# The Physics of Tumble Dryers

1. Jun 10, 2014

### BOAS

Hello,

I have a question regarding a tumble dryer and the conditions that must be met for it to work as described in the following question.

I'm not totally convinced by my reasoning so would like to run it by somebody.

1. The problem statement, all variables and given/known data

In an automatic clothes dryer, a hollow cylinder moves the clothes on a vertical circle (radius r = 0.32m). The appliance is designed so that the clothes tumble gently as they dry. This means that when a piece of clothing reaches an angle θ above the horizontal, it loses contact with the wall of the cylinder and falls onto the clothes below. How many revolutions per second should the cylinder make in order that the clothes lose contact with the wall when θ = 70°?

2. Relevant equations

3. The attempt at a solution

I have drawn diagrams to illustrate my answer, but hopefully my working is clear to you without them.

Centripetal force $$F_{c} = \frac{mv^{2}}{r}$$

Weight of the clothes $$W = mg$$

If you imagine a triangle with the hypotenuse being Fc and the vertical component equal to the weight of the clothes, with an angle of 70° between the horizontal and the hypotenuse then we can say that $$mg = F_{c} \sin \theta$$.

The condition for which the clothes will fall is when $$mg > F_{c} \sin \theta$$.

$$\frac{mv^{2}}{r} \sin \theta = mg$$

Cancel the mass and rearrange for v to find;

$$v = \sqrt{\frac{gr}{\sin \theta}} = 1.85$$ms-1

The circumference of the cylinder is ~2m, so the number of revolutions per second must be less than 2m/1.85ms-1 = 1.08s-1.

I'm just concerned that my reasoning is incorrect... I have used the value for g = 10 ms-2.

Thanks!

2. Jun 10, 2014

### BvU

For what it's worth: reasoning is correct, final conclusion is not: if you divide meters by meters/second you cannot expect to get s-1. (By the way: I get 1.09 seconds/revolution -- but remember: they want revolutions per second !)

3. Jun 10, 2014

### BOAS

Oh of course, thanks!