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Tangential acceleration, Inertia, and Torque

  1. Dec 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A solid rod of mass M = 1.35 kg and length L = 83 cm is suspended by two strings, each with a length d = 71 cm (see Figure), one at each end of the rod. The string on side B is cut. What is the magnitude of the initial tangential acceleration of end B?
    http://gauss.vaniercollege.qc.ca/webwork2_files/CAPA_Graphics/Gtype20/prob13a.gif


    The string on side B is retied and now has only half the length of the string on side A. What now is the magnitude of the initial tangential acceleration of end B if it's cut again?
    http://gauss.vaniercollege.qc.ca/webwork2_files/CAPA_Graphics/Gtype20/prob13b.gif

    2. Relevant equations

    a(tangential)= rα
    α=Torque/I
    I=1/3MR^2


    3. The attempt at a solution
    Given: m=1.35kg
    L=.83m
    d=.70m

    I= (1/3)(1.35)(.83)^2=.31 kg*m^2
    Torque= Frsinθ
    F=ma
    F=(1.35*9.8)=13.23N
    Torque=13.23(0.83)=10.98N*m
    α=10.98/.31= 35.42kg*m^2
    a(tangential)= 0.83(35.42)=29.4 m/s^2

    I haven't tried b) yet because a) isn't the right answer
     
  2. jcsd
  3. Dec 15, 2012 #2

    TSny

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    Did you use the correct value for the "lever arm" for the force of gravity?
     
  4. Dec 15, 2012 #3
    83 cm converted to .83 m... also tries using half the mass because each rope held 1/2 the weight of the rod, it didn't work either
     
  5. Dec 15, 2012 #4
    Ohhh, do I have to use the length of the string?
     
  6. Dec 15, 2012 #5

    TSny

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    The rod is an extended object. Each particle of the rod has some weight. So, the weight doesn't actually act at just one point of the rod. Nevertheless, every object has a special point where you can imagine all the weight of the object to be concentrated for the purpose of calculating torque due to weight.

    Do you know what that special point is called? Do you know where that point is located on the rod?
     
  7. Dec 15, 2012 #6
    Center mass?
     
  8. Dec 15, 2012 #7

    TSny

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    Yes, center of mass (or center of gravity). If you support an object at the center of gravity, it will balance. Suppose you wanted to balance a rod on the tip of your finger. At what point of the rod would you place your finger in order to have the rod balance? That point is the center of gravity.
     
  9. Dec 15, 2012 #8
    Okay, but how do I apply this to the problem^ Do I use the length 0.83/2?
     
  10. Dec 15, 2012 #9

    TSny

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    Yes. You need to use the distance from the axis of rotation to the center of mass.
     
  11. Dec 15, 2012 #10
    Torque=13.23(.83/2)=5.49N*m
    alpha= 5.49/.31=17.71rad/s^2
    a(tangental)= 17.71*.83/2=7.35m/s^2

    Where did I go wrong again?
     
  12. Dec 15, 2012 #11

    TSny

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    I think you now have the correct answer for alpha. But to get the tangential accelertion of end B, you need to use the correct distance from the axis to end B.
     
  13. Dec 15, 2012 #12
    It worked! Thank-you so much!, I'm going to try b) now..
     
  14. Dec 15, 2012 #13
    So, for b), moment of Inertia would be the same? Then for the torque=Frsin(theta), do I find the vertical component of gravity?
     
  15. Dec 15, 2012 #14

    TSny

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    Gravity is already vertical. In ##\tau = Frsin\theta##, ##\theta## is the angle between ##r## and the force of gravity. ##r## is directed from the axis to the center of mass.
     
  16. Dec 16, 2012 #15
    I'm still not getting the right answer, is the center of mass located at the center or is it more to the left now?
     
  17. Dec 16, 2012 #16

    TSny

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    The center of mass of the rod is always at the center of the rod. So in [itex]\tau[/itex] = Frsinθ, r will still be the distance from the axis to the center of the rod. However, you need to figure out the value of θ, which is the angle between the r direction and the force direction.
     
  18. Dec 16, 2012 #17
    sin90/0.83=sin(theta)/0.355
    sin-1= 23.8deg
    then, I followed the same steps as I did for part a).
    Torque=1.35*9.8*.355*sin23.8
    Inertia is the same
    alpha=T/I
     
  19. Dec 16, 2012 #18

    TSny

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    You need to find the angle labeled θ in the attached figure. It looks like you calculated [itex]\phi[/itex]. (You might want to recheck the calculation - I think you plugged .335 into your calculator instead of .355).

    Anyway, once you have [itex]\phi[/itex], it should be easy to see what θ is.
     

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  20. Dec 16, 2012 #19
    I'm so sorry!!
    I got the angle theta to be 115.322deg
    1.35(9.8)(.355)sin115.322=4.2453
    alpha=4.2453/.31=13.6948
    a=13.6948*.83=11.3666
    I know it's wrong because tangential acceleration should be greater than the one found in part a)
     
  21. Dec 16, 2012 #20

    TSny

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    Are you sure the answer should be greater? Note that for the tilted rod, the "lever arm" distance of the weight force is smaller than for the horizontal rod. So, less torque for the tilted rod.
     
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