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I found this interesting question on some MIT problem sheet. Say a rope with mass is attached to a shaft and experiences angular velocity so that it is horizontal, find a its tension as a function or distance $r$, ignoring gravitational effects.

The two initial conditions that I found were, $T\left(L\right)=0$ and $T\left(r\right)=mL\omega^{2}$

So I came up with two solutions via two different methods, the first solution (just by eye-balling it) is a follows

$$T(r)=\left(L-r\right)\frac{mr\omega^{2}}{L}$$

The second solution was found using differential elements and is the correct answer based of the material I found:

$$T(r)=\frac{m\omega^{2}}{2L}\left(L^{2}-r^{2}\right)$$

Now, my confusion is that:

  1. Shouldn't the two methods produce the same result? Perhaps the result of using integration produces a more accurate model of the behavior.

  2. The second solution does not check out with the initial conditions which I set out. When $r=L$, $T\left(L\right)=0$, this checks out, However when $r=0$, $T\left(0\right)=\frac{1}{2}mL\omega^{2}$. This leads me to believe that my initial condition is wrong, but I simply cant see why the tension at the start of the rope is $\frac{1}{2}mL\omega^{2}$ and not $mL\omega^{2}$.

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  • $\begingroup$ Also, when I was studying finite-element analysis in engineering, we compared our results with the "Exact Solution" but I never got an answer to where it came from, is the use of differential elements here the method of finding that "Exact Solution". $\endgroup$ – Sar Dec 4 '17 at 11:55
  • $\begingroup$ The problem can be found from this link web.mit.edu/8.01t/www/materials/modules/chapter09.pdf $\endgroup$ – Sar Dec 4 '17 at 11:56
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  1. As you already saw, you can't just guess a function that fits the boundary conditions. You must solve Newton's laws of motion to get the right one.

  2. As for the solution, you are right saying $$T\left(r\right)=\frac{m\omega^{2}}{2L}\left(L^{2}-r^{2}\right) \tag{1}$$ and in fact you are wrong that $T\left(0\right)=m\omega^{2}L$. You have a mass $m$ that is spread from $r=0$ to $r=L$. You are treating it like all the mass is at $r=L$ when you write $T\left(0\right)=m\omega^{2}L$. The right way to look at it is to observe that because the mass is evenly spread over this distance, one can equivalently say that there is a mass $m$ at half the distance $r=\frac{L}{2}$ (that is exactly the center of mass). Now it is easier to see that $T\left(0\right)=\frac{m\omega^{2}L}{2}$ in agreement with Eq. $1$.

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  • $\begingroup$ Thanks for pointing out that error in my logic. What you say makes perfect sense. If I were to make a cut at r=0, drawing a free-body diagram and analyze the forces, I would have to replace the object (the rope) as a point particle to do it, ergo, the center of mass of said object, which effectively halves the length of mLw^2. I do have one more question, how did you manage to edit the equations that I wrote to the correct symbols? I am new to this and could not find the function that would allow me to write proper equations. $\endgroup$ – Sar Dec 5 '17 at 16:01
  • $\begingroup$ See this math.meta.stackexchange.com/questions/5020/…. $\endgroup$ – eranreches Dec 5 '17 at 16:09

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