1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

More with Centripetal Acceleration

  1. Oct 9, 2005 #1
    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?

  2. jcsd
  3. Oct 9, 2005 #2
    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]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: More with Centripetal Acceleration