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Rate of revolution for a dryer?

  1. Oct 13, 2007 #1
    Rate of revolution for a dryer?!?

    1. The problem statement, all variables and given/known data

    In a home laundry dryer, a cylindrical tub containing wet clothes is rotated steadily about a horizontal axis as shown in the figure below. So that the clothes will dry uniformly, they are made to tumble. The rate of rotation of the smooth-walled tub is chosen so that a small piece of cloth will lose contact with the tub when the cloth is at an angle of 62.5° above the horizontal. If the radius of the tub is 0.385 m, what rate of revolution is needed? (revs/min)



    2. The attempt at a solution

    First I found the velocity of the dryer tub:

    Cos(62.5) = (v^2)/gR
    Cos(62.5) = (v^2)/9.8(.385)
    v = 1.32m/s

    Then I used that velocity to find the rate of revolution:

    (v/2(pi)(.385)) * 60sec = 32.74 Revolutions/min

    I submitted this answer and it said it was wrong. I can't seem to see what I did wrong. Any help or suggestions would be appreciated. Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 13, 2007 #2

    Doc Al

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    Staff: Mentor

    Where does the Cos(62.5) come from?
     
  4. Oct 13, 2007 #3
    mgCos(62.5) = m(v^2)/R
    mgCos(angle) is the horizontal component of mg

    The masses would cancel and divide g by both sides to get:

    Cos(62.5) = (v^2)/gR
     
  5. Oct 13, 2007 #4

    Doc Al

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    Since mg acts downward, it has no horizontal component. Find the component of the weight in the radial direction--that's where the centripetal acceleration points.
     
  6. Oct 13, 2007 #5
    Cos(62.5) = (v^2)/gR
    That was the equation that our professor lead us up to in class to find velocity. Wouldn't the component of the weight in the radial direction just equal mgCos(62.5)?
     
  7. Oct 14, 2007 #6

    Doc Al

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    I believe your professor made an error. 62.5 degrees is the angle between the radial and the horizontal, not between the radial and the weight.
     
  8. Oct 14, 2007 #7
    so to find the velocity i would take the square root of g*R?

    v = sqrt(9.8*.385)
    v = 1.9424 m/s

    then,
    (v/(2pi*R)) * 60 = revs/min

    (1.9424/(2pi*.385))*60 = 48.178 revs/min
     
  9. Oct 14, 2007 #8

    Doc Al

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    No. Follow my advice in post #4.
     
  10. Oct 14, 2007 #9
    sin(angle) = v^2 / gR ?
     
  11. Oct 14, 2007 #10

    Doc Al

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    Yes.
     
  12. Oct 14, 2007 #11
    Ok. The answer was correct. Thanks Doc Al!
     
  13. Oct 14, 2007 #12

    Doc Al

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    Good! (Mark this problem "solved".)
     
  14. Oct 14, 2007 #13
    There is no option in my "thread tools" to mark problems as solved. That's where the option should be right...?
     
  15. Oct 14, 2007 #14

    Doc Al

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    Yes, that's where it should be. You are not the first to state that can't find that option. (Question: Was this thread moved from another forum into this one? That might explain it.)
     
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