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Homework Help: Gravitational Torque about point?

  1. Jul 21, 2010 #1
    A 2.0 kg, uniform, horizontal rod has a point .25 m from the left and .75 from the right. Find the gravitational torque about this point.

    I'm confused I'm getting -6.7 because grav torque =mgd so for the the portion to the left of the point has a center of gravity of .125 m and to the right has .25 m. So the mass towards the left is .5 kg and mass towards the right is 1.5 kg. That means torque for the left portion is .5 *.125 * 9.8=.6125 and for the right is 1.5*9.8 * .5=7.35. So since .6125 - 7.35 = -6.7 N-m. The answer is -2.1 Nm what am I doing wrong?
     
  2. jcsd
  3. Jul 21, 2010 #2

    Doc Al

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    Yes, the distance from pivot to the center of the left portion is .25/2 = .125 m. But the distance from pivot to the center of the right portion is not .25 m.

    While it's perfectly OK to treat the rod as having two pieces--as long as you do it right, of course--but you can also just treat gravity as acting at a single place on the rod: it's center of mass. How far is that from the pivot?
     
  4. Jul 26, 2010 #3
    OK, thank you Doc Al but I'm still not getting the correct answer. Here is how I redid the problem:

    For piece 1 the center of mass is .25-0=.125 and .25-125=.125 so there .125 is the distance from the pivot to the center of mass. Now for the mass, so the mass must be .25 * 2.0 kg=.5 kg. Therefore since torque is mgd then it's .5 * 9.8 * .125=.6125.

    Now for the torque of the right portion. Center of mass is .75/2=.375
    The distance from the center of mass to the pivot is .375-.25=.125

    The mass is .75 * 2=1.5 kg

    So the torque is 1.5 * 9.8 * .125 is 1.8375 but the weight tends to move this object clockwise so it's
    -1.8375.

    Net torque = torque of left portion - torque of right portion=.6125-1.8375=
    -1.2 N-m. The book's answer is -2.1 N-m.

    Someone please tell me why I'm doing wrong! :(
     
  5. Jul 26, 2010 #4
    Anyone please help!!!!
     
  6. Jul 26, 2010 #5

    Doc Al

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    This is where you made your mistake. The length of the right portion is 0.75, so the distance to the center will be 0.75/2 = 0.375. That's the distance from the center of mass to the pivot.

    (Once you correct your error and get the right answer, I'll show you the really easy way to solve this.)
     
  7. Jul 26, 2010 #6

    Doc Al

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    That answer seems incorrect. What book are you using?
     
  8. Jul 26, 2010 #7
    Doc Al yes I'm sorry the Torque for the right portion Is 1.5 * 9.8 * .375
    The book I'm using is College Physics by Randall Knight.
     
  9. Jul 26, 2010 #8

    Doc Al

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    Good.
    I have that text at home. Tell me the problem number and I'll check it out later.
     
  10. Jul 26, 2010 #9
    #19 back of chapter 7 problems. Thank you kindly for your help!
     
  11. Jul 26, 2010 #10

    Doc Al

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    Unfortunately, I spoke too soon. I have a different text by Knight, not the College Physics one.

    In any case, assuming you've presented the problem exactly as in the book, the book's answer is incorrect. Show me the answer that you get.
     
  12. Jul 26, 2010 #11
    My answer is -4.9 N-m
     
  13. Jul 26, 2010 #12

    Doc Al

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    I agree with that answer.

    Here's the 'easy way' that I promised. The center of mass of the rod is right in the middle, which is 0.25 m from the pivot. So the torque due to gravity is just m*g*d = 2*9.8*0.25 = 4.9 N-m. (Instead of breaking the rod up into sections--which is perfectly OK--we can treat the rod as if gravity acts totally at its center of mass.)
     
  14. Jul 26, 2010 #13
    Doc Al thank you! But now I'm stuck on a problem where the mass is not uniform. If you have time can you please take a look at it:

    Consider a rod in which a 1 m portion (10 kg) is combined with a 2 m portion (40 kg) that make one rod together.

    Here is my approach:



    Here is how I approached it:*

    The origin is the part that divides the 1 m portion from the 2 m portion. *So the center of mass for the 1 m portion is .5 * and for the 2 m portion is 1 relative to the origin. * So that means the distance from the center of mass to the pivot for the 2 m portion is 2 *(1 m + 1 m) and for the 1 m portion is .5
    **But now how Do I determine the center of gravity of the whole object? **

    And for part B for the net torque I get*

    For the 1 m portion:

    10 * 9.8 * .5 = 49 N but the object will move clockwise so it's -49 N*

    40 * 9.8 * 2= 784 but this portion will also move clockwise so it's -784*

    So for net torque = -49 + (-784)= -833 N-m*

    Is this correct?*

    And how do I find the center of mass for the entire object?
    Thank you kindly for all your help!*
     
  15. Jul 26, 2010 #14

    Doc Al

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    Assuming you are finding the torque about the end of the rod (the 10 kg end), then it's perfectly correct.

    Use the general definition of center of mass:
    Xcm = (m1X1 + m2X2)/(m1 + m2)

    Treat the two sections as if all of their mass were concentrated at their centers. Pick an origin and figure out the coordinates of each center of mass.

    Tip: Once you find the center of mass of the entire rod, use it to find the torque in one step. (The easy way!)
     
  16. Jul 26, 2010 #15
    Thank you so much for your kind help, Doc Al!
     
  17. Jul 26, 2010 #16
    My answer is different from the book again!!

    For the center of gravity:

    M1= 10 kg
    X1= .5
    M2= 40 kg
    X2= 1

    So (( .5 * 10) + (1 * 40))/ (10 +40) = .9 but the book says the answer is 1.7. Why is this wrong? :(
     
  18. Jul 26, 2010 #17

    Doc Al

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    Measure the coordinates of the center of mass of each piece from the same reference point. Assuming you're measuring from the end of the rod, X2 ≠ 1.
     
  19. Jul 26, 2010 #18
    Thank you again Doc Al.

    X2 should be 2 and then

    ((.5 * 10) + (2* 40))/(10 + 40)) = 1.7 :)
     
  20. Oct 14, 2012 #19

    ELT

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    I have the same book (College Physics, Knight) and the answer in the back of my text for this problem is: -4.9N*m.
    I think your book has a misprint.
     
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