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Circular motion of a steel block

  1. Oct 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A .5 kg steel block rotates on a steel table, attached to a 1.20m-long hollow tube. Air is fed through the tube and is ejected from the block, giving it a thrust force of 5.21N perpendicular to the tube. Max tension the tube can withstand without breaking is 50N. Assume coefficient of kinetic friction between steel block and steel table is 0.60. If the block starts from rest, how many revolutions does it make before the tube breaks?

    upload_2014-10-22_21-11-54.png

    2. Relevant equations
    a_r = mv^2/r = omega^2 * R

    3. The attempt at a solution
    I have solved for the angular velocity and got 9.125. I'm not sure where to go from here or which equations to use. I think it might have something to do with the tangential acceleration, but I don't know how to find that.
     
  2. jcsd
  3. Oct 22, 2014 #2

    Simon Bridge

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    Angular velocity when?
    Have you drawn a free-body diagram for the block in motion?
     
  4. Oct 22, 2014 #3
    I thought I solved for the *initial* angular velocity.

    For the free-body diagram I have the tension force (not sure of the value of it), the frictional force, the thrust force, and then the normal force and gravity (which should cancel out). Am I missing something?
     
  5. Oct 22, 2014 #4

    Simon Bridge

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    From the problem statement post #1:
    ... (my emph.)
    How did you get 9.125rad/s from "at rest"?

    Have you drawn a free body diagram for the block in motion?
     
  6. Oct 22, 2014 #5
    Ok, so if the initial angular velocity is 0 and the final angular velocity is 9.125 rad/s, how can I calculate theta from the kinematics equations? With Theta_initial = 0 rad.

    The free body diagram I described above is the free body diagram I have for the block in motion. Am I missing something?
     
  7. Oct 22, 2014 #6
    Fnet_r = m*omega^2*r
    50 = .5 * omega^2 * 1.2
    omega = 9.13 rad/s

    Fnet_t = m*a_t
    2.27 (Thrust minus frictional force) = 0.5 * a_t
    a_t = 4.54 m/s^2

    Not sure where to go from here....
     
  8. Oct 22, 2014 #7

    Simon Bridge

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    You start by using a free body diagram - if you don't take advise I cannot help you.
    ##\sum\tau = I\alpha##

    What is the relationship between angular and tangential acceleration.
     
  9. Oct 22, 2014 #8
    Figured it out! Neglected the fact that alpha = a_t / R -- once I realized that I was able to use the equations. Got theta final = 10.987 rad, comes out to 1.75 revolutions. Thanks for the help!
     
  10. Oct 22, 2014 #9

    Simon Bridge

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    No worries.
    It often helps just to talk it out even if you don't get the answers directly from feedback.
    It does help us to help you quickly if you include your reasoning with your working.
    Cheers.
     
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