The dynamics of a swinging payload suspended from a stationary crane, an unwanted phenomenon on a construction site, can be described as a simple pendulum. However, an experienced crane operator can deliver a swinging payload and have it stop dead on target in a finite amount of time by carefully modulating the speed of the trolley. Generally, a series of precisely timed stop and go movements of the trolley are implemented to damp out the kinetic energy of the simple harmonic oscillator. Here, this mysterious crane operator's trick will be revealed and ultimately generalized to capture the case where the load is initially swinging. Finally, this modus operandi is applied to a torsion balance used to measure G, the universal gravitational constant responsible for the swinging of the crane's payload in the first place.

Many engineers have developed robust control algorithms for the automation of industrial cranes to solve the problem of damping the swinging motion of a suspended load.1–3 Still, few have analyzed the capabilities of the most classic controller—the human crane operator. While the pendulum provides a simple yet fascinating system to study physics,4 most articles on damping or excitation achieve their goal by changing the effective lengths of the pendulum.5–7 Moreover, the papers that describe damping rarely take advantage of the Laplace transformation, with12 being an exception.8–11 Here, we study a pendulum with a suspension point that is allowed to move horizontally via a trolley. Using the Laplace transform to obtain the pendulum's motion, we show that damping can be achieved simply by timing the suspension point's motion. In short, the authors combine their experience of driving overhead cranes (SS, LC) with their love for physics (all authors) to unveil the principles behind one of the tricks used by crane operators. This trick consists of two stop-and-go motions of the trolley. These can damp out the undesirable swinging motion of the pendulum so that it stops dead on target in a finite amount of time. These movements will be analyzed and explained in Sec. III. Finally, this technique is applied to a torsion pendulum physics experiment at the National Institute of Standards and Technology (NIST) for measuring the Newtonian constant of gravitation G, showing the ubiquity of simple harmonic motion, ranging from construction sites to the modern laboratory.

Figure 1 shows an idealized diagram of a typical tower crane with two degrees of freedom: the horizontal motion provided by the trolley and the vertical motion by the hoist. For this analysis, the load is assumed to have already been lifted to the proper height, and vertical motion during liftoff and touchdown is ignored. A point load with mass m is suspended by a rope with fixed length l and negligible mass and can swing at small angles with a period To. The horizontal position of the load (mass) is given by $x m$ and that of the trolley (on the crane) by $x c$. The corresponding velocities are abbreviated by $v m = x ̇ m$ and $v c = x ̇ c$, respectively. The angle θ of the rope with respect to vertical can be obtained from $sin θ = ( x m − x c ) / l$. With the trolley at rest, the velocity and acceleration of the load tangential to its arcuate motion are $l θ ̇$ and $l θ ¨$. For small excursions ($θ ≪ 1$), the velocity and acceleration can be approximated as $x ̇ m$ and $x ¨ m$, respectively. Hence, the linearized differential equation of motion is
(1)
where g is the local gravitational acceleration and c is a damping coefficient that accounts for the fluid friction force $c x ̇ m$, which is proportional to the velocity of the load. The damping term $c x ̇ m$ is included in Eq. (1) for completeness, but, from here on, we neglect it to avoid complicating the equations and overburdening the physical insight with mathematics. In addition, the experimental oscillator discussed in Sec. V exhibits almost no damping so that this approximation is valid in that case also. Without damping, Eq. (1) can be rewritten as
(2)
Appendix A shows the derivation of Eq. (2) with the Lagrange formalism. Using the Laplace transform,13 a popular technique used in control theory for solving ordinary differential equations, Eq. (2) can be transformed to the s-domain,
(3)

The Laplace transform is similar to the Fourier transform but uses the complex variable s instead of $i ω$ for converting between the time and frequency domains. We denote, as is usual, the variables in the Laplace domain with upper case letters and the ones in the time domain with lower case letters. A table in  Appendix B gives the transformations that are used in this article.

Finally, the response function, the ratio of output to input, in this case the load position to the trolley position, is given by
(4)

Let us start with a simple case that does not require any shenanigans. For t < 0, both the load and the trolley are initially at rest with $x m = x c = 0$. A series of two movements are examined: at t =0, the trolley begins to move at constant velocity (Sec. II A), and at $t = τ$, it comes to a stop (Sec. II B). Following these two motions, the crane operator's trick to stop the load from swinging is revealed (Sec. II C).

We assume that for t < 0, the trolley is at rest, and at t =0, it instantaneously starts to move with constant velocity $v c$. Hence, $x c$, determined by the Heaviside function u(t) multiplied with $v c t$, is given by
(5)
Note that in the formal definition, $u ( 0 ) = 0.5$, but this point is inconsequential here because $u ( t ) t = 0$ for t =0. The corresponding Laplace transformation $X c ( s )$ is obtained using  Appendix B so that the position of the load in the Laplace domain, $X m ( s )$ can be inferred from Eq. (4):
(6)
Equation (6), when converted back into the time domain for t > 0, yields
(7)
The horizontal velocity of the load, $v m$, is the time derivative of $x m$
(8)
Equation (8) shows that the average speed of the load, $v m$, is equal to the steady speed of the trolley, $v c$. However, its speed oscillates with an amplitude of $v c$, so, with respect to the construction site, the load's speed varies from 0 to $2 v c$. At $t = k T o$, the speed of the load is zero. Here, as throughout the text, k denotes a positive integer. If the operator stops the trolley at those instants, both the velocity of the trolley, $v c$, and the of load, $v m$, will be zero. Furthermore, as is apparent from Eq. (7), the position of both the load and the trolley will then be $v c k T o$. The mass will hang straight down, with θ = 0, from the resting trolley. Thus, the pendulum will be at rest and will remain at rest until the next stimulus arises.
In this section, we use the Laplace transformation to find the motion of the load after the trolley is brought abruptly to a stop, and we check the claim that stopping the trolley at times t = kTo will result in the load hanging at rest. As before, the trolley starts moving at t =0 with a velocity $v c$, but now comes to a stop at $t = τ$. In the Laplace domain, this motion is given by
(9)
where the first term represents the motion starting at t =0, and the second term represents the stop at a later time τ. In the Laplace domain, a time shift by τ requires multiplication by $exp ( − τ s )$. The transformation back to the time domain gives Eq. (7) added to a similar term, but with a negative sign and shifted in time by τ. Hence, the distance traveled by the load is given, for $t > τ$,14 by
(10)
Furthermore, the velocity of the load at the pendulum frequency is determined by the temporal derivative of Eq. (10):
(11)
Expanding the second term, and subsequent grouping of $cos ( ω o t )$ and $sin ( ω o t )$ terms yields
(12)
Using half-angle identities for $ω o τ$, one obtains
(13)
The expression in the square bracket is $sin ( ω o t − ω τ / 2 )$. Hence,
(14)
The equation is different from Eq. (8) as it is the outcome of two changes in the trolley velocity instead of one. The second sine in Eq. (14) describes the free pendulum motion with a phase shift. Its amplitude is modulated by the first sine that is dependent on τ. Thus, the load velocity becomes zero when $ω o τ = k 2 π$, or rather $τ = k T o$, exactly as conjectured above.
Section II B describes a trajectory where the load is at rest both initially and finally. However, during the translation, the horizontal displacement of the load with respect to the trolley can be quite large, up to $v c / ω o$. To avoid pendulum motion during transport, the trolley starts at t =0 with $v c / 2$ and then, at $t = τ$, the velocity is increased to $v c$. The crane operator's secret is in calculating the value for τ. In the s domain that stimulus is
(15)
The solution for $x m$ in the time domain for $t > τ$ is
(16)
Thus, if $ω o τ = ( 2 k + 1 ) π$, or $τ = T o / 2 + k T o$, the pendulum motion is cancelled. The crane operator's trick is to start the trolley with half the desired velocity. Half a period later, the load passes directly below the trolley with a velocity that is twice as fast as the trolley. At that point, the crane operator doubles the trolley speed to match the speed of the load. After this maneuver, or set of movements, the load glides along with constant velocity.

Since all good things must come to an end, the trolley must be stopped before the end of the track. Stopping with zero swing can be accomplished by performing the same trick in reverse: the trolley's speed is reduced to $v c / 2$ and then, an odd multiple of $T o / 2$ later, reduced to zero.

The analysis above has one shortcoming. It assumes that the trolley speed can go from 0 to $v c$ or $v c / 2$ instantaneously, which, unfortunately, is impossible. Instead, assume that, at t =0, the trolley starts accelerating with $a c$ over a duration of δt. A second acceleration with the same duration begins at $t = τ$. In the s domain, the stimulus is
(17)
For $t > τ + δ t$, the response in the time domain is
(18)
The term in the first line describes the motion with constant velocity and is identical to that of the trolley. Note that, for $t > τ + δ t$, the velocity of the trolley is $v c = 2 a c δ t$. For $τ = T o / 2 + k T o$, the first and third cosines as well as the second and fourth cosines are the same with opposite sign and cancel. Hence, all oscillatory terms vanish for $τ = T o / 2 + k T o$, and the load tracks the trolley.

In summary, both crane operator tricks are valid even if the trolley's acceleration is finite. Note that we assume that both accelerations, at t =0 and at $t = τ$, are identical and applied for the same duration, δt. This assumption may not hold in a real-world situation, such as if the motor moving the trolley outputs a constant mechanical power.

The scenario discussed in this section may be uncommon in the world of crane operators, but is applicable to the physics experiment discussed later. As previously, we assume the trolley to be initially at rest at $x c = 0$, but the load is now assumed to be swinging. Without loss of generality, we set t =0 as the time when the load swings directly under the trolley at $x m = 0$ with positive velocity $v m ( 0 ) = v o$. One could, for instance, imagine that a gust of wind causes the load to swing at the beginning of the crane operator's shift in which the task is to move the load from $x m = 0$ to $x m = x g$, so that it arrives with the smallest possible pendulum amplitude. To avoid unnecessary complication, we assume that the trolley can move with infinite acceleration but has a maximum speed of $v c$.

The load's motion will be damped using two identical trolley movements performed at a known speed vc each of duration δt. The combined distance traveled by the trolley is $x g = 2 v c δ t$. To succeed in the task, the crane operator has to quickly solve the equations of motion and determine the times t1 and $t 1 + τ$ when these trolley operations are to be executed, as the ground recipients are notoriously impatient.

The two moves can be described in the Laplace domain similar to Eq. (15). However, how should the initial conditions be taken into account? The easiest way is to assume that the pendulum is at rest and add a delta-function impulse term to the stimulus, which reproduces the initial conditions. This term is virtual because the crane operator does not actually have to physically execute this step. If the trolley experiences an initial impulse given by $x c = δ ( t )$ at t =0, then the pendulum at rest has, by definition of the response function,
(19)
To obtain the desired initial condition, i.e., a velocity vo, the virtual stimulus must, therefore, be $x c = ( v o / ω o 2 ) δ ( t )$. The combination of the virtual and real trolley motions in the Laplace domain is
(20)
Multiplying this by the response function and transforming back into the time domain gives for $t > t 1 + τ + δ t$
(21)
Since we are interested in the pendulum motion of the load after the maneuver, we take the temporal derivative of the previous equation, which yields
(22)
(23)
For the pendulum to be at rest, the terms within the two square brackets must be equal to zero. Inspecting the terms inside the second set of square brackets yields two possible ways the product can be zero. First, one could choose $ω o τ = π$. However, for that case, the result would be
(24)
As we have seen in Sec. II C, making two moves an odd multiple of half a period apart will not change the energy stored in the pendulum's motion, and the load's amplitude will not be damped. The second way to make the terms vanish would be
(25)
Then, the velocity of the pendulum for $t > t 1 + τ + δ t$ is
(26)
The sign before $4 v c$ is positive if k in Eq. (25) is even and negative if it is odd. The pendulum motion can be damped to zero as long as $v o ≤ 4 v c$. For the case $v o = 4 v c$, the choices for τ and δt would be $τ = T o / 2 + k 1 T o$ and $δ t = 2 k 2 T o$, with $τ > δ t$ and k1, k2 positive integers. Usually, δt is determined by the distance $x g = 2 v c δ t$ over which the load needs to be transported. For a given δt, the largest possible amplitude that can be damped to zero is
(27)
To accomplish this, τ must be chosen as
(28)
Furthermore, if
(29)
then the load will not be at rest after the maneuver. However, the energy of the pendulum can be minimized by setting $cos ( τ ω o / 2 ) = 1$, or in other words
(30)

In this section, we examine two examples based on the equations derived previously. We assume $l=20 m$, resulting in a pendulum period of 8.97 s for small angles. The goal is to move the load by 6 m horizontally to the right with a trolley speed of $v c=1 m s − 1$, yielding $δt=3 s$.

In the first example, it is assumed that the load has a velocity amplitude of $v o=1 m s − 1$. According to Eq. (28), $τ=5.32 s$ which, as an input for Eq. (25), yields $t 1=2.57 s$. Figure 2 shows a numerical simulation of this example where the solid traces represent the position $x m$ (bottom panel) and the velocity $v m$ (top panel) of the load while the dotted traces represent the position $x c$ and velocity $v c$ of the trolley. To better visualize the initial swing, t1 begins a full period late, i.e., $t 1=8.97 s+2.57 s=11.54 s$.

The calculation has been made by numerically solving the differential equation (1) with the Runge–Kutta method. The program is written in Python and is freely available at https://github.com/schlammis/pendulum. The simulation also allows solving the nonlinear differential equation of motion and the equation when friction is nonzero ($c ≠ 0$). Both topics would go beyond the scope of this article.

The second example is a case where the initial velocity amplitude of the load $v o=5 m s − 1$ is larger than four times the trolley speed. According to Eq. (30), $τ=17.95 s$, which yields $t 1=0.74 s$. Figure 3 shows the position and velocity of both the load and the trolley in this case. Similar to the previous example, t1 begins a full period late at $t 1=9.72 s$. The wait time between the two movements is twice the pendulum period, $τ=2 T o=17.95 s$. Due to the large initial velocity of the load, the motion can only be attenuated and not fully removed after the maneuver.

The equations derived in Sec. III do not only apply at a construction site. They can also be helpful in a physics laboratory, especially one involving a harmonic oscillator, which is not so unusual in science and engineering. In the present case, the above-described crane shenanigans are directly applicable to a torsion balance that is used to measure the gravitational constant G.15

As shown in Fig. 4, the torsion balance is comprised of four cylindrical copper test masses, each of mass approximately 1 kg, resting on a disk suspended from a weak torsion spring. This assembly is inside a vacuum vessel. An autocollimator monitors the small angle $φ t$ the disk makes with respect to a fixed direction via one of four mirrors attached to the disk. The measurement range of the autocollimator is $±0.17°$. A source-mass assembly consisting of four cylindrical source masses on a carousel is located outside the vacuum chamber. The operator can rotate the carousel around the symmetry center to an angle $φ s$ with the help of a stepper motor.

Depending on the difference $φ s − φ t$, a gravitational torque n can act on the pendulum. The torque can be written under the form $n = G Γ ( φ s − φ t )$, where $Γ ( φ )$ is a function that depends on the angle difference, but also on the mass distribution of the experiment (itself depending on the various distance of the masses to the rotation center, their height, and densities). Experimentally, numerical integration over the test and source mass volumes is necessary to determine Γ with relative uncertainties of $∼ 1 × 10 − 6$.

The experiment is performed as follows. The source masses are moved to the angle where the maximum clockwise torque acts on the pendulum, $φ s = φ s , max$. Due to the gravitational torque, the pendulum oscillates around an equilibrium position $φ t , max$ given by $κ ( φ t , max − φ t , 0 ) = G Γ ( φ s , max − φ t , max )$. The torque produced by the torsion strip, $κ ( φ t , max − φ t , 0 )$, counteracts the gravitational torque, and $φ t , 0$ denotes the unknown equilibrium position of the pendulum in the absence of external torque. The torsion pendulum is then observed with an autocollimator for a few periods, to determine $φ t , max$, after which the source masses are moved counterclockwise to the position $φ s , min$ where the torque acting on the pendulum is minimal, i.e., maximal counterclockwise torque. The pendulum is observed again, and we obtain $κ ( φ t , min − φ t , 0 ) = G Γ ( φ s , min − φ t , min )$. The procedure is repeated to achieve a good precision on $φ t , max − φ t , min$. By using the difference, the unknown angle $φ t , 0$ drops out. From these measurements, the gravitational constant can be obtained as
(31)
where the numerator is obtained from the successive measurements discussed above and the denominator from the numerical integration of Γ. The torsional constant, κ, is obtained by measuring the resonant frequency of the pendulum, fo, and numerically calculating the pendulum's moment of inertia, I: $κ = I ( 2 π f o ) 2$. For a more complete description of the setup, see Ref. 15.

For our geometry, the source mass positions of $φ s , max=− φ s , min≈18°$ produce an angle difference of $φ t , max− φ t , min=0.008 74°=153 μrad$. One detail worth mentioning is that the pendulum is constantly swinging. Hence, to determine the equilibrium position of the pendulum with the source masses in either state, a sine function must be fitted to the data obtained with the autocollimator, offset by $φ t , max$ or $φ t , min$. The uncertainty in the equilibrium position is smaller for smaller pendulum amplitude, for, in this case, the nonlinearities of the autocollimator affect the results to a lesser degree.

The differential equation describing the system is
(32)
where $n ( φ s − φ t ) = G Γ ( φ s − φ t )$ is the gravitational torque that is applied by the operator via moving the source mass assembly to $φ s$. Note that the omission of damping is justified here since the quality factor of the pendulum is very large ($≈100 000$). Here, $φ s ≫ φ t$, hence, $n ( φ s − φ t ) ≈ n ( φ s )$. In the Laplace domain, the response function is
(33)
which closely resembles Eq. (4).
The same crane operator is in charge of maneuvering the source masses, and, like before, only two movements per maneuver are allowed. The task is to optimize the timing of both movements, given by the parameters t1, δt, and τ to damp the pendulum as much as possible. One difference here is that the torque does not change linearly with respect to the source mass position but depends on $n o sin ( ( φ s / φ s , max ) ( π / 2 ) )$, where $n o=1.56×1 0 − 8 N m$. Note that this expression corresponds to the first order expansion, valid if the masses are point masses. It is a reasonable accurate approximation for the cylindrical masses used in the experiment. The problem can then be solved with a method similar to what has been described for the crane. The exact differential equation of motion and their solutions are beyond the scope of this text but can be found on the GitHub page mentioned earlier. Spoiler alert: the crane operator's trick similar to the one described in Sec. III but adapted to torques can be written in the Laplace domain
(34)
with $ν = π / ( 2 δ t ) , t 2 = t 1 + δ t , t 3 = t 2 + τ$, and $t 4 = t 3 + δ t$.

Figure 5 shows the pendulum angular position $φ t$ as a function of time. When the source mass is set at $+ φ s , max$, the corresponding $φ s$ and $φ t$ curves are in magenta, whereas when it is set at $− φ s , max$, the corresponding $φ s$ and $φ t$ curves are in cyan. The pendulum response to the source mass maneuvers are plotted with thin dotted lines. To understand the source mass motion plotted in Fig. 5, recall that there is a maximum amplitude that can be damped in one maneuver, see Eq. (29). For the torsion pendulum, this amplitude corresponds to 146 μrad. To achieve this reduction, the wait time between the two moves in one maneuver is $τ−δt=107 s$. This wait time is long, almost a full period. Alternatively, by setting $τ = δ t$, the pendulum swing amplitude is reduced by 137 μrad instead of 146 μrad. Thus, the amplitude reduction per maneuver is worse, but the average reduction per unit time is greater. Hence, for amplitude larger than 137 μrad, τ is set to zero. For smaller amplitudes, the ideal τ is calculated. In the latter case, the wait time between two moves is clearly visible in the last two maneuvers in Fig. 5. More information on this topic can be found in the supplementary material.16

At the beginning of the data shown in Fig. 5, the pendulum had an amplitude of 1830 μrad. After 15 maneuvers, the amplitude has been reduced to 1.5 μrad, which shows how effective the crane operator's trick is. However, an attenuation of 99.9% seems to be the limit of this method for two main reasons: (1) there is some variation in δt, the time it takes to perform one movement, and (2) the torque on the pendulum is not exactly proportional to the sine of the angular position of the source masses as mentioned earlier. Other than that, this work would likely get the nod of approval from the Society of Meticulous Crane Operators.

From the physics point of view, the damping of the torsion pendulum is desirable because it ultimately reduces the uncertainty of the angular position during the measurement. The values of $φ t , max$ and $φ t , min$ are determined by measuring the pendulum motion over one period. Fitting a large amplitude curve to determine the equilibrium position is more prone to systematic effects caused by nonlinearities of the autocollimator. Here, the authors employ the crane operator's tricks to achieve some serious damping: the amplitude of the oscillations in $φ t$ after an hour of such maneuvers is about 1200 times smaller than the initial amplitude. In contrast, a wait time of the order of months would be necessary before the oscillations naturally decay via dissipation in the torsion strip to the same levels if no trick is applied.

In this paper, we have taken a light-hearted look at a practical application of Newtonian mechanics, using Laplace transformation. We started with a simple example. The trolley of a crane starts moving with constant velocity. We calculated the effect that this impetus has on a classical harmonic oscillator: the load suspended from that trolley. The translation property of the Laplace transformation allows us to easily consider the effect of two such impulses in opposite directions, i.e., starting and stopping the trolley. With a few trigonometric manipulations, we showed that if the two impulses are separated in time by an integer number of pendulum periods, no energy is added or subtracted to the pendulum motion of the load. In the classroom, the formalism can be shown on the whiteboard, with a simulation using the code provided on GitHub, and experimentally by moving the top of a plumb bob by hand. The technique can be extended to include the initial swinging condition of the pendulum. We use a variation of the crane operator's trick on a torsion balance in our current research, determining the gravitational constant.

We believe that the ideas presented in this paper can be valuable for the classroom. First, the math is not too complicated and can be followed on the whiteboard. Second, the freely available python code can be downloaded, and students and teachers alike can play with different parameters, including damping. Finally, the motivated reader may endeavor to build an actual experimental realization of the systems discussed here—either a crane or a torsion balance. The topic can be approached from many different angles and is, therefore, fun and educational.

Applying the techniques discussed above, we found it satisfying that the torsion pendulum follows precisely the simple, albeit uncommon, mathematical formalism. We hope that the students and their teachers can find the same satisfaction in this system, even if only by simulation. This satisfaction can inspire a lifelong passion for understanding the world using mathematical and physical reasoning.

The horizontal position of the trolley is given by a twice differentiable function $x c ( t )$. The vertical position of the trolley, $y c = l$, and the length of the rope l are fixed. Then, the coordinates of the load are given by
(A1)
The corresponding velocities are
(A2)
The Lagrangian is
(A3)
The exact Lagrange equation for θ is
(A4)
To first order, we approximate $cos θ ≈ 1$ and $θ ≈ sin θ = ( x m − x c ) / l$. With $g / l = ω o 2$, we obtain
(A5)
which is given in the main text as Eq. (2).
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