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Math Help

  1. Oct 5, 2005 #1
    I'm suppossed to find the average acceleration of the mass during the time interval from -t to t.
    The formula I'm using is [tex]a(-t,t)=\frac{(v(t)-v(-t))}{(t-(-t))} [/tex]

    I've already derived v(t) and v(-t) but I'm not sure how to simplify them after subtracting them.

    [tex]v(t)=-R\omega sin(\omega t)*\hat{x}+R\omega cos(\omega t)\hat{y}[/tex]
    [tex]v(-t)=-R\omega sin(\omega (-t))\hat{x}+R\omega cos(\omega (-t))\hat{y}[/tex]

    I got as far as grouping the x and y terms...after that I don't know how else to simplify

    [tex][(-R \omega \hat{x}) ((sin(\omega t)+sin(\omega (-t))] + [(R \omega \hat{y}) ((cos \omega t) -cos (\omega (-t))] [/tex]

    Any help is appreciated.
    thanks!
     
  2. jcsd
  3. Oct 5, 2005 #2
    from what you have, i get that the velocities all cancel out. [tex] sin(\omega t) + sin (-\omega t) = sin(\omega t) -sin(\omega t)=0[/tex] because [tex] sin(-\omega t)= -sin(\omega t)[/tex]

    ok, well just the sine's cancel...

    the cosines simplify [tex] R\omega cos(\omega t) - R\omega cos(-\omega t)= R\omega(cos(\omega t) - cos(-\omega t))= R\omega( cos(\omega t) + cos(\omega t))= 2R\omega cos(\omega t)[/tex]

    and thats all in the y hat direction. so all your acceleration is vertical.
     
    Last edited: Oct 5, 2005
  4. Oct 5, 2005 #3
    Thanks Gale!

    I didn't know this rule: sin(-x) = -sin (x) :redface:
     
  5. Oct 5, 2005 #4
    hah, don't feel so bad... i didn't either, but i drew it out. when you look at the picture, obviously they're equal.
     
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