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Homework Help: Rotation of a rod about a fixed end

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data

    A 1 m long rod, 200g, is hinged at one end and held out horizontally, then released. What is the speed if the tip of the rod as it hits the wall?

    2. Relevant equations

    Initial energy = final energy
    rotational energy = .5Iω[itex]^{2}[/itex]
    gravitational potential energy = mgh
    kinetic energy (if needed?) = .5mv[itex]^{2}[/itex]

    3. The attempt at a solution

    Initial energy = mgh =mgL
    Final energy = .5Iω[itex]^{2}[/itex] = (for a rod) (6)[itex]^{-1}[/itex]mL[itex]^{2}[/itex]ω[itex]^{2}[/itex]

    set equal, cancel m and L:

    √(6g/L) = ω
    L=1, => 7.67≈ω
    ω(radius) = ω(L) = ω = |v|

    However the book states it as √(3g/L) = ω. It got this answer by using the center of mass of the rod to get the potential energy as mg(.5L) then solved ω as √(3g/L), then reasoned that the tip of the rod would be √(3g/L)(L) = √(3gL)≈5.4

    The crux of my problem is I don't understand why we must use the center of mass to derive ω. Does the tip somehow pick up some (non rotational) kinetic energy?
    Last edited: Aug 5, 2014
  2. jcsd
  3. Aug 5, 2014 #2

    Simon Bridge

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    Your argument is that gravitational PE lost by the rod gets stored as rotational KE.
    Where on the rod does the force of gravity act?
  4. Aug 5, 2014 #3
    I see what you're saying, we were indeed taught that gravity acts on the center of mass... but I'm not sure why one can't assume that gravity acts on all the tiny masses in the rod, and then, since all of the mass rotates in a periodic fashion around a fixed point all that energy be considered to rotational kinetic energy and not translational kinetic energy?
  5. Aug 5, 2014 #4


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    So if I understand correctly, you're wondering why it doesn't work to just look at the change in gravitational energy of a small (point) mass on the end of the rod, and then equate it to the kinetic energy of that small mass on the edge of the rod?


    And that is why you say the (Δ) gravitational energy is mgL?


    If that is your idea, then why did you use the rotational inertia of rod? Wouldn't you want the rotational energy of a point (at a distance L) which would lead to Final energy = [itex]\frac{1}{2}mL^2ω^2[/itex] (which = [itex]\frac{1}{2}mv^2[/itex])
  6. Aug 5, 2014 #5
    Gotcha, so the moment of inertia of a point about a fixed axis in mL[itex]^{2}[/itex], of course this makes sense as I = ∫r[itex]^{2}[/itex]dm, r[itex]^{2}[/itex] = L[itex]^{2}[/itex] and dm is just m.

    The only thing that bothers me is that rotational and translational kinetic energy can be defined differently, sort of rubs me the wrong way. Anyway, thanks!
  7. Aug 5, 2014 #6


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    They're defined differently merely for convenience. Rotational kinetic energy is not different from translational kinetic energy. But if you have something rotating like a rod, then the kinetic energy is different at each point on the rod, so you can't just use the center of mass.
    That is where the rotational kinetic energy formula comes in; it is a convenient formula that saves you from analyzing each rotational problem with calculus.

    Let's suppose I didn't know the rotational kinetic energy formula.
    This is how I would solve for the rod's kinetic energy using the translational kinetic energy formula:

    I know that the speed of each point on the rod depends on the distance from the rotation axis (let's call it r).
    Namely, v=ωr
    So the kinetic energy of a point as a function of it's distance, r, would be [itex]\frac{1}{2}dm(ωr)^2[/itex] (where dm is the mass of the point)
    That is simply the translational kinetic energy formula.
    We can say [itex]dm=\frac{M}{L}dr[/itex] (I hope that makes sense) and then we can integrate from zero to L and we get:
    [itex]\int^L_0(\frac{M}{2L}(ωr)^2)dr=\frac{1}{6}mL^2ω^2[/itex]= the kinetic energy

    So you see, rotational and translational kinetic energy are not really different at all! The apparent difference is just for convenience.
    (In translational kinetic energy, each point has the same speed as the center of mass, so it is more simple.)

    Hopefully it doesn't rub you the wrong way anymore!
    Last edited: Aug 5, 2014
  8. Aug 5, 2014 #7


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    Anyway I forgot to say, even if you use the correct rotational inertia of the point on the edge, you don't get the correct answer!

    You get [itex]v=\sqrt{2gL}[/itex] which is too small. (So your question is still intriguing!)

    The point mass on the edge of the rod somehow lost some of it's energy.

    Hmm, it's as if the edge of the rod shared some of it's energy with the rest of the rod by "pulling it down"?
    Maybe the rest of the rod was rotating more slowly than the point on the edge would have?

    It's interesting to think about :smile:
    Last edited: Aug 5, 2014
  9. Aug 5, 2014 #8

    Simon Bridge

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    (Just to make sure) ... that is an entirely valid approach.
    You just didn't do it that way. Instead you treated all the mass as if it were concentrated at the tip of the rod. Consider:

    The rod has mass m and length L so it has a linear mass density of m/L
    Measuring x along the rod from the pivot, the mass element dm between x and x+dx falls a total distance x so it's change in GPE is dE(x)=gx.dm (from mgh) and dm = (m/L).dx

    The total change in GPE is all the dE elements added together along the length of the rod: $$E=\frac{mg}{L}\int_0^L x\; dx = mg\frac{L}{2}$$... i.e. the change in GPE for the whole rod as a unit is the same as if the entire mass were concentrated at the center of mass for the rod.

    The key to understanding this calculation is to realize that different parts of the rod fall a different distance - so they have a different contribution to the overall change in GPE.

    The same thing works for the relationship between rotational and translational kinetic energy - the thing to realize is that different parts on the rod have different (translational) kinetic energies - to find the total kinetic energy, you have to add them up (i.e. integrate over the length of the rod). The result of the integral works out the same as the special treatment for rotational kinetic energy. You can see it right away if you work out the rotational KE for a point mass moving in a circle radius r with constant speed v, then simplify.

    I'm going to encourage you to do the integrals.
  10. Aug 7, 2014 #9
    Simon, thanks, I did your integral on the board and it definitely makes sense, as well as substituting the rotating point mass into the .5Iω[itex]^{2}[/itex] and ascertaining that it is indeed just .5mv[itex]^{2}[/itex]. Biggest mistake was probably assuming I could use I of a rod for just a point mass. I shant be so quick to plug and chug next time.

    Nathanael, yes that has me quite confused as well @_@. E[itex]_{f}[/itex]=E[itex]_{i}[/itex], and if rotational energy can be written in the same form as translational kinetic how can dmgL =/= .5dmv[itex]^{2}[/itex]?? Where else could it possibly go? Maybe it has something to do with holding the chemical bonds of the material together? *wild speculation*
  11. Aug 7, 2014 #10


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    I think the reason it's not okay to apply conservation of energy to the point on the end of the rod because there are external forces acting on it (from the rest of the rod) which do not net to zero.

    The advantage of looking at the entire rod as a whole is that all the inter-molecular forces (or whatever) net to zero because of newton's third law.

    But that doesn't really explain why the tip of the rod loses energy.

    I don't really have a worthy explanation except to point out the following:
    If you were to remove the tip of the rod and attatch it to a massless string (or rod) of the same length as the rod and then compare it's angular acceleration with that of "the rest of the rod" then you find that the tip of the rod is accelerating downwards (angularly) slower than the rest of the rod. (If that made sense?)
    So when you look at "the tip of the rod" and "the rest of the rod" as two seperate objects which are attatched, you can say that rest of the rod is "trying" to accelerate (angularly) downwards quicker than the tip of the rod, and so it is "pulling" the tip of the rod downwards, "sharing" some of it's energy.

    This isn't much of an explanation, but it's all I got

    (Edited because I said it all backwards the first time :tongue:)
    Last edited: Aug 7, 2014
  12. Aug 7, 2014 #11

    Simon Bridge

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    You can do the integral for translational KE for each point on the rod as well - same approach as the potential energy change ... each mass element dm between x and x+dx has a different final speed.

    This is what I'm trying to encourage you to do.

    - well, to start with, not every dm falls a distance L, and nor does every dm have the same final speed v.

    The mass dm between x and x+dx falls a distance x, so loses gpe of gx.dm=(gm/L)x.dx.
    This you already saw in post #8 - which you said made sense.

    It also has a final speed of v(x)=xω ... see where you tripped up?

    So you are wondering why you cannot write: $$\frac{gm}{L}x\;dx = \frac{m}{2L}\omega^2x^2\;dx$$ ... to see why not, mathematically, try solving for ##\omega##.
    (Note: ω should not depend on x.)

    However, if you integrate the LHS and the RHS along the length of the rod, you will end up with the correct overall conservation of energy rule. You'll see the correct moment of inertia of the rod drop out automatically.

    This is hardly surprising since: $$\frac{m}{L}\int_0^L x^2\;dx$$ ...is the definition of "the moment of inertia of the rod". But I think you should do the integral yourself to really get it.
    Last edited: Aug 7, 2014
  13. Aug 7, 2014 #12

    Simon Bridge

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    The physics you are after is that conservation of energy only works for closed systems

    However - the individual mass elements dm are not closed systems, they do not move independently: they are attached to the other mass elements in the rod. So you don't expect conservation of energy to work - energy lost in one part of the rod may be gained in another part.

    Thus: your intuition about intermolecular forces was correct - well done.
    Last edited: Aug 7, 2014
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