More with Centripetal Acceleration

[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!

 

[clive]

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

 

[quicknote]

I would suggest using the correct units in your calculations to avoid confusion. In this case, it would be helpful to convert the revolutions per second into radians per second, as one revolution is equal to 2π radians. So, 1.9 rev/s would be equivalent to 1.9 x 2π = 3.8π rad/s.

With this conversion, your calculation for velocity would be:

v = 2π(.1 m)(3.8π rad/s) = 3.8 m/s

And your calculation for centripetal acceleration would be:

Ac = v^2/r = (3.8 m/s)^2/.1 m = 14.44 m/s^2

I hope this helps and remember to always double check your units to ensure accurate calculations.