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Principle of work energy

  1. May 11, 2014 #1
    1. The problem statement, all variables and given/known data
    When Θ=0, the assembly is held at rest, and the torsional spring is untwisted. If the assembly is released and falls downward, determine its angular velocity at the instant Θ=90°. Rod AB has a mass of 6kg, and disk has a mass of 9kg.

    Rod is 450mm and disk has a radius of 75mm
    So there is a pin holding the assembly upwards which is when Θ=0 and at the pin there is a torsional spring with constant of k=20N m/rad. One end of the rod is attached to the pin and the other is attached to the disk.



    2. Relevant equations
    T1+∑U=T2



    3. The attempt at a solution
    So first I found the center of mass of the combined mass
    I called mass of the rod M and mass of disk m.
    Center of mass = (.5(.45)6+(.45+.075)9)/(6+9)= .405m
    Then I found the moment of inertia
    I=(1/3)(6(.45)^2)+(3/2)(9)(.075)^2+9(.525)^2= 2.96

    T1=0 since the assembly started from rest.
    Then I found T2
    .5mv^2+.5Iω^2
    .5(M+m)(.405ω)^2+.5(2.96)ω^2=2.71ω^2

    Solved for the potential energy of the spring
    .5(20)(π/2)^2=24.674
    solved for ∑U= mgh-24.674= 34.861
    solved for ω
    ω=3.59rad/s
    actual answer: 4.9rad/s
    What am I doing wrong?
     
  2. jcsd
  3. May 11, 2014 #2
    I think a diagram would help.
     
  4. May 11, 2014 #3
    Dont have a camera but I drew a picture :)
    and the numbers are in mm
     

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  5. May 11, 2014 #4

    tms

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    You are assuming that [itex]\sum U[/itex] is 0 at the end. Is that true?
     
  6. May 11, 2014 #5
    Diagram works.

    What is I for a disk?
     
  7. May 11, 2014 #6
    I think its (3/2)mr^2
     
  8. May 12, 2014 #7
    I don't think that is correct.
     
  9. May 12, 2014 #8
    Ixx=Iyy=(1/4)mr^2
    Izz=(1/2)mr^2
    Iz'z'=(3/2)mr^2
    This is what the book gives me
     

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  10. May 12, 2014 #9
    No, I'm assuming it equals mgh-1/2ks^2
     
  11. May 12, 2014 #10
    What is the z'z' axis?
     
  12. May 12, 2014 #11
    I think its just saying that its rotating about that axis. Izz is rotating about the z axis and Ixx is rotating about the x axis and so on
     
  13. May 12, 2014 #12
    Yeah, I get that part. My question is what is the z'z' axis?
     
  14. May 12, 2014 #13
    I'm guessing its the pin?
     
  15. May 12, 2014 #14
    Not sure where the pin is but, based on your second diagram, you are not being consistent when applying the parallel axis theorem. I would take I about the z-z axis instead.
     
  16. May 12, 2014 #15
    ok so it would be (1/2)9(.075)^2+9(.525)^2 for the disk?
     
  17. May 12, 2014 #16
    Looks better. Does it give the right answer now?
     
  18. May 12, 2014 #17
    No, for I I got 2.91 and solved it like I how I did before and got 3.60rad/s
     
  19. May 12, 2014 #18
    Ok, no surprise I guess.

    Can you show your calc for mgh?
     
  20. May 12, 2014 #19
    15(9.81)(.405)= 59.6
     
  21. May 12, 2014 #20
    OK, I see your mistake now. Your kinetic energy included a translational term but all the kinetic energy here is rotational only. Try and see if this gives the right answer.
     
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