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Homework Help: Torque of a uniform rod being pivoted.

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform rod 1.1 m long with mass 0.7 kg is pivoted at one end, as shown in Fig. 9-46, and released from a horizontal position.
    Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod makes with the horizontal direction. (Use theta for θ.)
    http://www.webassign.net/fgt/9-43.gif


    2. Relevant equations
    T = rFsin(theta)

    3. The attempt at a solution
    My attempt has consisted of the following:
    1st attempt:
    Using the basic formula, and then applying it to obtain ~7.6sin(theta), trying the negative version as well.

    I think realized that sin might not be valid to use due to the nature of the direction of application of the force and tried 7.6cos(90-theta), which also did not work.

    2nd attempt:
    T = angular acceleration * moment of inertia
    I solved the moment of inertia to be ~.2823 (mL^2/3).
    I defined omega as d/(dt)[int(g)sin(theta)/r].

    I then plugged in the values and came up with .2823d/(dt)[int(9.8)sin(theta)/1.1]

    Any help would be appreciated.
     
  2. jcsd
  3. Apr 21, 2010 #2

    PhanthomJay

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    You don't need to get into angular accelerations and moment of inertias per attempt 2. The problem is just asking for the torque as a function of theta, per your first attempt equation. You have a couple of errors. In calcualting the torque, where is the gravity force (weight force) applied, and using the cross product definition of torque, T=rf sin alpha, what is the correct angle to use for 'alpha'?
     
  4. Apr 21, 2010 #3
    I only tried 2) because 1) did not work.

    I crossed .5Lcos(theta)i - .5Lsin(theta)j + 0k with 0i -mgj + 0k
     
  5. Apr 21, 2010 #4

    PhanthomJay

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    That cross product formula looks OK, but it's way too easy to mess up the calculation when you do it that way. Instead, try using use T=r X F = rF sin alpha , where r is magnitude of the radius (.5L), F is the magnitude of the weight, and alpha is the angle between the radius and weight vector. alpha = ??
     
  6. Apr 21, 2010 #5
    Well I double checked my work on a calculator and I'm positive that my cross product was right. I did realize that an error I made was that R is actually r/2 since r goes from the pivot to the center of mass, but this still does not give me the correct answer.
     
  7. Apr 21, 2010 #6
    I got T = 3.8cos(90-theta) as a replacement to 3.9sin(theta). I just used my last "guesses" on this problem though, so I will talk to my teacher tomorrow about it. I am positive that one of my 50 answers were right (30+ variations on rounding and truncating) of 3.773cos(90-theta) and I also tried +/- 3.773cos(90-theta).

    WebAssign is glitchy anyways, and often requires strange sig. figs or the like.
    Thank you for the prompt help!
     
  8. Apr 21, 2010 #7

    PhanthomJay

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    Well, sig figures aside, none of your answers are correct. I inadvertently gave you the correct answer, so i've deleted it.
     
  9. Apr 21, 2010 #8
    3.773 = .5 * 1.1 * .7 * 9.8, so actually, I don't understand why you are saying my answers are incorrect.
     
  10. Apr 22, 2010 #9

    PhanthomJay

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    That 3.773 (call it 3.8) is fine...it's the cos(90- theta) , (or sin theta), part of the equation, that's wrong. Theta is not the angle between the force and position vectors. I don't know how you got sin theta, even using your cumbersome cross product equation.
     
  11. Apr 22, 2010 #10
    ....Okay sorry I don't know what I was thinking yesterday I meant I tried cos(theta) and cos(90-theta)...
     
  12. Apr 22, 2010 #11

    PhanthomJay

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    Okay, I gather you are saying that the (magnitude of) the torque is either 3.8cos theta or 3.8 cos (90-theta); one of those answers is correct for the magnitude of the torque; the other is not. But in any case, the answer may be off by a minus sign in front of the 3.8. Usually, standard convention using the 'right hand rule' assumes that clockwise torques are negative. In that case, you need the minus sign.
     
  13. Apr 22, 2010 #12
    Yeah, I meant my original thought was cos(theta) for some reason I kept writing sin. It was probably just kind of late. Apparently Webassign demanded that we round and truncate making the answer 4cos(theta) (shady if you ask me) but hey.
     
  14. Apr 22, 2010 #13

    PhanthomJay

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    Yes, shady, but correct I guess. Since the least number of significant figures in the given variables is 1 (0.7 has one sig figure), then the answer can have only one sig figure, hence the 4. That's the rule. But while 3.773 would be wrong, I would think that 3.8 ought to be accepted by webassign. I'm surprised they accepted the plus sign. I think a human teacher, instructor, tutor, professor, homework helper, or the like, would be a bit more forgiving, depending on how strict they are.
     
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