Analyzing Acceleration in a Particle's Spiral Motion

[amstrf89]

Homework Statement

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?

Homework 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}

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.

 

[learningphysics]

What are the values of a and A?

 

[amstrf89]

What are the values of a and A?

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

 

[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).

 

[amstrf89]

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).

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}

 

[learningphysics]

Is angular acceleration the same as radial acceleration? Also, what is the equation for tangential acceleration?

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 because I couldn't remember it.

 

[amstrf89]

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

 

[learningphysics]

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

That's tangential acceleration.

This is radial acceleration:
(\ddot r - r\dot\theta^2)\hat{\mathbf{r}}

 

[amstrf89]

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

 

[learningphysics]

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

Cool. no prob.