# Polar Coordinates Problem

1. Aug 7, 2008

### steven10137

1. The problem statement, all variables and given/known data

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

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

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

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

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.

Steven.

2. Aug 7, 2008

### steven10137

bump.
Does anybody have any suggestions?

3. Aug 8, 2008

### 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.