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Tangential speed of a rod pivoted from one end

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin, uniform rod of length L and mass M is pivoted from one end. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. Suppose a small metal ball of mass m = 2M is attached to the rod a distance d from the pivot. The rod and the ball are released from rest in the horizontal position. (a) Show that when the rod reaches the vertical position, the speed of its tip is:

    [tex]v_{t} = \sqrt{3gL} \sqrt{(1 + 4(\frac{d}{L}))/(1+6(\frac{d}{L}^{2})[/tex]

    2. Relevant equations



    3. The attempt at a solution
    I've managed to figure out how they got the [tex]\sqrt{3gL}[/tex] bit, but the rest I have no idea. It looks like they're following the [tex]v_{t} = \omega \cdot R[/tex] formula, because [tex]\sqrt{3gL}[/tex] is the angular velocity of the rod as it passes the vertical, but how [tex]\sqrt{(1 + 4(\frac{d}{L}))/(1+6(\frac{d}{L}^{2})[/tex] equals the radius from the pivot to the tip of the rod, I have no idea.

    What is the new center of mass of this system when you add the ball to it? The center of mass of just the rod is L/2, so is the center of mass after [tex]\frac{2}{3}(d-\frac{L}{2})[/tex]?


    In the formula for the tangential velocity of the tip, shouldn't the radius r just be the length L? Or do you have to use the distance to the center of mass here?

    I'll keep working on it. Tips would be appreciated.
     
    Last edited: Nov 11, 2008
  2. jcsd
  3. Nov 11, 2008 #2

    Hootenanny

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    Let's start by thinking about the centre of mass (COM). Looking at your proposed COM what would happen if d<(L/2), i.e. if the ball was placed before the middle of the rod? If this is the case then your expression above would be negative, which would mean that your COM of mass is not on the rod. Does that sound realistic?
     
    Last edited: Nov 11, 2008
  4. Nov 11, 2008 #3
    Yes, I seem to have problems calculating the centre of mass. The formula for the centre of mass I'm using is:

    [tex]COM= \frac{m1 X1 + m2 X2}{m1 + m2}[/tex]

    Which in this case is [tex]COM= \frac{M \cdot \frac{L}{2} + 2M \cdot d}{3M}[/tex]

    The masses cancel and it leaves L/2 + d. Does that seem more reasonable?

    Also, I made an algebraic mistake with the angular velocity of the rod before the mass is added. It's [tex]\omega = \sqrt{\frac{3g}{L}}[/tex], not [tex]\omega = \sqrt 3gL[/tex]. So then I need to find how it changes when adding a mass 2M at a distance d, and then get their equation for the tangential velocity.
     
    Last edited: Nov 11, 2008
  5. Nov 11, 2008 #4
    It seems to me that you should be able to represent the motion of the rod as potential energy before = rotational kinetic energy after + potential energy after. Does this formula make sense?

    [tex]E_{i} = 3(MgL)[/tex]

    [tex]E_{f} = \frac{1}{2}I \omega ^{2} + Mg\frac{L}{2} + 2Mgd[/tex]
    [tex]I =\frac{1}{3}ML^{2}[/tex]

    [tex]3(MgL) = \frac{1}{2}(\\frac{1}{3}3ML^{2}) \omega ^{2} + Mg\frac{L}{2} + 2Mgd[/tex]

    Then you'd solve that for omega, and multiply omega by L (the radius) to get the tangential velocity... But when I work it out I don't get an equation close to the one they get...
     
    Last edited: Nov 11, 2008
  6. Nov 11, 2008 #5

    Doc Al

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    OK.

    The last term is problematic. "d" is the distance from the pivot, not the other end.

    That's just the rotational inertia of the rod. What about the ball?

    Redo this after you correct the terms above.

    Good.
     
  7. Nov 11, 2008 #6
    Am I really expected to include the moment of inertia of the ball in this? The question doesn't say whether the ball is solid or not. I assume that the axis of rotation goes through the center of the ball, but again the question doesn't say. So there are three possible moments of inertia that the ball could have...

    Also, I don't know the radius of the ball, and the moments of inertia of a sphere depend on R.

    [tex]E_{f} = \frac{1}{2}I \omega ^{2} + Mg\frac{L}{2} + 2Mgd[/tex]
    This becomes [tex]E_{f} = \frac{1}{2}I \omega ^{2} + Mg\frac{L}{2} + 2Mg(L-d)[/tex]?
     
    Last edited: Nov 11, 2008
  8. Nov 12, 2008 #7

    Hootenanny

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    Since the radius of the ball is small, you may treat it as a point mass. So you need to determine the moment of inertia of a rod of length L with a point particle located at a point d along the length of the rod.
    Better :approve:
     
  9. Nov 12, 2008 #8
    Are they expecting me to write the distance d as [tex]\frac{d}{L}[/tex] instead of L - d, though, to get their formula? Would that work?

    Here is the new equation with the moment of inertia of the particle:

    [tex]3(MgL) = \frac{1}{2}(\frac{1}{3} ML^{2}) \omega ^{2} + \frac{1}{2}(2M(\frac{d}{L}^{2})) \omega ^{2}+ Mg\frac{L}{2} + 2Mg(\frac{d}{L})[/tex]
     
    Last edited: Nov 12, 2008
  10. Nov 12, 2008 #9

    Doc Al

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    No, that makes no sense. d/L is a fraction, not a distance. Don't worry about getting the final version of the formula until you solve for v.

    Fix those terms where you have d/L. :yuck:
     
  11. Nov 12, 2008 #10
    Okay, so you get: [tex]3(MgL) = \frac{1}{2}(\frac{1}{3} ML^{2}) \omega ^{2} + \frac{1}{2}(2M((L-d)^{2})) \omega ^{2}+ Mg\frac{L}{2} + 2Mg(L-d)[/tex]

    But to be honest, I don't know where to begin simplifying this to solve for omega. Where do I even start?
     
  12. Nov 12, 2008 #11

    Doc Al

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    Start by correcting the 2nd term. What's the distance between the ball and the axis?
     
  13. Nov 12, 2008 #12
    Ah.
    [tex]3(MgL) = \frac{1}{2}(\frac{1}{3} ML^{2}) \omega ^{2} + \frac{1}{2}(2M(d)^{2}) \omega ^{2}+ Mg\frac{L}{2} + 2Mg(L-d)[/tex]

    Now is there anything *else* that's wrong with the formula? :frown:
     
    Last edited: Nov 12, 2008
  14. Nov 12, 2008 #13

    Doc Al

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    Good! Now you can start simplifying.
     
  15. Nov 12, 2008 #14
    [tex]6(MgL) = (\frac{1}{3} ML^{2}) \omega ^{2} + (2M(d)^{2}) \omega ^{2}+ Mg L + Mg(L-d)[/tex]

    Divide [tex]2M(d^{2})[/tex] out:

    [tex]\frac{3gL}{(d^{2})} = \frac{1}{6} \frac{L^{2}}{(d^{2})} \omega ^{2} + \omega ^{2}+ \frac{g L}{(2(d^{2})} + \frac{gL}{d}[/tex]


    ... Like I said, I don't know how to simplify this.
     
  16. Nov 12, 2008 #15

    Doc Al

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    Isolate ω.
     
  17. Nov 13, 2008 #16
    I realised that I had to isolate omega, as is obvious by the fact that I put in my posts precisely that:

    What I did not know is how to begin simplifying this equation.

    There is a difference between gently nudging someone toward the right answer and being deliberately vague and unhelpful.
     
  18. Nov 13, 2008 #17

    Doc Al

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    Then do it! That's the first step.

    Your right hand side has two terms that don't contain omega. Move them!

    The physics is done. Now comes a little algebra.
     
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