# Differential tension in a rope experiencing constant angular velocity

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 $mL\omega^{2}$.

• 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". – Sar Dec 4 '17 at 11:55
• The problem can be found from this link web.mit.edu/8.01t/www/materials/modules/chapter09.pdf – Sar Dec 4 '17 at 11:56

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. $1$.