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Homework Help: Polar Coordinates Problem

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data
    http://img127.imageshack.us/img127/2695/coord2pq5gm6.jpg [Broken]

    2. Relevant equations
    [tex]\underline v = \dot r\;\underline e _r + r\dot \theta \;\underline e _\theta[/tex]
    [tex]\underline a = \left( {\ddot r - r\dot \theta ^2 } \right)\underline e _r + \left( {r\ddot \theta + 2\dot r\dot \theta } \right)\underline e _\theta[/tex]

    3. The attempt at a solution
    Hi guys, been stuck on this question for a while now.
    Here are the diagrams I made:

    http://img183.imageshack.us/img183/9205/coord2pq5diag1rx9.jpg [Broken]

    http://img359.imageshack.us/img359/3100/coord2pq5diag2xt7.jpg [Broken]

    So from the second diagram:
    [tex]\begin{array}{l}
    \cos 15 = \frac{{ - \dot r}}{v} \\
    \therefore \dot r = - v\cos 15 = - 50\cos 15 = - 48.296\;ms^{ - 1} \\
    \end{array}[/tex]
    then
    [tex]\begin{array}{l}
    \sin 15 = \frac{{ - r\dot \theta }}{v} \\
    \therefore \dot \theta = \frac{{ - v\sin 15}}{r} = \frac{{ - 50\sin 15}}{{770}} = - 0.017\;\deg s^{ - 1} \\
    \end{array}[/tex]

    We are told in the question that the acceleration of P is zero, hence the components of acceleration must be zero:
    [tex]\begin{array}{l}
    0 = r\ddot \theta + 2\dot r\dot \theta \\
    \Rightarrow r\ddot \theta = - 2\dot r\dot \theta \\
    \therefore \ddot \theta = \frac{{ - 2\dot r\dot \theta }}{r} = \frac{{ - 2 \times - 48.296 \times - 0.017}}{{770}} = - 0.002\deg s^{ - 2} \\
    \end{array}[/tex]

    This isn't the right answer.
    I think it could be do with my angles used because I thought angular velocity was meant to be measured in radians. All help will be greatly appreciated.


    Thanks in advance.
    Steven.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Aug 7, 2008 #2
    bump.
    Does anybody have any suggestions?
     
  4. Aug 8, 2008 #3
    Oh my god, I answered my own question.

    I changed all my angles to radians and put them into the the same equations, gave my answer in rad/s^2 and bingo!

    [tex]{\dot r = - 50\cos \left( {\frac{\pi }{{12}}} \right) = - 48.296\:ms^{ - 1} }[/tex]

    [tex]{\dot \theta = \frac{{ - 50\sin \left( {\frac{\pi }{{12}}} \right)}}{{770}} = - 0.017\:rad\;s^{ - 1} }[/tex]

    [tex]{\ddot \theta = \frac{{ - 2\dot r\dot \theta }}{r} = \frac{{ - 2 \times - 48.296 \times - 0.017}}{{770}} = - 0.002\;rad\;s^{ - 2} }[/tex]

    thanks anyways.
     
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