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Homework Help: Torque: what am I doing wrong

  1. Mar 17, 2004 #1
    The question shows a horizontal uniform rod with a pivot point on the left end and a weight tied to the right end.

    The rod and weight have mass 7 kg each. The rod is 35m long

    It asks what the inital torque of the system on the pivot point will be when dropped with gravity being 9.8 m/s^2

    What I tried to do was solve for the torque of the weight and the rod then add the two results for the total torque on the pivot point.

    For the weight since it is perpendicular force then T = r * F so I did 35m * -9.8 m/s^2 * 7 kg = -2401 kg (m/s)^2

    For the rod I tried to use the theorem that involved parallel axis from the center of mass in the book. The book also gave several Formulas for common objects for inertia of center of mass. Since I didn't know how to integrate r^2 with respect to dm (only in calc II)this is what I did.

    I = I (com) + mh^2 (h is the distance from the axis perpendicular com)

    so.. I = 1/12 ML^2 (from the book pg. 227 for rod) + 1/4 M L^2 (since h is L/2)

    so I = 1/3 ML^2 = 1/3 7 kg (35m)^2 = 2858 kg*m^2 for the inertia

    Therefore Torque equals alpha * I so I had T = 2858 * (-9.8 m/s^2) / 35m = -800.3

    so I had -800.3 kg*m^2/s^2 - 2401 kg*m^2/s^2 = -3201 kg*m^2/s^2

    I tried this and several other ideas but to no avail.. Can you please help me. What mistake am I making?
  2. jcsd
  3. Mar 17, 2004 #2

    Doc Al

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    If all you need to do is find torque, you are over complicating things.

    Torque = distance X Force (perpendicular)

    Torque from weight = 35*mg
    Torque from rod = 35/2 *mg

    Add them. Units would be Newton-meters.
  4. Mar 17, 2004 #3
    shouldn't this be negative since it is clockwise. If so I already tried that answer. Didn't try it postive.
  5. Mar 17, 2004 #4


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    Your work looked good up to this line, I'm not sure if this line is true.
    Are you sure that is what alpha is? I'm not too sure that it is.
  6. Mar 17, 2004 #5
    alpha * r = acceleration

    I'm not sure exactly what r should be in this case. I'm pretty sure it would be 35.. but I tried 35/2 also.. it didn't work.

    maybe you are right actually that's for the tangential component.. the other component being V^2 / r or w^2 r.

    I'm not sure how this helps though because the gravity should be the tangential acceleration right?? therefore I divide by r to get alpha it says.. I think
    Last edited: Mar 17, 2004
  7. Mar 17, 2004 #6


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    I'm just thinking that the acceleration at the right end of the rod will not be 9.8m/s^2. The weight at the end, if in freefall, would have that acceleration. The problem is that the mass moment of inertia for the bar (which you calculted) is actually fighting against the acceleration of gravity. That's what moment of inertia does, it fights against angular acceleration.
  8. Mar 17, 2004 #7

    Doc Al

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    For this problem, find the torque by direct calculation as I showed above. The sign depends on your sign convention; if clockwise is positive, the torque is positive.

    Do not try to use [itex]\tau = I\alpha[/itex] to find the torque, since you don't know [itex]\alpha[/itex].

    Note: g is the acceleration due to gravity for a freely falling object. This object is being constrained by a pivot!
  9. Mar 17, 2004 #8
    it gives me gravity in this problem and in part 2 and 3 the system swings along the pivot point due to gravity so I'm sure it comes into play. Also talking about how alpha is being related to the end etc etc.. isn't that the point of solving for the moment of inertia?

    I may try your method positive but that wouldn't really make sense since gravity being negative and popular convention seem to go against this answer.
  10. Mar 17, 2004 #9
    You were right, thanks, even though my book clearly stated that counter-clockwise motion is negative and also it does not ask for magnitude. Thanks.
  11. Mar 17, 2004 #10

    Doc Al

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    What exactly is the problem you are asked to solve? In your first post you said the problem was to find the torque. If that's true, there is no reason to calculate the moment of inertia.

    As far as I can see, they gave you "gravity" just so you know what to use for "g".
    The method I gave is the method. What convention does your instruction use for torque? Clockwise equals positive or negative?

    I just checked your value for torque: Never mind the sign, the magnitude you have is incorrect. Do it over.

    The only reason to calculate moment of inertia is to find [itex]\alpha[/itex], which is a perfectly reasonable next question, once you've found the torque.
  12. Mar 17, 2004 #11

    Well again, I solved it. I think I understand that I should only use inertia to solve for alpha since that is the net torque and I was leaving out the block or something.

    Anyway my answer was wrong cause I solved it the wrong way. 3601 was the answer I got, when I got it CORRECT the way you told me!

    Although I sent a letter to my professor because it said in the book "The SI unit for torque... ... ...A torque t is positive if it tends to rotate a body at rest counterclockwise and negative if it tends to rotate a body at rest clockwise."

    Anyway thanks again.
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