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.

## I. INTRODUCTION

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, with^{12} 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.

*m*is suspended by a rope with fixed length

*l*and negligible mass and can swing at small angles with a period

*T*. 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 \u0307 m$ and $ v c = x \u0307 c$, respectively. The angle

_{o}*θ*of the rope with respect to vertical can be obtained from $ sin \u2009 \theta = ( x m \u2212 x c ) / l$. With the trolley at rest, the velocity and acceleration of the load tangential to its arcuate motion are $ l \theta \u0307$ and $ l \theta \xa8$. For small excursions ( $ \theta \u226a 1$), the velocity and acceleration can be approximated as $ x \u0307 m$ and $ x \xa8 m$, respectively. Hence, the linearized differential equation of motion is

*g*is the local gravitational acceleration and

*c*is a damping coefficient that accounts for the fluid friction force $ c x \u0307 m$, which is proportional to the velocity of the load. The damping term $ c x \u0307 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

^{13}a popular technique used in control theory for solving ordinary differential equations, Eq. (2) can be transformed to the

*s*-domain,

The Laplace transform is similar to the Fourier transform but uses the complex variable *s* instead of $ i \omega $ 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.

## II. STARTING WITH THE LOAD AT REST

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 = \tau $, 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).

### A. Trolley starts moving

*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

*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):*

*t*> 0, yields

*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.

### B. Trolley starts moving and then stops

*t*=

*kT*will result in the load hanging at rest. As before, the trolley starts moving at

_{o}*t*=

*0 with a velocity $ v c$, but now comes to a stop at $ t = \tau $. In the Laplace domain, this motion is given by*

*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 \u2009 ( \u2212 \tau 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 > \tau $,

^{14}by

*τ*. Thus, the load velocity becomes zero when $ \omega o \tau = k 2 \pi $, or rather $ \tau = k T o$, exactly as conjectured above.

### C. The crane operator's trick

*t*=

*0 with $ v c / 2$ and then, at $ t = \tau $, 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

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.

### D. Finite acceleration of the trolley

*t*=

*0, the trolley starts accelerating with $ a c$ over a duration of*

*δt*. A second acceleration with the same duration begins at $ t = \tau $. In the

*s*domain, the stimulus is

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 = \tau $, 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.

## III. STARTING WITH A SWINGING LOAD

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 *v _{c}* each of duration

*δt*. The combined distance traveled by the trolley is $ x g = 2 v c \delta t$. To succeed in the task, the crane operator has to quickly solve the equations of motion and determine the times

*t*

_{1}and $ t 1 + \tau $ when these trolley operations are to be executed, as the ground recipients are notoriously impatient.

*t*=

*0, then the pendulum at rest has, by definition of the response function,*

*v*, the virtual stimulus must, therefore, be $ x c = ( v o / \omega o 2 ) \u2009 \delta ( t )$. The combination of the virtual and real trolley motions in the Laplace domain is

_{o}*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 \u2264 4 v c$. For the case $ v o = 4 v c$, the choices for

*τ*and

*δt*would be $ \tau = T o / 2 + k 1 T o$ and $ \delta t = 2 k 2 T o$, with $ \tau > \delta t$ and

*k*

_{1},

*k*

_{2}positive integers. Usually,

*δt*is determined by the distance $ x g = 2 v c \delta t$ over which the load needs to be transported. For a given

*δt*, the largest possible amplitude that can be damped to zero is

*τ*must be chosen as

## IV. EXAMPLES BY SIMULATION

In this section, we examine two examples based on the equations derived previously. We assume $l=20\u2009m$, 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\u2009m\u2009 s \u2212 1$, yielding $\delta t=3\u2009s$.

In the first example, it is assumed that the load has a velocity amplitude of $ v o=1\u2009m\u2009 s \u2212 1$. According to Eq. (28), $\tau =5.32\u2009s$ which, as an input for Eq. (25), yields $ t 1=2.57\u2009s$. 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, *t*_{1} begins a full period late, i.e., $ t 1=8.97\u2009s+2.57\u2009s=11.54\u2009s$.

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 \u2260 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\u2009m\u2009 s \u2212 1$ is larger than four times the trolley speed. According to Eq. (30), $\tau =17.95\u2009s$, which yields $ t 1=0.74\u2009s$. Figure 3 shows the position and velocity of both the load and the trolley in this case. Similar to the previous example, *t*_{1} begins a full period late at $ t 1=9.72\u2009s$. The wait time between the two movements is twice the pendulum period, $\tau =2 T o=17.95\u2009s$. Due to the large initial velocity of the load, the motion can only be attenuated and not fully removed after the maneuver.

## V. APPLICATION IN THE LABORATORY

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 $ \phi 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 $\xb10.17\xb0$. 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 $ \phi s$ with the help of a stepper motor.

Depending on the difference $ \phi s \u2212 \phi t$, a gravitational torque *n* can act on the pendulum. The torque can be written under the form $ n = G \Gamma ( \phi s \u2212 \phi t )$, where $ \Gamma ( \phi )$ 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 $ \u223c 1 \xd7 10 \u2212 6$.

*κ*, is obtained by measuring the resonant frequency of the pendulum,

*f*, and numerically calculating the pendulum's moment of inertia,

_{o}*I*: $ \kappa = I ( 2 \pi f o ) 2$. For a more complete description of the setup, see Ref. 15.

For our geometry, the source mass positions of $ \phi s , max=\u2212 \phi s , min\u224818\xb0$ produce an angle difference of $ \phi t , max\u2212 \phi t , min=0.008\u200974\xb0=153\u2009\mu rad$. 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 $ \phi t , max$ or $ \phi 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.

*t*

_{1},

*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 \u2009 sin \u2009 ( ( \phi s / \phi s , max ) ( \pi / 2 ) )$, where $ n o=1.56\xd71 0 \u2212 8\u2009N\u2009m$. 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

Figure 5 shows the pendulum angular position $ \phi t$ as a function of time. When the source mass is set at $ + \phi s , max$, the corresponding $ \phi s$ and $ \phi t$ curves are in magenta, whereas when it is set at $ \u2212 \phi s , max$, the corresponding $ \phi s$ and $ \phi 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 *μ*rad. To achieve this reduction, the wait time between the two moves in one maneuver is $\tau \u2212\delta t=107\u2009s$. This wait time is long, almost a full period. Alternatively, by setting $ \tau = \delta t$, the pendulum swing amplitude is reduced by 137 *μ*rad instead of 146 *μ*rad. Thus, the amplitude reduction per maneuver is worse, but the average reduction per unit time is greater. Hence, for amplitude larger than 137 *μ*rad, *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 *μ*rad. After 15 maneuvers, the amplitude has been reduced to 1.5 *μ*rad, 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 $ \phi t , max$ and $ \phi 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 $ \phi 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.

## VI. CONCLUSION

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.

### APPENDIX A: DERIVATION OF THE DIFFERENTIAL EQUATION

*l*are fixed. Then, the coordinates of the load are given by

*θ*is

### APPENDIX B: LAPLACE TRANSFORMATIONS USED IN THIS ARTICLE

Name . | f(t)
. | F(s)
. |
---|---|---|

Impulse | $ \delta ( t )$ | 1 |

Unit step | u(t) | $ s \u2212 1$ |

Unit ramp | $ u ( t ) \u2009 t$ | $ s \u2212 2$ |

Unit acceleration | $ u ( t ) \u2009 t 2 / 2$ | $ s \u2212 3$ |

Sine for t > 0 | $ u ( t ) \u2009 \u2009 sin \u2009 ( \omega t )$ | $ \omega / ( s 2 + \omega 2 )$ |

Cosine for t > 0 | $ u ( t ) \u2009 \u2009 cos \u2009 ( \omega t )$ | $ s / ( s 2 + \omega 2 )$ |

Translation in time | $ f ( t \u2212 t o )$ | $ exp \u2009 ( \u2212 t o s ) F ( s )$ |

Derivative | $ d f ( t ) / d t$ | sF(s) |

Integral | $ \u222b 0 t f ( \tau ) d \tau $ | $ s \u2212 1 F ( s )$ |

Name . | f(t)
. | F(s)
. |
---|---|---|

Impulse | $ \delta ( t )$ | 1 |

Unit step | u(t) | $ s \u2212 1$ |

Unit ramp | $ u ( t ) \u2009 t$ | $ s \u2212 2$ |

Unit acceleration | $ u ( t ) \u2009 t 2 / 2$ | $ s \u2212 3$ |

Sine for t > 0 | $ u ( t ) \u2009 \u2009 sin \u2009 ( \omega t )$ | $ \omega / ( s 2 + \omega 2 )$ |

Cosine for t > 0 | $ u ( t ) \u2009 \u2009 cos \u2009 ( \omega t )$ | $ s / ( s 2 + \omega 2 )$ |

Translation in time | $ f ( t \u2212 t o )$ | $ exp \u2009 ( \u2212 t o s ) F ( s )$ |

Derivative | $ d f ( t ) / d t$ | sF(s) |

Integral | $ \u222b 0 t f ( \tau ) d \tau $ | $ s \u2212 1 F ( s )$ |

## REFERENCES

*Mathematical Methods for Physicists: A Comprehensive Guide*