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Help with a curve problem

  1. Mar 25, 2004 #1

    I'm confused about the above problem because I don't know why there would be an a,b, or greek letter "w" in there. Anyone?
  2. jcsd
  3. Mar 25, 2004 #2

    matt grime

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    They're constants. That's all. Position is still a function of t. w might well work out to be something like angular momentum.
  4. Mar 25, 2004 #3
    Looks like helical motion... the kind of motion you would if you gave a particle an inital velocity inside a magnetic field, when there's angle between the velocity and the field.

    In the XY plane the particle is going in a circle, the radius of which is [itex]a[/itex], and the angular velocity of the particle in that plane is [itex]\omega[/itex] (omega). The velocity in that plane is:
    [tex]v_r = \omega a[/tex]

    Along the Z axis, the particle is traveling at a constant speed of [itex]b[/itex]. Since the velocities in the Z axis and in the XY plane are perpendicular, the magnitude of the velocity of the particle is:
    [tex]V = \sqrt{b^2 + v_r^2} = \sqrt{b^2 + \omega ^2a^2}[/tex]

    As for the acceleration, it only exists in the XY plane (since the motion along the Z axis is at a constant speed). The acceleration in a circular motion is given by:
    [tex]a_r = \omega ^2a[/tex]

    The angle [itex]\theta[/itex] between the velocity and the acceleartion, is the angle between the velocity and the XY plane.
    Last edited: Mar 25, 2004
  5. Mar 25, 2004 #4


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    'w' (that fancy w) is angular velocity, and is measured in radians per second. 'w = 2*pi*f', where f is the frequency in Hz.
  6. Mar 25, 2004 #5
    Thanks for the info... it's funny this problem is nothing like any of teh other problems i've encountered in this section of my math text book.
  7. Mar 25, 2004 #6
    I didn't notice this is a math problem... it does include some physics but you can also prove it mathematically.

    For example, you know that the change of position is velocity right? So if:
    [tex]x = a\cos (\omega t)[/tex]
    [tex]y = a\sin (\omega t)[/tex]
    Derive both to get the velocity in each axis:
    [tex]x' = v_x = -\omega a\sin (\omega t)[/tex]
    [tex]y' = v_y = \omega a\cos (\omega t)[/tex]
    To find the "total" velocity in the XY plane, use pythagoras:
    [tex]v_{xy}^2 = v_x^2 + v_y^2 = \omega ^2a^2(\sin ^2(\omega t) + \cos ^2(\omega t) = \omega ^2a^2[/tex]
    [tex]v_{xy} = \omega a[/tex]
    Just like I said above, but here you actually proved it. :smile:
  8. Mar 25, 2004 #7
    thanks. I'm aware of how to obtain the velocity, speed and acceleration vectors, however I just didnt know why he would include a,b,w. doesnt make much sense to me.
  9. Mar 26, 2004 #8
    question, i understand the way you worked this, however, why would i have to calculate the velocities separately for each axis?

    wouldnt the position vector be r(t) = acos(wt)i + asin(wt)j + btk ?

    then i would just do dr/dt to find the velocity vector wouldnt i? and dv/dt to find the acceleration?

    im only questioning this because if i do it the way i mentioned above, for speed i end up with

    sqroot( w^2 a^2 + b^2 ) which doesnt seem right since it would make the rest of the questions quite messy. but i've checked it and double checked it and thats what I come up with.
    Last edited: Mar 26, 2004
  10. Mar 27, 2004 #9
    The speed is indeed [tex]\sqrt{\omega ^2a^2 + b^2}[/tex], you got it exactly right (this is also what I said in post #3).
  11. Mar 27, 2004 #10
    can i ask what programs you guys use to write out your mathematical formulas? i realize they are images.. so you must be making them some where.

    edit: by the way, thanks chen
    Last edited: Mar 27, 2004
  12. Mar 27, 2004 #11
  13. Mar 27, 2004 #12


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    We use LaTex. There are stickies around which tell you how to format. These aren't images but are created inline thus [ tex ] 2\pi r^2 [ / tex] becomes, if you remove the blanks, [tex] 2\pi r^2 [/tex]
  14. Mar 27, 2004 #13
    oh ok, thanks
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