# When does the bead fly off the rod?

1. Aug 25, 2015

1. The problem statement, all variables and given/known data
A rod of length L is fixed at one end, and rotates in the X-Y plane with angular velocity ω. (To be clear, it is sweeping out an area of $π (L/2)^{2}$.) A bead starts at position $r(0)=L/2$ with $\dot{r}(0)=0$. Find $r(t)$ and the time it takes for the bead to fly off the end of the rod.

2. Relevant equations
$F=ma$

3. The attempt at a solution
First, I wanted to find an expression for the velocity of the bead in general:
$v(t)=\dot{r}\hat{r}+rω\hat{φ}$

Then I find the acceleration of such a situation:
$a(t)=(\ddot{r}-ω^{2}r)\hat{r}+(2ω\dot{r})\hat{φ}$

Then I need to apply the 2nd law and solve the diff eq. My question at this point is: Is this problem readily solvable in this coordinate system, or do I need to switch to something else like $r(φ)$ first?

I've played with a few attempts, and my best guess right now is:
$\dot{r}(t)=(L/2)e^{(ω/m)t}$
or
$\dot{r}(t)=(L/2)e^{(2ω/m)t}$

(and of course the position is just the integral of that)

But I'm not really confident on that answer...

2. Aug 25, 2015

### TSny

OK

Using polar coordinates is good. No, you do not need to express r as a function of angle.

This solution does not satisfy $\dot{r}(0)=0$. Also, the argument of the exponential should be dimensionless.

Can you state the differential equation that you need to solve?