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?

 

[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