Artificial gravity in a washing machine?

Question

If a washing machine's drum has a radius of 25 cm and spins at 3.9 rev/s, what is the strength of the artificial gravity to which the clothes are subjected? Express your answer as a multiple of g.

Answer 1

r = 0.25 m, T = 1/3.9
so v = 2πr/T = 2π(0.25)(3.9) = 6.13 m/s
a = v^2/r = 6.13^2 / 0.25 = 150.1 m/s^2
a/g = 150.1 / 9.81 = 15.3.

So the clothes are subjected to an artificial gravity of 15g.

Answer 2

The clothes are subjected to an "artificial gravity" of magnitude

~ 15.30... times g.

The angular speed w = 3.9 rev/s = 2pi*3.9 radians/s =
24.504... rad/s.

The value of "artificial gravity"is the inward centripetal acceleration w^2 r where3 r is the radius. So:

w^2 r = (24.504...)^2 * 25 cgs units = 15011.7... cgs units.

So in terms of g (981 cgs units), this is 15011.7... /981 = 15.30... times g.

Live long and prosper.