# More with Centripetal Acceleration

1. Oct 9, 2005

### rockmorg

Hey again all - I think I have another problem where my math skills just aren't cutting it --

Kitchen gadget that dries lettuce leaves by spinning them, the radius of the container is 10 cm and it is rotating at 1.9 revolutions per second. Magnitude of centripetal acceleration at the wall?

r = 10 cm (.1 m)
rps = 1.9
Ac = ?

1.9 rev/1 sec = 1.9 sec?

So I use v = 2Pir/T

v = 2Pi(.1 m)/1.9 s
v = .330 m/s

Ac = v2/r = (.330 m/s)2/.1 m = .1089/.1 = 1.1 m/s2

I have a feeling the problem is the revolutions per second, like I'm not converting that into the correct amount for one revolution...

Any thoughts?

Thanks!

2. Oct 9, 2005

### clive

Hi rockmorg,
rev/s is a frequency so if $$\nu=1.9 s^{-1}$$ the period (T) will be $$1/1.9$$ because of $$\nu=1/T$$