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Mechanics - Pivoting rod and angular momentum

  1. Nov 21, 2006 #1
    A uniform rod of length l is freely pivoted at one end. It is intially held horizontally and then released from rest.

    Write down expressions for the angular momentum L(theta) of the rod and the torque T(theta) about the pivot when the rod is inclined at an angle theta to the horizontal.

    Hence show that the angular velocity at this instant is given by:

    theta[double-dot]^2 = (3*g*sin(theta))/l


    I apologise in advance for the horribleness of my equation; I'll figure out how to use that Latex thing in a little while=)

    Anyway, I think I've solved the first two bits of the problem; for the angular moment I get L = (1/3)*m*l^2*theta[dot], using the angular velocity and the moment of inertia of the rod.

    Differentiating gives the torque N = (1/3)*m*l^2*theta[double-dot] and the torque is also given by N = l*m*g*sin(theta) because of the gravitational force acting on the rod. Equating those two expressions for N and solving for theta[double-dot] gives me:

    (3*g*sin(theta))/l

    which is the result that I'm supposed to get for the squared angular velocity. Now, I have the feeling that I went wrong somewhere, I just can't figure out where! What have I done wrong?
     
  2. jcsd
  3. Nov 21, 2006 #2

    OlderDan

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    I haven't checked everything you did carefully, but you seem to have the right idea. However, where is the force mg acting on the rod?
     
  4. Nov 21, 2006 #3
    I brought in the force when calculating the torque N(theta) because N = F*l; the force is then m*g*sin(theta), and equating that to the other result for N gives me the result above. I think that I'm close to the result but I must have missed out some crucial detail.
     
  5. Nov 21, 2006 #4

    OlderDan

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    What if I told you that if F = mgsinθ as you you said, then N does not equal F*l?
     
  6. Nov 21, 2006 #5
    Hm... But what then? I thought that N depends on l and the force acting perpendicular to it- is that not mg*sin(theta)?
     
  7. Nov 21, 2006 #6

    OlderDan

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    Draw a diagram of the rod at an angle θ and put a vector to represent the weight of the rod, mg. What is the distance between the pivot point and the line of the force?
     
  8. Nov 21, 2006 #7
    Now I get l*cos(theta) for the distance. That still doesn't help me with the end result though; I'm still getting:

    theta[double-dot] = (3*g*cos(theta))/l
     
  9. Nov 21, 2006 #8

    OlderDan

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    I actually had not noticed the incorrect sinθ instead of cosθ because it depends on which angle you are using (horizontal or vertical) and I was thrown off by the problem statement, which has an error
    theta[double-dot] is angular acceleration not angular velocity. The angular velocity is theta[dot]

    You are correct that the problem specified the angle from the horizontal so the torque is proportional to cosθ. The angular velocity should be maximum at θ = 90°, so the problem statement must really mean velocity, not acceleration.

    What you are still missing is that the force of gravity does not act at the end of the rod. It acts at the CM in the middle of the rod. At θ = 0 the torque is not mgl as your new equation suggests. It is only half that.
     
  10. Nov 21, 2006 #9
    head ---> desk

    So sorry, that was my mistake! The problem is for theta[dot], so it does ask for the angular velocity. I'm so sorry if I've confused you=(

    And I'll work through that CM of mass bit- let's see if that works any better.

    Thank you for help already though, because that problem has been bugging everybody in my year for days now!
     
  11. Nov 21, 2006 #10
    So, now I get:

    theta[double-dot] = (3/2)*cos(theta)*g/l

    Integrating that with respect to t changes the theta[double-dot] to (1/2)*theta[dot]^2 on the l.h.s. and changes cos(theta) to sin(theta) on the r.h.s, giving me the result as specified in the original question.

    Is that right?
     
  12. Nov 21, 2006 #11

    OlderDan

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    To summarize, you are saying

    L = Iω = (1/3)*m*l²*ω

    N = dL/dt = (1/3)*m*l²*α = Iα
    and
    N = (1/2)*m*g*l*cosθ

    (1/3)*m*l²*α = (1/2)*m*g*l*cosθ

    α = (3/2)*(g/l)*cosθ

    So yes, that looks good. How are you doing that integral? The derivative of ½ω² is ωα and the derivative of sin(t) is cos(t) but you don't have cos(t); you have cosθ(t). There is a way to get there, but I'm wondering how you did it.
     
  13. Nov 21, 2006 #12
    At the moment I'm purely guessing that integrating *will* give me the right result; I suppose it's got something to do with the chain rule, but I'll have a closer look at it tomorrow. I think that integrating theta[double-dot] will give me 1/2 theta[dot]^2; but I'm not too sure about cos(theta) because I don't think we've done that in our maths course yet.
     
  14. Nov 21, 2006 #13

    OlderDan

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    The chain rule is the key. On the left you will want a dω and on the right you will want a dθ.
     
  15. Nov 22, 2006 #14
    Are there sites on the web that explain integrating while using the chain rule? I've never done this before, so while I roughly know how it should work, I'm not too clear on the details.
     
  16. Nov 22, 2006 #15

    OlderDan

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    I'm sure you can find it somewhere out there. In this case it is just

    α = dω/dt = (dω/dθ)(dθ/dt) = ω(dω/dθ) = (3/2)*(g/l)*cosθ

    Multiply through by dθ and integrate both sides. As an indefinite integral, there is a constant of integration that must be chosen to satisfy the inital conditions of the position and angular velocity of the rod.
     
  17. Nov 22, 2006 #16
    Thank you so much, that really helps! Wouldn't the constant of integration be omega at t=0?
     
  18. Nov 22, 2006 #17

    OlderDan

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    You have taken the explicit time dependence out of the problem and solved for ω in terms of θ. The initial condition is ω = 0 when θ = 0. If you wanted to find ω(t) and θ(t) you would have some more work to do.
     
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