its generality?
     
  7. Jan 5, 2006 #6

    Tide

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    All you have to do is integrate the equation

    [tex]\frac {d^2 (\vec r - \vec r_0)} {dt^2} = \frac {\vec F}{m}[/tex]

    when [itex]\vec F[/itex] is a constant vector. Your result should follow with little effort.
     
  8. Jan 5, 2006 #7

    George Jones

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    The assumption z0 = 0 is not being made - you're equation implies that when t = 0, r' = 0, so z'0 = 0. Notice the prime.

    It seems that Tide has interpreted your question the same way that I initially did. Let's see if I understand correctly now - you are *not* asking how to show how r' = F/(2m) t^2 + p/m t follows if F is constant.

    Regards,
    George
     
  9. Jan 5, 2006 #8

    Tide

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    George,

    Yes, I did interpret the problem that way. I see now that stunner is trying to prove that the path is a parabola.

    One way to do it is to show there exists a vertex and a directrix such that points on the trajectory are equidistant from the vertex and corresponding points along the directrix.
     
  10. Jan 5, 2006 #9
    i think i like tide's suggested method. First question i asked myself was how would you show something was a parabola - using a vertex and a directrix
     
  11. Jan 5, 2006 #10

    George Jones

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    Let me expand on my previous comments.

    First, choose an orthonormal basis [itex]\left\{ e_{1}, e_{2}, e_{3} \right\}[/itex] for space that exploits the symmetries of the problem.

    [itex]\vec{r}' \cdot \left( \vec{F} \times \vec{p} \right) = 0[/itex] gives that the motion takes place in the plane spanned by [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex], so choose one of the basis vectors, [itex]e_{3}[/itex] say, perpendicular to this plane. Consequently, the other 2 basis vectors are in the plane spanned by [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex].

    It would be nice to choose the other basis vectors to be in the directions of [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex], but we aren't given that [itex]\vec{F}[/itex] and [itex]\vec{p}[/itex] are orthogonal to each other. The best we can do is choos one of the basis vectors along either [itex]\vec{F}[/itex] or [itex]\vec{p}[/itex].

    Which one? Think of the physical situation - motion with a constant force. An example of this is motion under the influence of gravity (and neglecting air resistance) near the EartH's surface. A thrown ball follows a parabola, with the symmtry axis of the parabola parallel to the force of gravity. Thus, choose [itex]e_{2}[/itex] to be in the direction of [itex]-\vec{F}[/itex], and [itex]e_{1}[/itex] to be perpendicular to both [itex]e_{2}[/itex] and [itex]e_{3}[/itex]. Hence, motion is confined to the plane spanned by [itex]e_{1}[/itex] and [itex]e_{2}[/itex].

    Since [itex]\vec{p}[/itex] is in the plane spanned by [itex]e_{1}[/itex] and [itex]e_{2}[/itex], we can write

    [tex]\vec{p} = p \mathrm{cos} \theta e_{1} + p \mathrm{sin} \theta e_{2},[/tex]

    where [itex]\theta[/itex] is the constant angle between (the initial momentum) [itex]\vec{p}[/itex] and [itex]e_{1}[/itex].

    We now have

    [tex]\vec{r}' = x' e_{1} + y' e_{2} + z' e_{3} = - \frac{F}{2m} t^2 e_{2} + \frac{p}{m} t \mathrm{cos} \theta e_{1} + \frac{p}{m} t \mathrm{sin} \theta e_{2}.[/tex]

    Note that [itex]z' = 0[/itex]. Equate components of [itex]e_{1}[/itex] on both side, and solve for [itex]t[/itex] in terms of [itex]x'[/itex]. Use this to eliminate all t's in the expression.

    What do you get?

    Are you at U of T or York? My wife graduated from both.

    Regards,
    George
     
    Last edited: Jan 5, 2006
  12. Jan 5, 2006 #11
    so the basic idea of this problem is to show that y is proportional to x of the second order, right? Because such motion describes a parabola.

    i am at york

    thank you for the help
     
  13. Jan 5, 2006 #12

    George Jones

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  14. Jan 12, 2006 #13
    oh i have a question... why is it -F?? Is it a typo or is there a specific reason for it ? Convenience?
     
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