# Conveyor belt and friction- Determine slip time

1. Sep 18, 2010

1. The problem statement, all variables and given/known data
A series of small packages is moved by a conveyor belt that passes

over a 300mm radius pulley. The belt starts from rest at t=0 and its

speed increases at a constant rate of 150 mm/s2. The

coefficient of friction is μ=0.75. Determine the time the first

package slips.

2. Relevant equations
$$V_{0}=0$$
$$r_{pulley}=300mm$$
$$a_{pulley}=150mm/s^{2}$$
$$\mu=0.75$$

These are equations for polar coordinates, which I'm thinking I may need to use to solve the problem.
$$F_{r}=ma_{r}$$
$$F_{\theta}=ma_{\theta}$$
$$F_{net}=F_{r}{\bf e}_{r}+F_{\theta}{\bf e}_{\theta}$$

$$m(\ddot{r}-r\dot{\theta}^{2})=F_{r}$$
$$m(r\ddot{\theta}+2\dot{r}\dot{\theta})=F_{\theta}$$

3. The attempt at a solution
First of all, I'm trying to decide if I should use cartesian, polar, or path coordinates for this problem.

If cartesian coordinates work, then I would set the problem up as below. Also, I don't think the radius of the pulley would matter...?

Sum of forces x-dir = -Fs+Fpulley=ma
Sum of forces y-dir = N-mg=0 -----> N=mg

The maximum static friction would be Fssmg.

For polar or path coordinates, I wouldn't know where to start.

2. Sep 19, 2010

I'm getting closer... I think.

I actually think I can solve this without using polar or path coordinates.

So, for constant linear acceleration, we have:

vf2=vi2+2a(x2-x1)

Because x2-x1 is really 2πr (arc length), we have:

vf2=vi2+2a(2πr)

I can solve this for vf and then plug that into vf=vi+at to get time... But I'm missing something with the whole Fssmg=ma... Like it should factor into the acceleration value for the two formulas above. Any help?