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Rod and angular acceleration

  1. May 9, 2008 #1

    rock.freak667

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    [SOLVED] Rod...and angular acceleration

    1. The problem statement, all variables and given/known data
    A uniform rod AB, of length 2a and mass m, has a particle of mass 1
    2m attached to B. The rod is
    smoothly hinged at A to a fixed point and can rotate without resistance in a vertical plane. It is released
    from rest with AB horizontal. Find, in terms of a and g, the angular acceleration of the rod when it
    has rotated through an angle of [itex]\frac{\pi}{3}[/itex]


    2. Relevant equations



    3. The attempt at a solution

    My idea is that the torque due to the weight and the mass is the net torque and that should be the same as [itex]I \alpha[/itex] where I=moment of inertia of the entire thing.

    I got [itex]I=\frac{10ma^2}{3}[/itex] but I don't know how to incorporate that angle it is rotated through.
     
  2. jcsd
  3. May 10, 2008 #2

    alphysicist

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    Hi rock.freak667,

    How did you find the moment of inertia? It looks like you used that the moment of inertia of the particle is 2 m a^2, but I don't think that is correct. (Is the mass of the particle 2m? If so, there is an extra 1 in your post.)

    To find the angular acceleration, draw a diagram when it's at that angle, and use the formula (net torque = I alpha) that you mentioned. From the diagram you can calculate the torques at that particular angle.
     
  4. May 10, 2008 #3

    rock.freak667

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    The MOI through the centre is given by [itex]I_c=\frac{1}{12}ML^2[/itex]

    so for the rod in question, the MOI would be (through the centre)

    [tex]I_c=\frac{1}{12}(mg+\frac{mg}{2})(2a)^2=\frac{1}{2}mga^2[/tex]

    and then by the parallel axis theorem to get the MOI about the end.

    [tex]I=I_c+Mr^2[/tex]

    [tex]I=\frac{1}{2}mga^2 + (mg+\frac{mg}{2})(a)^2=2mga^2[/tex]

    is that correct moment of inertia about the end?

    And I typed part of the question wrong. It should read

     
    Last edited: May 10, 2008
  5. May 10, 2008 #4

    Doc Al

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    What's the moment of inertia of a rod about one end? Of the particle about that end? Add them.

    (The moment of inertia depends on mass distribution not weight, so g should not appear.)
     
  6. May 10, 2008 #5

    rock.freak667

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    ok well then. Using the parallel axis theorem

    [tex]I=\frac{1}{12}m(2a)^2+ma^2=\frac{4}{3}ma^2[/tex]

    and then adding the MOI of the mass(=[itex]\frac{m}{2}(2a)^2=2ma^2[/itex]

    then

    [tex]I_{total}=\frac{4}{3}ma^2+2ma^2=\frac{10ma^2}{3}[/tex]

    So then when I draw the forces at that angle (weight of rod+weight of the mass at B) of [itex]\frac{\pi}{3}[/itex]. How would I get the torque at that angle? Would it just be the total torque*sine of the angle?
     
  7. May 10, 2008 #6

    Doc Al

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    Perfect.

    You can find the torque due to each weight separately and add them up:
    [tex]\tau = \vec{r}\times\vec{W} = rW\sin\theta[/tex]
     
  8. May 10, 2008 #7

    rock.freak667

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    Torque due to Weight of rod=mgasin(pi/3)=[itex]\frac{mga\sqrt{3}}{2}[/itex]
    Torque due to weight of mass=(mg/2)(2a)sin(pi/3)=[itex]\frac{mga\sqrt{3}}{2}[/itex]

    Total torque = [itex]mga\sqrt{3}[/itex]

    so that

    [tex]I_{total} \alpha=mga\sqrt{3} \Rightarrow \alpha=\frac{mga\sqrt{3}}{I_{total}}[/tex]

    are the torques correct?
     
  9. May 10, 2008 #8

    Doc Al

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    Looks good!
     
  10. May 13, 2008 #9

    rock.freak667

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    Apparently not...I found that the answer is really [tex]\alpha= \frac{3g}{10a}[/tex]

    and I get

    [tex]\alpha=\frac{9\sqrt{3}g}{20a}[/tex]
     
  11. May 13, 2008 #10

    Doc Al

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    Let me look it over one more time.
     
  12. May 13, 2008 #11

    Doc Al

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    I see the problem.
    In the above expression for torque, [itex]\theta[/itex] is the angle between [itex]\vec{r}[/itex] and [itex]\vec{W}[/itex], which is not [itex]\pi/3[/itex] but its complement.
    I must have been asleep at the wheel and assumed you were taking the sine of the correct angle. Sorry about that! :redface:

    Correct that and you'll get the expected answer.
     
  13. May 13, 2008 #12

    rock.freak667

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    ahhh..thanks..my diagram was wrong...I pit pi/3 between the rod and the hinge as the complement...thank you!
     
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