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Angular acceleration of a rod

  1. Oct 22, 2016 #1
    • Thread moved from the technical forums, so no Homework Help Template is shown
    1. The problem statement, all variables and given/known data
    https://holland.pk/uptow/i4/ae8d3da6c3ce3cad10eb98dd3208a955.png [Broken]

    2. Relevant equations
    τ=I⋅α

    3. The attempt at a solution

    I want to discuss in part (a) of the question
    Tension in 5 kg block string=mg=5*9.8=49N
    Taking moment at the pivot, a net torque acting on 15 kg post is 49*(1.75-0.5)=61.25Nm
    Torque total=I*α
    61.25=((15)(1.75)^2)/3*α
    α=4 rad s^-2

    But the answer is wrong, I found another solution from other forum
    http://physicshelpforum.com/advanced-mechanics/1978-angular-rotation-question.html
    I don't know why we need to consider the moment of inertia of the 5 kg block also.
    Can anyone help me please! Including a graph for explanation is better if you can:cry:
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Oct 22, 2016 #2
    In your calculation, you use the tension in the string before the cable breaks. Once the cable breaks, the tension in the string changes. Because the 5 kg block starts to accelerate, the tension in the string has to be less than the weight of the block. If the tension in the string remained the same before and after the cable broke, the 5 kg block would not accelerate.

    P.S. Welcome to PhysicsForums.
     
  4. Oct 22, 2016 #3

    kuruman

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    To expand on what @TomHart said: You need to draw two free body diagrams, one for the rod and one for the hanging mass. You will get two equations from them when you write Newton's second law. Use T for the tension, do not use the hanging weight 49 N. You will have three unknowns, the tension T, the angular acceleration α of the rod and the linear acceleration of the mass, a. Find a third equation relating α and a and you will have a system of three linear equations and three unknowns to solve.
     
  5. Oct 22, 2016 #4

    CWatters

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    Tom and Kuruman describe the conventional approach.

    What do people think about the approach "Parvez" on that other thread takes? Eg treating the 5kg mass as if it's a point mass fixed to the rod where the string is connected. Seems ok to me.
     
  6. Oct 22, 2016 #5

    kuruman

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    I don't think Parvez is saying that the 5 kg mass can be treated as if it were fixed on the rod. If that were the case, there would be no tension to exert a torque on the rod. His observation that "the angular acclns of the post and that of the pully will be same" is incorrect. The linear acceleration, a, of the hanging mass must be the same as the linear acceleration of any point along the string and therefore the same as the linear acceleration of the point on the rod where the string is attached. This allows the formulation of the equation αrod = ah.m. / d, where "h.m." stands for "hanging mass" and d is the distance from the pivot to the point of attachment. If Parvez were correct and the angular acceleration of the pulley and the rod were the same, then α = apulley / rpulley = arod / d in which case for the linear acceleration to be the same (apulley = arod), one must have rpulley = d which is clearly not the case.
     
  7. Oct 22, 2016 #6
    Parvez's solution seemed very strange to me and not valid. (However, I think Parvez was at a great disadvantage in that he did not originally see the diagram that we got to see here.) Like @kuruman said above, if the 5 kg block is drawn as a point mass positioned on the post at the point where the string connects, it will not produce a torque. So it is not valid to place a point mass at that position. And if the 5 kg block was being considered as a point mass, where would it be positioned to accurately represent the string/pulley/block configuration that we currently have?

    Parvez used the same torque as was done in the OP's attempted solution here in this forum - namely, the weight of the block times the perpendicular distance to the pivot point = (5 kg)(9.8 ms-2)(1.25 m) = 61.25. However, because Parvez added in the moment of inertia of the 5 kg block to the moment of inertia of the post, that ended up offsetting the incorrect torque calculation, resulting in the correct answer. :/

    To summarize, it was a confusing attempt at a solution and I am still not clear about it. :/ If I could give advice to @dennislam, I would say, "Forget the idea of considering the moment of inertia of the 5 kg block, and follow the steps recommended by @kuruman."
     
  8. Oct 22, 2016 #7
    Thank you for all of you help!
    It seems like I can get the correct solution now:kiss:
    mg-T=ma
    a=rα
    1.25T=Iα
    just solve the above equations and I can get α=2.65 rad s^-2:biggrin:
     
  9. Oct 22, 2016 #8

    haruspex

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    Reading that thread, it seems Parvez had not understood the set-up, and declared all his working wrong when he saw the diagram.
    But, yes, treating the inertia of the mass as equivalent to that of a point mass attached to the rod where the string is attached does work. I have used similar tricks in mass-and-pulley questions in the past, but would not encourage students to do so - too easy to apply it wrongly.
     
  10. Oct 23, 2016 #9
    Could you please explain how that works? I've never heard of that before. Thank you.
     
  11. Oct 23, 2016 #10

    haruspex

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    consider a simpler set-up. Same diagram, but now in a horizontal plane, so no gravity. Ignore the mass of the rod.
    The mass slides on a smooth table. The string is attached to the rod at distance r from the pivot.
    If you exert an anticlockwise torque τ at the pivot, the acceleration of the mass is a=τ/(mr). The angular acceleration of the rod is a/r=τ/(mr2. So the mass effectively contributes a moment of inertia about the pivot of mr2.
     
  12. Oct 24, 2016 #11

    CWatters

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    He's not physically moving the mass to the rod...

    The 5kg mass (m) is accelerating (at unknown a) so the tension isn't mg. The classic approach to account for this is to reduce the tension by subtracting the acceleration using a term m(g-a).

    What Parvez is saying is that you can keep the tension at mg and increase the moment of inertia of the rod. He does that by pretending the mass M is fixed to the rod where the string is attached.

    So although he says he misunderstood the problem it looks like his solution works.
     
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