- #1
steven10137
- 118
- 0
Homework Statement
http://img127.imageshack.us/img127/2695/coord2pq5gm6.jpg
Homework 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]
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:
[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.
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