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Pendulum pulled through hole adiabatically

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data

    A string with a lead ball of mass ##m## is slowly pulled upward through a small hole. Consider the work done on the system during this process and find the change in energy and the frequency of this pendulum during this "adiabatic process", assuming the amplitude of the pendulum to be small.

    3. The attempt at a solution

    If the process is adiabatic, meaning quasi-static, then the pendulum should be in mechanical equilibrium at each instant in the direction along the string, as if it were just swinging side to side from a string tied to a fixed ceiling with tension balancing the component of gravity along the string. In other words, adopting polar coordinates centered on the hole with the radial direction ##\hat{r}## along the string and the polar direction ##\hat{\theta}## tangent to the pendulum swing, the string should have zero radial velocity ##\frac{dr}{dt}\rightarrow 0## when being pulled up because it is being pulled quasi-statically. In the present case, the force applied in pulling the string should be along the string and such that it equals the force of gravity along the string i.e. ##\vec{F} = -mg\cos\theta \hat{r}## where ##\theta## is the angle the pendulum makes with the vertical. Let ##l## be instantaneous length of the pendulum below the hole. Due to adiabaticity, the energy of the pendulum at any given instant will be ##E = \frac{1}{2}m l^2\dot{\theta}^2 - mgl\cos\theta##. The work done on the pendulum in pulling it through the hole is ##W = \int_{l_0}^l \vec{F}\cdot \vec{ds} = \int_{l_0}^{l} mg\cos\theta dr = mg\cos\theta(l - l_0)## so ##\delta W = mg\cos\theta \delta l## hence ##dE = \delta W = mg\cos\theta \delta l \approx mg\delta l - \frac{1}{2}mg\theta^2\delta l## since ##\cos\theta \approx 1 - \frac{1}{2}\theta^2## for small amplitudes.

    The book on the other hand says the answer should be ##dE = -\frac{1}{2}mg\bar{\theta^2}\delta l## where ##\bar{\theta^2}## is the time average of ##\theta^2## over one period of the pendulum. Incidentally, the quantity ##\bar{\theta^2}## is the time average over one period but the period of the pendulum is changing at each instant since it is being pulled through the hole so I'm not even sure how this time average is well-defined; perhaps it doesn't matter in the end for the necessary calculation but it isn't obvious to me. More importantly and pertinently, I honestly have no idea how the book even got this answer, where and why a time average even comes into the problem, and where my own solution goes wrong. Any help is appreciated, thank you.
     
  2. jcsd
  3. Jul 17, 2014 #2

    TSny

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    There is more to ##\vec{F}## than ##-mg\cos\theta \hat{r}##. Does the pendulum bob have any centripetal acceleration?
     
  4. Jul 18, 2014 #3
    Ah yes you're right. We should really have ##F = -(mg\cos\theta - ml \dot{\theta}^2)\hat{r}## where ##ml\dot{\theta}^2## is the magnitude of the centripetal acceleration on the pendulum at any instant, again since ##\frac{dl}{dt} \rightarrow 0## in the quasi-static limit. Then the work done on the pendulum during this process would be ##W = \int_{l_0}^l \vec{F}\cdot d\vec{s} = \int_{l_0}^{l} (mg\cos\theta - ml \dot{\theta}^2)dr##.

    However I cannot proceed past this point for the following reason: as the string is pulled through the hole, at any instant of time during the process the string will be at a different angle to the vertical than the previous instant of time. In other words, as the length of the string decreases below the hole, its angle will also be changing as the pendulum is still swinging from side to side. As a result, ##\cos\theta## would be a function of ##r = l## the instantaneous length of the string. The same goes for ##\dot{\theta}## since the tangential velocity of the pendulum changes as it swings about and so it will be different from instant of time, or one value of the instantaneous length ##l##, to the next. Am I correct in that regard? If so I'm not entirely sure how to move ahead since we don't have ##\cos\theta## or ##\dot{\theta}## explicitly as functions of ##r = l##.
     
  5. Jul 18, 2014 #4

    TSny

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    OK, but I believe there is a sign error in the equation. Carefully write out ##\sum F_r = m a_r## for the pendulum bob.

    This is where the time averaging comes in. The string is pulled so slowly that the pendulum undergoes many oscillations during the time it takes to shorten the string by a small amount ##\delta l##. Thus, when calculating the work done by the tension force for an infinitesimal shortening of the string, you can use the time average of the magnitude of the tension force over a complete oscillation.

    So, you can express the work done by the tension for an infinitesimal shortening of the string in terms of time averages of ##\cos \theta## and ##\dot{\theta}^2##.
     
  6. Jul 19, 2014 #5
    Ah yes sorry. We have, taking as usual the outward radial direction emanating from the hole to be positive, ##mg\cos\theta - F = -ml\dot{\theta}^2## so ##F = mg\cos\theta + ml\dot{\theta}^2## hence ##\vec{F} = -(mg\cos\theta + ml\dot{\theta}^2)\hat{r}## since the applied force on the string is radially inwards.
    I guess my confusion here is, the period of oscillation itself depends on how long the string is below the hole. So say I take the time average ##\bar{F}## over the period of the pendulum when it has length ##l## i.e. I calculate ##\bar{F} = \frac{1}{T}\int_0 ^T F dt## where ##T = 2\pi \sqrt{l/g}##. At ##l - \delta l## it will have a new period of oscillation with change ##dT = \frac{\pi}{g}\sqrt{g/l}\delta l##. Why then will ##\bar{F}## be independent of ##l## given that the period, as mentioned, depends on ##l##? The goal is to make ##\bar{F}## independent of ##l## is it not?
    I'm a bit confused by this as well. As of now the work done on the system in pulling the string by an infinitesimal amount is ##\delta W = mg\cos\theta \delta l + ml \dot{\theta}^2 \delta l##. Because the process is quasi-static, the pendulum undergoes many oscillations during the act of pulling it by the amount ##\delta l## and we have the luxury of time-averaging the applied force over these oscillations, or equivalently over just one period of oscillation. In turn we get a time averaged infinitesimal work ##\delta \bar{ W} = mg\langle{\cos\theta}\rangle \delta l + ml \langle \dot{\theta}^2\rangle\delta l##; I've used ##\langle{\cos\theta}\rangle## and ##\langle \dot{\theta}^2\rangle## to denote the time average just to make it easy on the eyes. But why will ##\delta\bar{ W}## give us the change in energy ##\delta E## as opposed to just some average change in energy ##\delta\bar{ E}##? Presumably it is ##\delta E = \delta W## that we want and not ##\delta\bar{ E}## unless ##\delta\bar{ E} = \delta E##, the truth of which isn't immediately obvious to me. Furthermore if we're only interested in the infinitesimal work done, wherein no integrals over ##dl## need to be evaluated, then why do we need to time average at all? Couldn't I just write down ##\delta W = mg\cos\theta \delta l + ml \dot{\theta}^2 \delta l## and work with that? Thanks for the help so far.
     
  7. Jul 20, 2014 #6

    TSny

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    OK

    ##\bar{F}## will depend on ##l##. But, you are trying to get an expression for the work done by ##F## for an infinitesimal displacement ##\delta l## so that you can get a result for ##dE## that is accurate to first order in ##\delta l##. The work done, ##dW = - \bar{F} \delta l##, is proportional to ##\delta l##. So, ##dW## will be accurate to first order in ##\delta l## if you treat ##\bar{F}## as independent of ##l## during the infinitesimal displacement.

    Consider an infinitesimal shortening ##\delta l## of the string. The pendulum will have a certain definite energy before the shortening and a certain definite energy after the shortening. So there will be a definite value of ##dE## for the pendulum. There is no need to consider a time average of ##dE##. However, to calculate ##dE## from the work done by the force ##F##, you need to use the time averaged force during the shortening of the string. That's because we are considering a quasi-static shortening where the pendulum undergoes many oscillations during the shortening ##\delta l##.
     
  8. Jul 22, 2014 #7
    Bear with me here because this is the part I dont understand. Why does the quasi-static shortening necessitate the need for a time average of the tension when calculating the infinitesimal work done? In other words, why can't we make sense of the infinitesimal work done by the exact force, not its time average, during infinitesimal shortening? Why does the adiabaticity and presence of many oscillations during the shortening prevent us from using the exact force in getting the infinitesimal work?

    As an aside I went ahead and tried to calculate the work done with the time average. I used the virial theorem ##\langle V\rangle = \langle T\rangle## for the pendulum to get ##-mgl\langle \cos\theta \rangle = \frac {1}{2} ml^2 \langle \dot {\theta}^2 \rangle## hence ##\delta W = -mg\langle \cos\theta \rangle \delta l \approx -mg \delta l + \frac {1}{2} mg\langle \theta^2\rangle## which isn't what the book claims is the correct answer, see post #1, but I am not sure where my mistake is.

    Thanks.
     
  9. Jul 22, 2014 #8

    TSny

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    I suspect I can't satisfy you here (which shows my own limitations!). My way of thinking about an adiabatic (quasi-static) shortening of the string by an "infinitesimal" amount ##\delta l## is that it is done so slowly that the time ##\Delta t## that it takes to change the length by ##\delta l## is "large". In other words, if the length is changed at a constant rate ##\dot{l}##, then ##\dot{l}## is so small that $$\Delta t = \frac{\delta l}{\dot{l}} $$ contains many oscillations. So, there isn't one instantaneous value of the force that you can use for the shortening by ##\delta l##. Rather, you have to use an average value of the force during the large time ##\Delta t##. This is certainly "hand-wavy". I haven't thought about it any deeper. I imagine it's all been made rigorous.

    The virial theorem result ##\langle V\rangle = \langle T\rangle## is only applicable if you choose the "zero of potential energy" so that ##V \propto \theta ^2##. That is, the zero of potential energy should be at the bottom of the swing (θ = 0).
     
  10. Jul 23, 2014 #9

    WannabeNewton

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    Here's one way to think about the time averaging, building off of what TSny said. Intuitively, because the process is quasi-static, the work done by the applied force over a given period of oscillation and in fact over many periods will be zero. The work done will only be non-zero over extremely large time scales ##\delta t \gg T## where ##T## is one period or even several periods. In other words, oscillations in the applied force ##F## due to the oscillation of the pendulum itself will be on time scales that are much smaller than the time scale on which work is non-zero. Thus when you compute the work done you will not be sensitive to these oscillations because you are only measuring the change in length of the pendulum over ##\delta t##, not ##T##; rather, you will only be sensitive to the average force.

    An analogy that might be useful to consider is that of an ideal gas in a cylinder with a piston inserted wherein you move the piston inwards quasi-statically. You want to compute the work done on the gas in moving the piston inwards quasi-statically from one position to another. Now if we were interested in the work done on extremely short time scales, say the time between collisions of gas particles with the piston, then we would have to consider the individual forces exerted by the gas particles on the piston which will in general vary on these time scales. But in a quasi-static compression, the work done will be zero on these time scales. We are only interested in the work done over the extremely large time it takes to move the piston quasi-statically from one position to another and on such time scales the extremely complicated time variation in force on the piston due to collisions with the gas particles will not be discernible. We can only discern the average force i.e. the pressure.

    Here is a more formal argument. Now technically the frequency ##\omega## with which ##F## oscillates varies over each period because the pendulum is being pulled in but the pendulum is being pulled so slowly that the time derivative ##\dot{\omega}## of the frequency is essentially zero, ##\dot{\omega} \approx 0##, and we can therefore basically take it to be constant. We can then expand ##F## in a Fourier series: ##F(t) = \bar{F} + \sum a_n \cos n\omega t + \sum b_n \sin n\omega t##. Here ##\bar{F}## is the time average of ##F## over a single period and the Fourier coefficients ##a_n,b_n## are independent of time. Let's say the pendulum is pulled from an initial length of ##l_0## at time ##t = 0## to a length ##l## at time ##\Delta t##. Then ##W = \int_{l_0}^l F(l) dl## but we can invert ##l(t)## to get ##t(l)## and write ##W = \int_{0}^{\Delta t} F(t) \frac{dl}{dt}dt##.

    Now not only is ##\frac{dl}{dt}## extremely small but it is also approximately constant in order for the process to be quasi-static. Hence ##W = \frac{dl}{dt}\int_{0}^{\Delta t} F dt##; let ##\frac{2\pi n}{\omega}## be the number of periods of oscillation of the pendulum nearest to ##\Delta t## so that ##\frac{dl}{dt}\int _{0}^{\frac{2\pi n}{\omega}} F dt = \frac{dl}{dt}\int_{0}^{\Delta t} Fdt + \frac{dl}{dt}\int_{\Delta t}^{ \frac{2\pi n}{\omega}}Fdt##. But that second term is just the work done on the pendulum between ##\Delta t## and ##\frac{2\pi n}{\omega}## which we know is zero because the length of the pendulum essentially does not change over entire periods let alone in between a single period.

    Therefore ##W = \frac{dl}{dt}\int _{0}^{\frac{2\pi n}{\omega}} F dt = \frac{\Delta l}{\frac{2\pi n}{\omega}}\frac{2\pi n}{\omega}\bar{F} + \frac{dl}{dt}\sum a_n \int _{0}^{\frac{2\pi n}{\omega}}dt \cos n\omega t + \frac{dl}{dt}\sum b_n \int _{0}^{\frac{2\pi n}{\omega}}dt \sin n\omega t = \bar{F}\Delta l ## where I've used the fact that ##\frac{dl}{dt}## is essentially constant so that ##\frac{dl}{dt} = \frac{\Delta l}{\frac{2\pi n}{\omega}}##. Taking the limit ##\Delta l \rightarrow 0## we get ##\delta W = \bar{F} \delta l##. The essential point here is that the oscillatory terms ##\cos n\omega t,\sin n\omega t## integrated over many periods will end up vanishing.

    Now I said this was a "more" formal argument but as you can see it has quite a bit of hand-waving in it. I personally couldn't come up with anything more rigorous. Hopefully TSny can comment on whether or not he was convinced by it :smile:
     
  11. Jul 23, 2014 #10

    TSny

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    Hello, Wannabe. I think we are essentially on the same page. I need to take more time to look at your post. In your Fourier expansion of F(t), I'm not convinced that you can take the expansion coefficients an, bn as time independent. But they would be very slowly varying in time. But, I will think some more about it. In the meantime, I constructed the following somewhat laborious argument which has some parts similar to yours. Not sure if it helps at all.

    At any specific time t the tension force is $$F = mg \cos \theta+ml \dot{\theta}^2 \approx mg-\frac{mg}{2} \theta^2 + ml\dot{\theta}^2$$ For adiabatic shortening, the angle ##\theta## varies with time to good approximation as ##\theta(t) = A(t) \cos [\omega(t) t]##, where ##A(t)## and ##\omega(t)## are very slowly varying functions of time.

    Then, the time dependence of F can be expressed generally as $$F(t) = F[A(t), \omega(t), t]$$
    Explicitly, I think you get ##F = mg[1+A(t)^2-\frac{3}{2}A(t)^2\cos ^2( \omega(t) t)]## but we won’t actually need an explicit expression.

    Suppose we want to know how much work is done by the tension force over one oscillation; say, between ##t = t_0## and ##t = t_0 + T## where ##T## is the time to execute one oscillation starting at time ##t_0##. Then, $$\delta W = -\int_{l_0}^{l_0+\delta l}F[A(t), \omega(t), t] dl = -\int_{t_0}^{t_0+T}F[A(t), \omega(t), t] \dot{l} dt$$

    We may assume that the rate of shortening ##\dot{l}## is constant, so $$\delta W = -\dot{l} \int_{t_0}^{t_0+T}F[A(t), \omega(t), t] dt = -T \dot{l} \;\; \frac{1}{T} \int_{t_0}^{t_0+T}F[A(t), \omega(t), t] dt = -\delta l \langle F \rangle$$
    ##\dot{l}## is assumed to be so small that the change in length ##\delta l## of the pendulum during one oscillation is a small, first order quantity.

    So, we have ##\delta W = -\delta l \langle F \rangle##, where ##\langle F \rangle## is the time average of the force over the one oscillation while the string is being shortened.

    For calculating <F> over one oscillation, the force at time t can be expanded as $$F[A(t), \omega(t), t] \approx F[A(t_0) + \dot{A}(t_0)(t-t_0), \; \omega(t_0) + \dot{\omega}(t_0)(t-t_0), \; t]$$ $$ \approx F[A[t_0) , \omega(t_0), t] + \frac{\partial F}{\partial A}\dot{A}(t_0)(t-t_0) + \frac{\partial F}{\partial \omega}\dot{\omega}(t_0)(t-t_0)$$
    where ##\dot{A}## and ##\dot{\omega}## are assumed to be small, first order quantities.

    ##\delta W## will still be accurate to first order in ##\delta l## if we drop the ##\dot{A}## and ##\dot{\omega}## terms (since ##\delta W## already has a first-order factor ##\delta l##). Hence, for ##\langle F \rangle## we can use $$\langle F \rangle = \frac{1}{T} \int_{t_0}^{t_0+T}F[A(t_0), \omega(t_0), t] dt \equiv \overline{F}$$ where ##\overline{F}## is the time average of the force over a cycle if the length of the string were to be kept constant during the oscillation. So, for the average force over one oscillation that is needed to find ##\delta W##, we do not need to take account of the fact that the string is being shortened. We can just use the time average of the force over a cycle assuming the string length remains constant at whatever length ##l_0## it had at the beginning of that oscillation.

    So, we get ##\delta W = - \overline{F} \delta l##.
     
  12. Jul 23, 2014 #11

    WannabeNewton

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    Hi TSny! Aren't the Fourier coefficients time-independent by definition? They are given by ##a_n = \frac{1}{T}\int _T F(t) \cos n \omega t## and ##b_n = \frac{1}{T}\int _T F(t) \cos n \omega t## so they are just numbers. Of course this assumes that ##\omega = \text{const.}## whereas in the OP's problem, ##\omega = \omega(t)## but ##\frac{d\omega}{dt} \sim \frac{dl}{dt}## since the time variation of frequency comes solely from the time variation of length. So we have ##\frac{d\omega}{dt}\ll 1## and we can approximately take ##\omega## to be constant in the problem, so as to Fourier expand ##F## in the usual way. In the limit as ##\frac{dl}{dt}\rightarrow 0## this should become exact. Presumably this is what you meant when you said that the Fourier coefficients are slowly varying in time.

    But your argument is definitely more elegant than mine without a doubt and it also involves far less hand-waving so it satisfies my own curiosity :) thanks!
     
  13. Jul 23, 2014 #12

    TSny

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    F(t) is not a "fixed" function of time. As the string is pulled shorter, the frequency and amplitude of the motion will change (very slowly). So, the coefficients of the expansion would change slowly over time. But I think you realized that. I believe you are considering short enough time intervals in your argument so that the changes in the coefficients are negligible. So, it's OK.
     
  14. Jul 23, 2014 #13

    TSny

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    Actually, the force would contain additional terms with factors of ##\dot{A}(t)## and ##\dot{\omega}(t)##. But for adiabatic shortening, these terms can be neglected.
     
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