# Particle along a spiral

1. Sep 8, 2007

### amstrf89

1. The problem statement, all variables and given/known data
A particle moves outward along a spiral. Its trajectory is given by r=Aθ, where A is a constant. A=1/π m/rad. θ increases in time according to θ=αt^2/2, where α is a constant.
Show that the radial acceleration is zero when θ=1/√2 rad. At what angles do the radial and tangential accelerations have equal magnitude?

2. Relevant equations

$$\frac{d\mathbf{r}}{dt} = \dot r\hat{\mathbf{r}} + r\dot\theta\hat{\boldsymbol\theta}$$
$$\frac{d^2\mathbf{r}}{dt^2} = (\ddot r - r\dot\theta^2)\hat{\mathbf{r}} + (r\ddot\theta + 2\dot r \dot\theta)\hat{\boldsymbol\theta}$$

3. The attempt at a solution

I tried plugging things into the second equation. However, I don't know what t and α are, which is why I am stuck. I also tried substitution, but that didn't really get me anywhere either.

Last edited: Sep 9, 2007
2. Sep 8, 2007

### learningphysics

What are the values of a and A?

3. Sep 8, 2007

### amstrf89

a is supposed to be alpha, an unknown constant. A is equal to 1/(pi) m/rad.

4. Sep 9, 2007

### learningphysics

The angular acceleration is: $$\frac{d^2R}{dt^2} - R(\frac{d\theta}{dt})^2$$. substitute your values for R and $$\theta$$ into this equation and simplify as much as you can.

Finally substitute in the value for theta 1/sqrt(2).

5. Sep 9, 2007

### amstrf89

Is angular acceleration the same as radial acceleration? Also, what is the equation for tangential acceleration? I just changed my original comment. The relevant equations are

$$\frac{d\mathbf{r}}{dt} = \dot r\hat{\mathbf{r}} + r\dot\theta\hat{\boldsymbol\theta}$$
$$\frac{d^2\mathbf{r}}{dt^2} = (\ddot r - r\dot\theta^2)\hat{\mathbf{r}} + (r\ddot\theta + 2\dot r \dot\theta)\hat{\boldsymbol\theta}$$

Last edited: Sep 9, 2007
6. Sep 9, 2007

### learningphysics

Oops, I meant to write radial acceleration... yeah, that formula I posted is radial acceleration.

tangential acceleration is: 2*dr/dt*w+r*d(w)/dt, where w = dtheta/dt... I looked this up here: http://aemes.mae.ufl.edu/~uhk/DYNAMICS.html [Broken] because I couldn't remember it.

Last edited by a moderator: May 3, 2017
7. Sep 9, 2007

### amstrf89

What is this part of the equation: $$\((r\ddot\theta + 2\dot r \dot\theta)\hat{\boldsymbol\theta}$$?

8. Sep 9, 2007

### learningphysics

That's tangential acceleration.

$$(\ddot r - r\dot\theta^2)\hat{\mathbf{r}}$$

9. Sep 9, 2007

### amstrf89

Thanks for the help. I think I figured it out.

10. Sep 9, 2007

### learningphysics

Cool. no prob.