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Centripital acceleration problem

  1. Oct 20, 2004 #1
    This is talking about one of those kitchen gadgets where you put the lettuce in the container and spin it so all of the water gets spun off.

    The radius of the container is 12cm (0.012m). When the culinder is rotating at 2.0 revolutions per second, what is the magnitude of the centripital force at the outer wall?
  2. jcsd
  3. Oct 20, 2004 #2
    What have you done so far?
  4. Oct 20, 2004 #3

    Doc Al

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    Staff: Mentor

    I assume they want the magnitude of the centripetal acceleration. So, what's the formula for centripetal acceleration? (Check the book!)
  5. Oct 20, 2004 #4
    All I've done so far is looked at equations and wrote down what I know:


    What I know:
  6. Oct 20, 2004 #5
    Well you have rev/sec. How much distance will an object go in one rev? When you figure that out, you can find the "tangential" speed in m/s, and then your formula might make more sense.
  7. Oct 20, 2004 #6
    hey, 12cm is 0.12m.

    acc= (v*v0) / r

    cirumfrance = pi * 2 * 0.12 = 0.24pi

    so v = (0.24pi * 2)/1 = 0.48pi

    acc = 0.48pi / 0.12 = 4pi m/s*s

    i think!
  8. Oct 20, 2004 #7

    Doc Al

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    Staff: Mentor

    That's the equation you want. To use it you have to figure out the speed.

    Careful. 12 cm = 0.12 m, not 0.012 m.
    Now figure out the speed using v = distance/time. It goes 2 revolutions per second. Each revolution is the circumference of a circle: [itex]c = 2 \pi r[/itex]. Figure out the speed in m/s.

    Forget the centripetal force (since you are not given a mass); you want to find the centripetal acceleration.
  9. Oct 20, 2004 #8
    Thanks guys, this helped a lot.
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