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I have a system with a mass $m$ attached to the end of a cable. The cable mass is assumed negligible. The cable is attached to the ground at the one end while the other, with the attached mass $m$, is moving vertically with some known velocity $v$. The cable is modelled as a 2nd order system with known values for the damping $b$ and spring $k$ coefficients. I'm trying to use the conservation of energy principle to determine the impulse force resulting in the cable (at the moving end specifically) as the cable becomes taut, but I'm having trouble accounting for the energy dissipated due to the cable damping. The cable is well damped and this significantly reduces the impulse force generated. I've setup up the following system of equations, does anyone have any advice on how I can go about solving them? I'm guessing that I need to follow some sort of iterative process? Or is there an alternative method that I could follow that would be simpler?

$$ E_1 = E_2 $$ $$E_1 = \frac{1}{2}mv_1^2$$ $$E_2 = \frac{1}{2}kx^2 +mgx + \int_0^t bv(t)^2 dt$$ $$v(t) = v_0 + \int_0^ta(t) dt$$ $$a(t) = F(t)/m$$ $$F(t) = kx(t) + bv(t)+mg$$

$E_1$ is the kinetic energy the instant before the cable begins to stretch, while $E_2$ is the energy of the system when the mass has come to rest with the cable having stretched vertically some distance $x$. The energy in the system at $E_2$ is equal to to the stored energy in the spring, the increase in potential energy due to the cable stretching a distance $x$ and the energy dissipated by the damper/dashpot. I'm trying to solve Equation 3 to calculate the change in cable length $x$, which I'll then use to calculate the impulse force from, $$F_{impulse} = kx$$

Does the answer perhaps lie with using these equations with the conservation of momentum principle?

$$mv_1 = F_{impulse}*t$$

The mass is also assumed to not rotate on impact and I neglect the energy dissipation due to the propagation of lateral or longitudinal waves in the cable.

Cable Model at E_1

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  • $\begingroup$ A couple of clarifying questions: is the cable considered massless? Is the damping occurring throughout the cable? Can you draw a diagram of the setup - where is the damping occurring, exactly? If you are asking about the instantaneous acceleration of the mass when it is released (with the rope in tension), this is just a solution of the differential equation for a damped harmonic oscillator with boundary conditions of v(0)=0. You should be able to find $dv/dt$ and thus the acceleration - from which the force follows. $\endgroup$ – Floris Sep 20 '15 at 21:09
  • $\begingroup$ This is unclear: a cable with damping properties? Do you mean some bungee rope with high degree of hysteresis? Diagram, please. $\endgroup$ – Gert Sep 20 '15 at 21:50
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    $\begingroup$ Hi @Floris, Gert I've added a simple diagram of the model at instant 1 just before the cable begins stretching. The mass is moving vertically (only, please excuse the diagram being a little skew) with a known initial velocity $v = v_1$. $\endgroup$ – Jennifer Sep 21 '15 at 10:34
  • $\begingroup$ To answer your question, @Floris, yes I do neglect the cable mass, I use lumped values for the spring and damping coefficients and assume they are constant throughout the cable. I'm trying to calculate the impulse force generated by the moving mass on the taut cable. I can get it from the instantaneous acceleration from release but then I require the change in length of the cable surely? This value is unknown. All I have known is the velocity $v_1$ and the cable properties. $\endgroup$ – Jennifer Sep 21 '15 at 10:34
  • $\begingroup$ @Gert, the cable would, I imagine have some hysteresis contribution but the cable is not very elastic and thus I model it as negligible. The damping is a result of internal friction forces in the cable which dissipate the energy as heat. $\endgroup$ – Jennifer Sep 21 '15 at 10:34
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If initially the mass is at $x=0$ and the initial velocity is $V$ then the (underdamped) position response is:

$$ x(t) = X \exp(-\beta t)\sin(\omega t) = \frac{V}{\omega} \mathrm{e}^{-\zeta \omega_n t} \sin(\omega t) $$

where $$\begin{aligned} \omega_n & = \sqrt{\frac{k}{m}} \\ \zeta & = \frac{d}{2 m \omega_n} = \frac{d}{2 \sqrt{k m}} \\ \omega &= \omega_n \sqrt{1-\zeta^2} = \sqrt{\frac{k}{m} - \frac{d^2}{4 m^2}} \end{aligned} $$

The force on the rope is $F=k x + d \dot{x}$ and the impulse is $J=\int F\,\mathrm{d}t$ defined over half a cycle of oscillation. Plugging the position response yields

$$ J = \int \limits_0 ^ {\frac{\pi}{\omega}} k x + d \dot{x} \,\mathrm{d} t = \\ = V m \left(1 + \mathrm{e}^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}} \right)$$

So without damping $\zeta=0$ and $J=2 V m$ for a perfect "bounce" and with critical damping $\zeta=1$ and $J=V m$ with a "plastic" response. The above can be redefined as a coefficient of restitution $\epsilon$ with

$$ \epsilon = \mathrm{e}^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}} $$

The kinetic energy is $E=\frac{1}{2} m \dot{x}^2$ and its value at the $n$-th half cycle of oscillation is $$E_n = \frac{1}{2} m V^2 \epsilon ^ {2 n} $$ and since the coefficient of restitution is $0 \leq \epsilon \leq 1$ then $E_n \leq E_0=\frac{1}{2}m V^2$

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  • $\begingroup$ You star, thank you! That looks pretty much spot on. I just have two questions. Firstly, can I find the solution to the integral - $J = \int_0^\frac{\pi}{\omega} kx + d\dot{x} dt$ - somewhere as a sanity check or did you calculate that yourself? Secondly, should the equation - $F = kx + d\dot{x}$ - not include the gravity component? $\endgroup$ – Jennifer Sep 21 '15 at 19:41
  • $\begingroup$ The damping part simplifies to zero $$d \int \dot{x}\,\mathrm{d}t = d \int \limits_0^{\frac{\pi}{\omega}} \frac{\mathrm{d} }{\mathrm{d}t} X \exp(-\beta t) \sin( \omega t ) \,\mathrm{d} t = 0 $$ $\endgroup$ – ja72 Sep 21 '15 at 20:26
  • $\begingroup$ And the stiffness part yields $$ k \int x \, \mathrm{d} t = k \int \limits_0^{\frac{\pi}{\omega}} X \exp(-\beta t)\sin(\omega t)\,\mathrm{d}t = \frac{k X \omega}{\beta^2+\omega^2} \left(1+\mathrm{e}^{-\pi \frac{\beta}{\omega}}\right) $$ $\endgroup$ – ja72 Sep 21 '15 at 20:31
  • $\begingroup$ A small tweak - the "equilibrium position" for the oscillation is the position where the spring force would oppose the force of gravity; except that since this is a rope, it can only pull, not push. The equation of motion is the same for the time that the rope is taut, but there is an additional force term that displaces the equilibrium position slightly. Otherwise, this is the analysis I was hinting at but didn't have time to write out - so props (and upvote) to you for doing it. $\endgroup$ – Floris Sep 22 '15 at 15:12
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It seems to me you are making this more complicated than it needs to be. When the cable first becomes taut, the spring force is not yet in play and the only force will be $v\cdot k$ - by the definition of the drag in the dash pot. You can compute the subsequent motion by solving the damped harmonic oscillator.

Let me know if this is enough?

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  • $\begingroup$ No I don't think that's correct @Floris. The internal friction will only begin absorbing energy as the cable lengthens. The cable will then simultaneously begin absorbing energy as elastic/strain energy due to the spring $E_s = \frac{1}{2}kx^2$, and the system will also gain potential energy (transferred from the initial kinetic energy of course) due to the subsequent increase in height. $\endgroup$ – Jennifer Sep 21 '15 at 14:39
  • $\begingroup$ Ultimately the impulse force will depend on both the viscous damper and the spring, with the contributions from each component varying with time (the damper contribution will be at a maximum at $t = 0+$, with $F_d = bv_{1+}$, while the spring contribution will be at a maximum when the mass comes to rest at $t = t$, with $F_k = kx$. What I'd like, ideally, is some way to estimate the energy dissipated by the damper as the mass slows. Then I'll remove that amount of energy from the system to estimate the distance $x$ that the cable stretches, and use that to estimate the average impulse force. $\endgroup$ – Jennifer Sep 21 '15 at 14:52
  • $\begingroup$ OK - so you are looking for the maximum height reached, and an estimate of the average force as the mass decelerates. And you think the displacement is sufficient that the influence of gravity cannot be ignored during that deceleration. Right? $\endgroup$ – Floris Sep 21 '15 at 15:42
  • $\begingroup$ Yes exactly. I did a rough calculation with and without the potential energy, although I neglected the damping, and it had quite a significant effect on the impulse force. Around 30-35%, although that will definitely decrease if I'm able to include an estimate for the damping somehow. $\endgroup$ – Jennifer Sep 21 '15 at 18:49

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