Polar Coordinates: Show Acceleration Angle of 30°

[devious_]

A particle P describes the curve with polar equation $r = a e^{\theta \sqrt{3}} \cosh 2\theta$ in such a manner that the radius vector from the origin rotates with uniform angular speed $\omega$. Show that the resultant acceleration of the particle at any instant makes an angle of 30 degrees in the radius vector.

Any ideas? :yuck:

 

[PhilG]

Use the polar unit vectors:

$$\hat{\mathbf{r}} = \cos \theta \hat{\mathbf{i}} + \sin \theta \hat{\mathbf{j}} \ \ \ \ \ \hat{\mathbf{\theta}} = -\sin \theta \hat{\mathbf{i}} + \cos \theta \hat{\mathbf{j}}$$

The nice thing about these is that you can start out with the position vector

$$\mathbf{r} = r \hat{\mathbf{r}}$$

and take derivatives of it to find the velocity and acceleration vectors. You are given r as a function of theta, and $$\dot{\theta} = \omega$$, so this should allow you to express all vectors in terms of theta.

 

[devious_]

I'm still not getting anywhere. Can you please show me the first couple of steps?

 

[devious_]

Right. I've tried some more and here's where I ended up:

$$\text{acceleration} = \langle \ddot{r} - r \dot{\theta ^2}, 2\dot{r}\dot{\theta} + r\ddot{\theta} \rangle$$

Now I've got to differentiate my expression for $r$ and use the fact that $\dot{\theta} = \omega$.

Is that correct, or is there a better method?

 

[OlderDan]

Right. I've tried some more and here's where I ended up:

$$\text{acceleration} = \langle \ddot{r} - r \dot{\theta ^2}, 2\dot{r}\dot{\theta} + r\ddot{\theta} \rangle$$

Now I've got to differentiate my expression for $r$ and use the fact that $\dot{\theta} = \omega$.

Is that correct, or is there a better method?

Looks good so far. Don't forget that $$\ddot{\theta} = 0$$ in this problem

 

[devious_]

I've obtained an expression for the magnitude of the acceleration; how do I show that the particle makes an angle 30 at the radius vector?

 

[PhilG]

compute dot product of acceleration and the radius vector two different ways, and set them equal.

 

[devious_]

Thanks