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Trajectories with Conserved Quantities

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle moves along a trajectory with constant magnitude of the velocity |[itex]\stackrel{→}{v}[/itex]|=[itex]\stackrel{→}{v0}[/itex] and constant angular momentum L⃗ = L⃗0. Determine the possible trajectories.

    2. Relevant equations

    d(L⃗)/(dt)=[itex]\stackrel{→}{N}[/itex] where [itex]\stackrel{→}{N}[/itex]=torque

    [itex]\stackrel{→}{N}[/itex]=[itex]\stackrel{→}{r}[/itex]x[itex]\stackrel{→}{F}[/itex]

    F=ma

    3. The attempt at a solution

    I thought that there would be 3 paths that it could take

    (i) a strait line if its in a strait line it can have constant velocity and no change in angular momentum, because torque is 0, because there is no Force.

    (ii) a circle because force is parallel to the radius, there is no torque, means no change in angular momentum, and uniform circular motion has a constant |v|.

    (iii) helix just like a combination of a strait line and a circle. both there is no change in angular momentum, so together there should be no change as well. (no force in any direction besides parallel to radius). also since there is constant |v| in the uniform circle, and now you are applying a constant v in a direction perpendicular to the v of the circle, the new |v| magnitude should be constant as well.

    So my question is, are these are correct? if so are they the only three? and also if so, how do I know there all no others? (is there anyway to prove these are the only 3)

    if it helps at all the title to this problem as my teacher gave is (Trajectory with conserved quantities)

    Thank you for your help!
     
    Last edited: Feb 18, 2014
  2. jcsd
  3. Feb 18, 2014 #2
    I would start by analysing the implications of constant angular momentum. What restrictions does that impose on the trajectory?
     
  4. Feb 18, 2014 #3
    I am thinking that implies that there is no torque, which would mean that the angular velocity remains constant? so that would mean that there either needs to me no external force or a force that is always in the direction of of the radius.
     
  5. Feb 18, 2014 #4
    Note that my question was about implications on the trajectory. How does that affect its shape?
     
  6. Feb 18, 2014 #5
    Well it would mean that the radius of the curve would either have to stay constant, or the velocity would need to change with it. So that combined with the constant velocity stipulation would mean that the particle would have to keep a constant radius from some origin.

    But I don't this isn't entirely true, because the angle between the radius and the velocity can change as the length of the radius changes, keeping rxmv a constant. I think this is the case for a strait line.
     
    Last edited: Feb 18, 2014
  7. Feb 18, 2014 #6
    I do not think it implies that "either - or" statement. I cannot see how you arrived at that conclusion, anyway. A more specific question, with a bit of a hint: what does that mean about mutual orientation of the position and velocity vectors?
     
  8. Feb 18, 2014 #7
    hmm. well the parallelogram that the vectors of position and velocity make would have to remain a constant area...

    if R*M*V*sin(theta) remains a constant that angular momentum is conserved. but wouldn't that mean that any of those 4 values could change if one of the other changes inversely to it. so the angle between the position and velocity vectors can change if say the velocity or radius changes. since V and M are constant in this example, then R*sin(theta)=constant so either R and theta stay the same, or the both change at the same rate.
     
  9. Feb 18, 2014 #8
    Remember that angular momentum is a vector. It has a direction. And because the vector is constant, the direction is constant. What does that mean about the position and the velocity? How are they related to the direction of the angular momentum?
     
  10. Feb 18, 2014 #9
    means that velocity vector and position vector always have to lie on the same 2 diminutional plane. which would mean angular momentum isn't constant in a helix.
     
    Last edited: Feb 18, 2014
  11. Feb 18, 2014 #10
    Exactly. (Dimensional is the correct word.) What does that tell you about the third options you enumerated originally?

    Now, there is a catch. What if angular momentum is zero? There is no direction in that case. Do we have a useful relationship between position and velocity in this case?
     
  12. Feb 18, 2014 #11
    so angular momentum would not be constant in the helix.
    If angular momentum is zero then the position and velocity vectors would either have to be parallel (or antiparallel) or one of them would have to be zero. doesn't make sense that a moving particles position or velocity vectors would be zero, so there fore they would have to be parallel.

    and since the origin location is arbitrary, for any strait line you can place the origin so the position vector and velocity vectors are always parallel.
     
  13. Feb 18, 2014 #12
    So, when angular momentum is zero, the trajectory is?

    Now let's go back to the case when it is not zero and we have a plane of motion. Because we have the plane, we can describe the trajectory in that plane using two coordinates, x and y. What is the magnitude of angular momentum in these coordinates? And what about the magnitude of velocity?
     
  14. Feb 18, 2014 #13
    Magnitude of velocity is Sqrt([itex]\dot{x}[/itex]^2+[itex]\dot{y}[/itex]^2)

    the magnitude of the angular momentum would be ([itex]\dot{x}[/itex]y- [itex]\dot{y}[/itex]x)m

    I got this just by taking the cross product of the velocity and position vectors

    v={[itex]\dot{x}[/itex],[itex]\dot{y}[/itex],0}

    r={x,y,0}
     
    Last edited: Feb 18, 2014
  15. Feb 18, 2014 #14
    Well, at this stage you should just try and solve the differential equations you got. If you get really stuck, come back.
     
  16. Feb 18, 2014 #15
    so i took the derivative of x'y-y'x and set it equal to zero. (no change in angular momentum) and got x''y=y''x.

    and the derivative of sqrt(x'^2+y'^2) and got ((x'x''+y'y'')/(sqrt(x'^2+y^2))=0

    the problem is i don't really know what I'm looking for. I am not sure what the equation can tell me about the trajectories or what value i am trying to solve for.

    how do I go about solving a diff equation were there are 2 variables both derived with respect to t?

    also wouldn't i need some type of initial condition? I really have no idea what to do at this point
     
    Last edited: Feb 18, 2014
  17. Feb 19, 2014 #16
    Very well. The final equation implies ## \dot x \ddot x + \dot y \ddot y = 0 ##. Its left hand side is the time derivative of ## \frac 1 2 (x^2 + y^2) ##.
     
  18. Feb 19, 2014 #17
    is that supposed to be ½([itex]\dot{x}[/itex]^2+[itex]\dot{y}[/itex]^2)? which would equal some constant.

    And I'm still unsure as to what that then tells me. Im really struggling to understand where this is headed. I'm assuming in the end I want some equation for r(x,y;t) Sorry for being so difficult lol
     
  19. Feb 19, 2014 #18
    In the end you should obtain an equation or equations linking ##x## and ##y## and some constants. That will give you the (families of) trajectories. I think you can test with the equations obtained in #13 that options (i) and (ii) of #1 are valid. But to prove that no other options are possible, you have to solve those equations.

    In my post #16, I made a mistake because I hoped very much to see ## x^2 + y^2 = c^2 ## (without dots) in your results. It can be derived from the basic equations of #13. And the steps you took - differentiating those equations - are steps in the right direction. You just need some more symbol gymnastics to get there.
     
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