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Physics: centripetal motion

  1. Oct 15, 2004 #1
    The question is: There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 14 cm. When the cylinder is rotating at 1.8 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

    My answer is 3.07 m/s^2, however it is wrong.

    What i did: A(c)=(4pi^2 x 0.14 m)/1.8 sec

    (1.8 sec)(A(c))=5.527

    A(c)=3.07 m/s^2

    What did I do wrong?
     
  2. jcsd
  3. Oct 15, 2004 #2
    sorry..i figured it out:

    just for anyone's curiosity, the solution:

    Had to convert period to frequency, or 1.8 to 1/1.8 (5/9).

    A(c)=(4pi^2 x 0.14 m)/(5/9)^2

    (25/81)A(c)=5.53

    A(c)=17.9 m/s^2
     
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