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Tension question (this should be basic)

  1. Feb 12, 2006 #1
    tension question on object moving in a circle (this should be basic)

    A .6 kg steel block rotates on a steel table while attached to a 1.20 m-long hollow tube. Compressed air fed through the tube and ejected from a nozzle on the back of the block exerts a thrust force of 4.10 N perpendicular to the tube. The maximum tension the tube can withstand without breaking is 50.0 N. (coefficient of kinetic friction:.6) If the block starts from rest, how many revolutions does it make before the tube breaks?

    [​IMG]

    im stumped...

    look 1 post down... im trying
     
    Last edited: Feb 12, 2006
  2. jcsd
  3. Feb 12, 2006 #2
    ah i found something:

    at= 6.833333 m/s^2

    Fr= mat - Ff

    Fr= 4.1N- Ff
    = 4.1-3.528
    =.572N

    a(c)= .953333m/s^2
     
    Last edited: Feb 12, 2006
  4. Feb 12, 2006 #3
    ok heres what i tried:

    at 50N, the object is going at:
    v= 10m/s

    so i plugged this in the T=(mv^2)/r equation and got: T=50s, but this looks wrong, and i dont know how to get revs from this...
     
  5. Feb 12, 2006 #4
    hmmm...

    angluar velocity= (at/r)t= 284.722rad/s= 2718.89 rpm

    2718.89= 45.315 revs/second

    45.315 x 50 = 2265.75 revolutions (wrong)
     
    Last edited: Feb 12, 2006
  6. Feb 12, 2006 #5
    guys please help me, theres only one hour left
     
  7. Feb 12, 2006 #6

    Tide

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    Science Advisor
    Homework Helper

    Here's a start:

    The angular acceleration is given by

    [tex]I \frac {d^2 \theta}{dt^2} = r (F_T - \mu mg)[/tex]

    where I is the moment of inertia (essentially just [itex]m R^2[/itex]). Use the equation solve for the angular velocity and the angular displacement as a function of time then determine the time at which the centripetal force reaches the breaking point.
     
  8. Feb 12, 2006 #7

    thank you , thank you for responding
     
  9. Feb 12, 2006 #8
    thanks so much, but i dont really get what youre saying. what does Ft mean? o, i didnt learn about inertia yet
     
    Last edited: Feb 12, 2006
  10. Feb 12, 2006 #9
    guys, i dont want you to waste your time. if you arent able to help in the next 30 minutes, dont even worry about it
     
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