1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Polar Coordinates Problem

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    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:

    [​IMG]

    [​IMG]

    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.
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Polar Coordinates Problem
Loading...