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Confused about torque (again)

  1. Dec 22, 2013 #1
    Tell me where I've gone wrong (here r(1,2) means position of point 1 with respect to point 2 etc):

    ω x r(1,2)=v(1,2)
    Differentiating both sides with respect to time:
    ω x v(1,2) + [itex]\alpha[/itex] x r(1,2) = a(1,2)
    =>ω x (ω x r(1,2)) + [itex]\alpha[/itex] x r(1,2) = a(1,2)

    Now let us imagine a uniform rod in free space of length l and mass m, lying with the origin at its centre. A force F(+[itex]\hat{j}[/itex]) is applied at one end (+l/2[itex]\hat{i}[/itex]). We observe the motion about centre of mass first at time 0:
    [itex]\tau[/itex]=+l/2[itex]\hat{i}[/itex] x F[itex]\hat{j}[/itex]=F*l/2[itex]\hat{k}[/itex].
    [itex]\alpha[/itex]=6F/(m*l) [itex]\hat{k}[/itex]
    Now since ω=0,
    a(cm, positive end)=6F/(m*l) [itex]\hat{k}[/itex] x l/2[itex]\hat{i}[/itex]= 3F/m [itex]\hat{j}[/itex]

    Now we observe the motion about the positive end:
    [itex]\tau[/itex]=0 (since the force is applied at that very point)
    => [itex]\alpha[/itex]=0
    Hence a(positive end, cm)=0
    =>a(cm, positive end)=0

    What have I done wrong? :(
    Edit: By positive end I mean the end which is on the positive side of the x axis.
    Last edited: Dec 22, 2013
  2. jcsd
  3. Dec 22, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Things aren't so simple when you take torques with respect to an accelerating point other than the center of mass. You cannot set net torque equal to Iα in such a case.
  4. Dec 22, 2013 #3
    Why? And isn't the centre of mass accelerated?

    Edit: Didn't read "other than the centre of mass" part. But still, why is it so?
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