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Spool of thread physics question

  1. Mar 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A spool of thread consists of a cylinder of radius R1 = 6.6 cm with end caps of radius R2 = 9.3 cm as depicted in the end view shown in the figure below. The mass of the spool, including the thread, is m = 290 g. The spool is placed on a rough, horizontal surface so that it rolls without slipping when a force Tarrowbold.gif = 0.690 N acting to the right is applied to the free end of the thread. For the moment of inertia treat the spool as being a solid cylinder of radius R1, as the extended edges are thin and therefore light.

    (a) What is the acceleration of the spool? Take positive to be to the right.

    http://www.webassign.net/serpse8/10-p-085.gif

    2. Relevant equations
    Solid Cylinder I = 1/2(M)(r)^2

    3. The attempt at a solution

    I need some help with this problem. I don't think my attempt is even correct. But my thought is since T = I*A. Then I just need to solve for A. A = T*I. The moment of inertia is what is stumping me. I'm not sure if I'm doing this correctly. I = 1/2(.3)(.066) = 0.0099. This does not solve the problem adequately. Thanks of the help!

    *Where T is force
    A is angular acceleration
    I is moment of inertia.
     
    Last edited: Mar 19, 2016
  2. jcsd
  3. Mar 19, 2016 #2

    ehild

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    What is the question?
     
  4. Mar 19, 2016 #3
    Updated in main post. Sorry about that!
     
  5. Mar 19, 2016 #4

    ehild

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    What are A and a? T is the force in the problem. Better to denote it F.
     
  6. Mar 19, 2016 #5
    Updated in main post. I left it as T because that is what the problem defines it by.
     
  7. Mar 19, 2016 #6

    rcgldr

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    The horizontal surface applies a force to the left at the edges of the end caps as the spool accelerates. This torque is partially opposed by the thread's force to the right at the edge of the inner part of the spool.
     
  8. Mar 19, 2016 #7

    ehild

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    The torque is defined with respect a point or axis. What is your point of reference?
     
  9. Mar 19, 2016 #8
    The question says to take positive to be the right. It is moving to the right as well.
     
  10. Mar 19, 2016 #9

    ehild

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    How do you calculate torque?
     
  11. Mar 19, 2016 #10
    T=I*A

    Where T is torque (Given)
    I is moment of inertia (Not given)
    A is angular Acceleration (Need to find)
     
  12. Mar 19, 2016 #11

    ehild

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    Yes, but how is the torque of a force defined?
     
  13. Mar 19, 2016 #12
  14. Mar 19, 2016 #13

    ehild

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  15. Mar 19, 2016 #14
    I assume all of it because it says it doesn't slip.
     
  16. Mar 19, 2016 #15

    ehild

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    Answer my question, please.
     
  17. Mar 19, 2016 #16
    0.690N?

    Not trying to irritate you or anything. It's my first time on this board.
     
  18. Mar 19, 2016 #17

    ehild

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    The torque is not force.
     
  19. Mar 19, 2016 #18
    F = M*A
    2.94N = 0.3Kg*9.8m/s/s?
     
  20. Mar 19, 2016 #19

    SteamKing

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    Check your arithmetic. You calculated an incorrect value for I.
     
  21. Mar 19, 2016 #20

    ehild

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    I asked the torque.
    I leave you now, I have to sleep. Read your notes or handbook or the page you cited about the torque. You might remember that you get it multiplying the force with its lever arm.
    http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html
     
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