What are the Polar Coordinates for this Problem in Homework Statement?

[steven10137]

Homework Statement

http://img127.imageshack.us/img127/2695/coord2pq5gm6.jpg

Homework Equations

\underline v = \dot r\;\underline e _r + r\dot \theta \;\underline e _\theta
\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

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

http://img359.imageshack.us/img359/3100/coord2pq5diag2xt7.jpg

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

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

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.

 

[steven10137]

bump.
Does anybody have any suggestions?

 

[steven10137]

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!

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

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

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

thanks anyways.