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Homework Help: Showing acceleration is constant

  1. Jan 30, 2009 #1
    I hope I posted in the right place. Sorry in advanced.

    1. The problem statement, all variables and given/known data
    A buzzing fly moves in a helical path given by the equation
    r(t) = ib sin [tex]\omega[/tex]t + jb cos [tex]\omega[/tex]t + kct[tex]^{2}[/tex]
    Show that the magnitude of the acceleration of the fly is constant, provided b, [tex]\omega[/tex], and c are constant.

    3. The attempt at a solution
    x = b sin [tex]\omega[/tex]t
    y = b cos [tex]\omega[/tex]t
    z = ct[tex]^{2}[/tex]

    In class we did a similar problem, but in that problem we had to find the trajectory in space. I'm a little slow, but it's just not helping me with this one. Same with the textbook. I'll go to my teacher if I have to.

    I'm not asking someone to do the problem, just get me started. Once that happens, I'll try to go over it here in case I have more questions. Thank you!
     
  2. jcsd
  3. Jan 30, 2009 #2

    Dick

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    The acceleration is the second derivative of r(t) with respect to t. What is that?
     
  4. Jan 31, 2009 #3
    The first derivative:
    ib[tex]\omega[/tex] cos [tex]\omega[/tex]t - jb[tex]\omega[/tex] sin [tex]\omega[/tex]t + 2kct

    Second derivative:
    -ib[tex]\omega[/tex][tex]^{2}[/tex] sin [tex]\omega[/tex]t - jb[tex]\omega[/tex][tex]^{2}[/tex] cos [tex]\omega[/tex]t + 2kc

    Is that right?
     
  5. Jan 31, 2009 #4

    Nabeshin

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    Now find the magnitude...
     
  6. Jan 31, 2009 #5

    gabbagabbahey

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    Sure; that;s correct :approve:....But very ugly!:yuck::wink: Try writing the entire equation inside the [ tex] or [ itex] tags instead:

    [tex]\mathbf{a}(t)=-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}[/tex]

    (You can click on the above equation to see the code that generated it)

    Now, as Nabeshin said, calculate the magnitude :smile:
     
  7. Feb 2, 2009 #6
    Thank you guys :)

    Now what is the first step in calculating the magnitude? I am used to plugging in numbers to do that.
     
  8. Feb 2, 2009 #7

    gabbagabbahey

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    You know what the x,y, and z-components of a are, so square them, add the squares, and take the square root as per usual.

    [tex]||\mathbf{a}||=\sqrt{a_x^2+a_y^2+a_y^2}[/tex]
     
  9. Feb 2, 2009 #8
    So I take [tex]\mathbf{a}[/tex] (the second derivative above), and factor in x for the first part. So it would look like this:

    [tex]\mathbf{a_x}(t)=[-b\omega^2\sin(\omega t)\mathbf{i}-b\omega^2\cos(\omega t)\mathbf{j}+2c\mathbf{k}*\mathbf{b}sin(\omega t)]^2[/tex], where [tex]\mathbf{b}sin(\omega t)[/tex] is x.

    Then do the same thing for y and z, add the terms up, and take the square root. Am I on the right track? Then I simplify as much as possible?
     
  10. Feb 2, 2009 #9

    gabbagabbahey

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    Huh?!:confused:

    No! [itex]a_x[/itex] is the x-component of a....that's just [itex]-b\omega^2\sin(\omega t)[/itex]....what are
    [itex]a_y[/itex] and [itex]a_z[/itex]?
     
  11. Feb 2, 2009 #10
    Thank you. I always over complicate things. I believe I know how to do it now.

    To answer your question, y is [tex]-b\omega^2\cos(\omega t)[/tex] and z is [tex]2c[/tex]
    Now I square them, add them up, and take the square root of that sum (this:[tex]||\mathbf{a}||=\sqrt{a_x^2+a_y^2+a_y^2}[/tex]). Correct?
     
  12. Feb 2, 2009 #11

    D H

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    Correct. So what is the result?
     
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