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Tension in Whirling Rope

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A uniform rope of mass ##M## and length ##L## is pivoted at one end and whirls with uniform angular velocity ##\omega##. What is the tension in the rope at distance ##r## from the pivot? Neglect gravity.
    scan0001.png

    2. Relevant equations
    $$\vec{F}=m\vec{a}$$

    3. The attempt at a solution
    Consider the differential element as shown in the figure.
    FBD of differential element.PNG

    I considered the setup of the tension acting on the differential element as such because I thought that net force that should be acting on the differential element should be in the direction of the centripetal acceleration. So, we get
    $$T+dT-T=(dm)\omega^2x\implies dT=\left(\dfrac{M}{L}\omega^2x\right)dx$$
    On, integrating the expression with proper limits we get
    $$\int^{T}_{T_0}{dT}=\int^{r}_{0}\left({\dfrac{M}{L}\omega^2x}\right)dx \implies T-T_0=\dfrac{M}{2L}\omega^2r^2$$
    Now, to find ##T_0##, we put the value of that ##T## which acts at the end of the rope. The value of ##T##, i.e. the value of the tension acting in the direction opposite to the direction of the acceleration of the element we see that the value of ##T## is ##0## as there is no tension acting in the outward direction of the rope.

    So, we get $$-T_0=\dfrac{M}{2L}\omega^2L^2\implies T_0=-\dfrac{M}{2L}\omega^2L^2$$

    $$\therefore T=T_0+\dfrac{M}{2L}\omega^2r^2=\dfrac{M}{2L}\omega^2\left(r^2-L^2\right)$$

    The answer that I get has an opposite sign than that of the book's, which is ##\dfrac{M}{2L}\omega^2(L^2-r^2)##. Is it because the direction in which I considered ##T## to act is opposite to what I had assumed. If that is so, then doesn't that mean that ##T+dT(\gt T)## is in the direction opposite to that of the centripetal acceleration, so how would the centripetal acceleration be supplied to the differential element if the net force is in the direction opposite to that of the centripetal accleration.
     
  2. jcsd
  3. Sep 16, 2016 #2

    Orodruin

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    You have defined the change in the x coordinate as the right minus left value of x while defining the change in the tension as the left value minus the right. This gives you a sign error.
     
  4. Sep 16, 2016 #3
    But shouldn't the gravity increase as we go towards the center of the circle in which the circle whirls. Does, it just mean that the expression ##dT=\left(\dfrac{M}{L}\omega^2x\right)dx## should be ##dT=-\left(\dfrac{M}{L}\omega^2x\right)dx##.

    Also, I had not read the "neglect gravity" part of the question so I was thinking that what should be the minimum velocity for which the whole of the rope rotates , i.e. the rope doesn't slacken anywhere while rotating. And I was not able to think of a way to approach this problem, if you can suggest how to attempt it would be very helpful.
     
  5. Sep 16, 2016 #4

    Orodruin

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    "Gravity" does not. The tension does. You must define dT and dx in compatible ways. You cannot take the difference for the x coordinate one way and the difference in the tension another and then expect to get the correct result.

    Yes.
     
  6. Sep 16, 2016 #5
    My bad. It was a silly mistake on my part I was thinking of what I had mentioned in the last part of the post, but I meant tension only.

    Also, can you tell me how to interpret the situation that I have mentioned in the last part of the previous post.
     
  7. Sep 16, 2016 #6

    haruspex

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    Do you mean, if gravity is not ignored? It might be like a catenary, except that the horizontal component of tension is not constant.
     
  8. Sep 17, 2016 #7
    If what you say about the shape of the rope being that of a catenary, then it would look like this am I correct.
    geogebra-export.png
    But as you can see in the figure too I am not able to get the length of the differential element of the rope.
    Hmm....isn't the vertical component also non-constant.
     
  9. Sep 17, 2016 #8

    haruspex

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    In a catenary, the horizontal component of tension is constant but the vertical component varies. Here they will both vary.
     
  10. Sep 17, 2016 #9
    I still can't seem to find a way to tackle this problem. Can you give another hint.
     
  11. Sep 17, 2016 #10

    haruspex

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    The equation including gravity is a very hard differential equation. I can help you get to that if you are determined, but let's deal with the problem as given first, ignoring gravity. Or have you solved that?
     
  12. Sep 17, 2016 #11
    Yes, I have solved it with the help that I received from Orodruin.
     
  13. Sep 18, 2016 #12

    haruspex

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    Ok.
    For the gravitational case, consider an element of horizontal length dx at horizontal distance x from the tether.
    Suppose the element makes angle θ to the horizontal at one end and θ+dθ at the other.
    Similarly, let the tension be T at one end and T+dT at the other.
    If it has linear density ρ and the rotation rate is ω, what is the mass of the element and what are the forces on it?
     
  14. Sep 18, 2016 #13
    Hmm....I was thinking of doing it by considering ##dx## and ##dy## as the differentials for the horizontal and the vertical heights respectively, but the ##d\theta## differential is always handy, dunno why I didn't think of it. Now then ,what I did is as shown as follows:-

    abrakadabra.png
    The horizontal length of the differential element as given in the figure is taken as ##dr##, so the length of the differential comes out to be ##dx\sec\theta##.

    So, we get the mass of the differential as ##dm=\dfrac{M}{L}\sec\theta dr##

    Now, dealing with the horizontal force equation of the differential we get
    $$T\cos{(\theta+d\theta)}-(T+dT)\cos\theta=dm(\omega^2r) \\
    \implies T\cos\theta-T\sin\theta d\theta-T\cos\theta-\cos\theta dT=\dfrac{M\sec\theta}{L}\omega^2rdr$$
    $$\implies T\sin\theta d\theta+\cos\theta dT=-\dfrac{M\sec\theta}{L}\omega^2rdr \tag{1}$$

    And doing the same for the vertical force equation we get
    $$T\sin{(\theta+d\theta)}=(T+dT)\sin{\theta}+dmg$$
    $$\implies T\cos\theta d\theta-dT\sin\theta=\dfrac{Mg\sec\theta}{L}dr\tag{2}$$

    $$(1) + (2) \implies T(\sin\theta+\cos\theta)d\theta +(\cos\theta-\sin\theta)dT=\dfrac{M\sec\theta}{L}(g-\omega^2r)dr$$

    Now only if there was no ##\sec\theta## in RHS it would have been a lot easier, but am I going the right track.
     
  15. Sep 18, 2016 #14

    haruspex

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    Yes, that's a good start. I would have put θ on the left and θ+dθ on the right to be consistent. As you have it you will have to be careful with signs.
    For ease of writing, I would write ρ for a linear density to replace M/L.

    There's not much point in adding the two equations together like that at the end. It loses information without eliminating a variable.
    Manipulate the equations to get an equation without T' (by which I mean dT/dr) into the form T=.... Differentiate that to get an equation for T'. You can then use those equations to eliminate T and T' from one of your two equations.
    Note that you will then have a second order differential equation. That's the price for eliminating a variable.
     
  16. Sep 18, 2016 #15
    I am sick of these differentials, what you think is the correct one is not correct because of the convention you are following:headbang:. It also makes the expression for ##\dfrac{dT}{dr}## a not so good as compared to what it had been before. Anyways, can you tell me the reason as to why you suggest that ##\theta## should be on the left and ##\theta + d\theta## on the right. And, how am I supposed to decide whether the convention for the angle that I have chosen is correct or not. Is it based on the concept that as I have decided that the differential ##dx## is directed in the positive direction, then so should be the differential ##d\theta##, i.e is in anticlockwise direction.
     
    Last edited: Sep 18, 2016
  17. Sep 18, 2016 #16

    haruspex

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    Yes. As you have it, dθ/dr>0 would correspond to the slope decreasing as r increases.
     
  18. Sep 19, 2016 #17
    I left as it is because it was a little difficult to deal with the term having two differentials multiplied. Would it change anything except the signs. o_O

    I did as what you instructed to and what I got was ##\sin\theta=c_1(r-\dfrac{L}{2})##, but on putting ##\dfrac{d\theta}{dr}=\dfrac{c_1}{\cos\theta}## in the expression
    $$\dfrac{dT}{dr}=\dfrac{\rho\sec\theta\omega^2r+T\sin\theta\dfrac{d\theta}{dr}}{\cos\theta} \\
    =\dfrac{\rho\omega^2r+Tc_1\sin\theta}{\cos^2\theta} \\
    \implies \dfrac{dT}{dr}+Tc_1\tan\theta\sec\theta=-\rho\omega^2r\sec\theta$$

    On solving the above linear differential equation some ridiculous relation between the variables ##T, \theta \text{and} r##

    $$T=\dfrac{\rho\omega^2}{2\sin^2\theta c_1^2}(1+\cos{2\theta})+k_1e^{-\frac{2c_1r\sin\theta}{\cos{2\theta}+1}}-\dfrac{\rho r \omega^2}{c_1\sin\theta}$$
     
  19. Sep 19, 2016 #18

    haruspex

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    I do not understand how you avoided getting a second order DE. You should have got ##T=\rho(g-\omega^2r\tan(\theta))/\theta'##. Differentiating that will give ##\theta''## terms.
     
  20. Sep 19, 2016 #19
    I did get a 2nd order differential equation b/w ##\theta \text{ and } r##, which was ##\dfrac{d^2\theta}{dr^2}=\tan\theta\left(\dfrac{d\theta}{dr}\right)^2##, which gives ##\sin\theta=c_1r+c_2##
     
  21. Sep 19, 2016 #20
    I thought that I should also show my work, so here goes:-

    $$T\cos(\theta+d\theta)-(T+dT)\cos{\theta}=(dm)\omega^2r \\
    \implies T\sin\theta\dfrac{d\theta}{dr}+\cos\theta dT= -\rho\sec\theta\omega^2rdr \qquad\qquad\qquad\qquad...(1)$$

    From ##(1)##, we get $$\dfrac{dT}{dr}=-\dfrac{\rho\sec\theta\omega^2r+T\sin\theta\frac{d\theta}{dr}}{\cos\theta}$$

    From the vertical force equation we see from post #13, that
    $$T\cos\theta \dfrac{d\theta}{dr}-\sin\theta dT=\rho g\sec\theta dr \\
    \implies T\cos\theta + \tan\theta \left(\rho\sec\theta\omega^2r+T\sin\theta\dfrac{d\theta}{dr}\right)=\rho g\sec\theta \\
    \implies T\sec\theta\dfrac{d\theta}{dr}=\rho\sec\theta(g-\omega^2r\tan\theta) \\
    \implies T\dfrac{d\theta}{dr}=\rho(g-\omega^2r\tan\theta)$$

    Now differentiating the above expression throughout w.r.t ##r##, we get
    $$\dfrac{dT}{dr} \dfrac{d\theta}{dr}+T\dfrac{d^2\theta}{dr^2}=\rho(-\omega^2r\sec^2\theta)\dfrac{d\theta}{dr} \\
    \implies T\dfrac{d^2\theta}{dr^2}=T\tan\theta\left(\dfrac{d\theta}{dr}\right)^2$$

    On solving the differential equation, we get
    $$\sin\theta=c_1r+c_2$$

    Now that I see it now it looks like I wrongly concluded the value of the integral constant ##c_2## as ##c_2=-\dfrac{c_1L}{2}##. Well but it was not going to the change the relation that I found in post #17.
     
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