Inclined large-angle pendulum may produce endless linear motion of a
cart when friction is negligible
DENNIS P. ALLEN, Jr.
17046 Lloyds Bayou Drive Apt 322
Spring Lake, MI 49456
USA
CHRISTOPHER G. PROVATIDIS
School of Mechanical Engineering
National Technical University of Athens
9 Iroon Polytechniou, 157 80 Zografou
GREECE
Abstract: - We present the mechanics for the oscillation of an inclined large-angle pendulum-drive attached to a
cart which is allowed to perform translation in one direction only. Neglecting the overall friction, the
application of Newton’s second law shows that the oscillation of the pendulum is continuously converted into
oscillating linear motion thus achieving a travel of infinite length. It is also shown that the frequency depends
on the usual data of any pendulum plus the mass of the cart on which it is attached. After the determination of a
novel effective pendulum length, a closed-form accurate analytical expression is presented for the amplitude of
the pendulum, whereas semi-analytical formulas are provided for the period as well as the time-variation of the
large azimuthal-like angle. Moreover, a simple expression was found for the position of the cart in terms of the
azimuthal angle of the pendulum and the elapsed time. The extraction of the analytical formulas was facilitated
by a computer model programmed in MATLAB®.
Keywords: - Large-angle pendulum, Motion conversion, Inertial drive, Frictionless model, Perpetual device
Received: November 19, 2021. Revised: October 26, 2022. Accepted: November 24, 2022. Published: December 31, 2022.
1. Introduction
The term ‘inertial propulsion’ was introduced
probably by the late Professor Eric Laithwaite
(1921–1997), who is well known from his
Christmas lectures on gyroscopes at Imperial
College London in the United Kingdom, [1].
Although most of his ideas on inertial propulsion
and gyroscopic thrust have been in advance rejected,
the interest is still alive and some of the relevant
mechanics are still covered by mystery. For
example, Wayte, [2], conducted an experimental
study in favor of Laithwaite while recently
Provatidis, [3], revealed some of the associated
mechanics and how a physicist may be cheated, [4].
Except for the aforementioned gyroscopes,
contra-rotating eccentrics (the latter called Dean
drive, [5]) as well as equivalent electromagnetic
means have been also studied jointly by academics
and the industry for possible suitability in
alternative propulsion, [6]. Nevertheless, since the
inertial forces are internal to the mechanical system,
the resulting net thrust over a time period is null
thus instead of propulsion at the best case we obtain
a ‘catapult’ or a marching device like the bumper of
a mobile phone. To become more specific, it is
indisputable that under certain conditions an inertial
drive may cause an initial velocity to an object so as
the center of mass of the mechanical system moves
as usual.
In this paper we study an alternative source of
inertial propulsion, i.e. an inertial drive which is
based on the oscillating motion of a pendulum and
particularly in conjunction with large angles. The
beneficiary characteristic is the very low friction
which occurs in a pendulum and makes it superior
(it takes much time to cease) compared to the
motion of fully rotating eccentrics. Within this
context, this study is probably the first publication
of this kind on the use of the pendulum as an inertial
drive.
WSEAS TRANSACTIONS on APPLIED and THEORETICAL MECHANICS
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From the pedagogical point of view, it is shown
that the initial kinetic and potential energy of a
pendulum can be easily converted into linear
(translational) motion when it is attached to a
movable cart. For the very theoretical assumption of
negligible friction, the initial velocity of the center
of mass causes the motion of the cart forever. Of
course, the physical reality is different, thus after a
few oscillations the appearance of friction on a real
prototype, [7], makes the translational motion of the
cart to cease. Therefore, this open system gives us
the opportunity to discuss the subject of the
conservation of energy and linear momentum, as
well as the non-conservation of angular momentum,
all of which are of major importance in the
education of physics and mechanics.
2. Basic theory
Let us consider an inertial Cartesian coordinate
system
fixed to the Earth, in which the
horizontal ground (on which the cart moves) is
parallel to the
-plane while is the vertical
axis. For the sake of simplicity, at the initial time
( ) we consider a chassis (cart) as a
concentrated mass exactly at the aforementioned
point, , so as the body-fixed coordinate system
 initially coincides with the fixed system

.
The mechanical system consists of the
abovementioned chassis (cart with its wheels) on the
horizontal ground (-plane) on which an inclined
pivot is attached at the moving point  (origin of
body fixed axis on the cart, with global coordinates
󰆓 󰇛󰇜, 󰆓=0 and 󰆓 ) while a large-
angle pendulum rotates about the aforementioned
body fixed axis  that is inclined by angle with
respect to the body fixed vertical axis , as shown
in Fig. 1a. In other words, the axis  is produced
by rotating the initial global system 
about
the axis
by angle , and then the resulting
system is free to translate in the -direction so as
the pivot  of the cart can shift from its initial
position to the final .
Fig. 1: (a) Coordinate system and (b) Set up of the
inclined pendulum.
Obviously, with respect to the global (Earth
fixed) system
, the upward unit vector
󰇍
along the axis of rotation  will be given by:
(1)
In accordance to the experiment, [7], the overall
mechanical system is governed by two degrees of
freedom (DOF), the former being the displacement,
󰇛󰇜, of the chassis (cart) and the latter is the angle,
󰇛󰇜, which is somehow related to the azimuthal
angle but not exactly. Clearly, instead of the usual
Euler angles, due to the constant inclination 󰇛
 󰇜, in this manuscript we have reduced
them in only one (see, Fig. 1), practically in the
interval  , as explained below.
Clearly, the position  corresponds to the
extension of the axis  in the negative direction,
while the position  corresponds exactly to
it. First of all, we divert the pendulum at an angle
in absolute value larger than  degrees (at
maximum in the negative extension of -axis), and
then we leave it rotate in the anti-clockwise direction
due to its weight. Therefore, when the axis of the
pendulum reaches the position  at time
, the accelerated downward motion gives an initial
angular velocity 󰇗 . At this position we may also
assume that the cart is at rest, i.e., and 󰇗
, at .
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Therefore, with respect to the fixed to the Earth
global system , the position of the oscillating
mass, which is attached to the point 󰇛 󰇜 at the
end of the bob, is given by:
( ) cos cos ,
sin ,
cos sin ,
b
b
b
x X t L
yL
zL




(2)
where is the length 󰇛󰇜 while the subscript ‘’
stands for the word ‘bob’.
Let and be the masses of the cart and the
bob, respectively. Between several alternative ways
to derive the equations of motion, we consider the
center of mass (subscript ‘’) of the mechanical
system “cart + bob”, of which the coordinates are:
,
, with 0
, with 0
b
c
b
c
b
c
mX Mx
xmM
mY My
yY
mM
mZ Mz
zZ
mM


, (3)
with 󰇛 󰇜 󰇛 󰇜 denoting the coordinates
of the cart in the global inertial system fixed to the
Earth, while 󰇛 󰇜 are the coordinates of the
bob given by Eq. (2).
According to Newton’s Second law, the sum of
all external forces in the -(horizontal) direction
equals the total mass times the acceleration of the
center of mass. Since the friction is zero
(temporarily), there is no external force in this
direction (otherwise is ), thus we have
󰇛 󰇜󰇘 . By virtue of Eq. (2) and (3), after
rearrangement of the terms this equation eventually
becomes:
( ) ( cos )(cos ) 0m M X ML

gg
. (4)
Equation (4) is a relationship between the second
temporal derivatives of the two DOF, 󰇛󰇜 and
󰇛󰇜, thus one of them may be eliminated. Although
the purpose of this paper is to derive the cart
displacement 󰇛󰇜, it seems that the primary
variable is the azimuthal-like angle 󰇛󰇜. Therefore,
after two successive integrations of Eq. (4) in time ,
we eventually derive the general solution:
00
0 0 0
(cos cos )
( sin ) ,
X X A
X A t


(5a)
in which, for convenience, we have introduced the
following constant :
cosML
AmM
. (6)
Taking the first derivative of 󰇛󰇜 with respect to
time , Eq. (5a) becomes:
0 0 0
( sin sin ).X X A
(5b)
One may observe that Eq. (5a) includes the initial
conditions ( 󰇗) for the cart, as well as the initial
conditions (󰇗) for the angle 󰇛󰇜 of the
pendulum, and obviously satisfies the ODE (4). But
the most interesting issue in Eq. (5a) is that 󰇛󰇜
consists of a bounded term of amplitude  plus a
linear term of the form  with the constant being
equal to
0 0 0
( sin )a X A


. This fact clearly
shows that the cart will perform an oscillatory
motion of amplitude  but it continuously moves
in the positive -direction.
Obviously, the above remark regarding the
continuous motion 󰇛󰇜 of the cart is not peculiar
and this happens simply because the center of mass
has an initial velocity, which due to the absence of
friction is preserved forever. From the practical
point of view, as also can be noticed in a video of a
prototype device, [7], the pendulum is diverted
(lifted) at an angle , at which (according to Eq.
(2)) the potential energy of the bob mass will be
equal to:
cos sin
lift
pot b
E Mgz MgL

(7a)
Setting in Eq. (7a) the technically maximum
possible value (most upper position of the
pendulum)   at which the cosine becomes
equal to (1), the aforementioned initial potential
energy (produced from the maximum possible
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rotation of the pendulum on the slant plane)
becomes:
max sin
lift
pot
E MgL
(7b)
Therefore, keeping the cart firmly fixed, as the
pendulum rotates in the anti-clockwise direction by
lowering the height and thus increasing its
angular velocity 󰇗󰇛󰇜, the energy conservation for
the pendulum gives:
22
0
( sin )cos
1
( sin )cos ( )( )
2
lift
MgL
MgL ML



(8a)
Setting   and
, Eq. (8a) gives
the maximum possible initial angular velocity at
, given by:
0 max 2 sin
() g
L
(8b)
Now, exactly when the pendulum passes through the
position
at any given initial angular
velocity 󰇗, not necessarily equal to that maximum
of Eq. (8b), the cart is suddenly left free to move
according to the dominating physical laws.
Henceforth, the problem is to determine the function
󰇛󰇜 and then, applying Eq. (5a), to determine the
position 󰇛󰇜 of the cart. In other words, despite the
motion of the cart, the problem decoupled as we
have to solve only the large angle oscillation of the
pendulum. Nevertheless, it does not fit the well-
known forms in physics (mechanics).
3. Solution of the mechanical model
For the sake of conservatism, we derive the equation
of motion considering the equation of energy
conservation. The mechanical system includes two
kinetic energies for the cart (of mass ) and the bob
(of mass ), respectively, and also the potential
energy of the bob mass.
In more detail, the kinetic energy of the cart is
simply given by:
2
1,
2
cart
kin
E mX
(9)
Also, according to Eq. (2) the velocity components
of the bob will be:
sin cos ,
cos ,
sin sin ,
b
b
b
x X L
yL
zL


(10)
thus the kinetic energy of the bob mass, with respect
to an inertial system fixed to the Earth is given by:
2 2 2
1()
2
bob
kin b b b
E M x y z
(11a)
Substitution of Eq. (10) into Eq. (11a), after
manipulation we eventually receive:
2 2 2
1( 2 sin cos )
2
bob
kin
E M X L LX
(11b)
In addition, the potential energy of the bob mass is
given by:
(0 cos sin )
bob
pot bob
E Mgz
Mg L


(12)
Summing Eqs. (9), (11b) and (12), after
rearrangement of the terms the total mechanical
energy of the system takes the form:
2 2 2
11
( ) ( )
22
( cos ) sin
( sin )cos .
total
E m M X ML
ML X
MgL

(13)
Substituting Eq. (5a) into (13), after elaboration the
latter takes the form:
20
B C E

(14)
with the variables and being functions of only
the degree of freedom (and the initial conditions
as well), as follows:
2 2 2
11
( ) sin
22
B m M A ML
(15)
and
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2
0 0 0
1( )( sin )
2
( sin )cos
C m M X A
MgL


(16)
Also, in Eq. (14) is the initial value of the total
energy, which according to (13) equals to:
2 2 2
0 0 0
0 0 0
0
11
( ) ( )
22
( cos ) sin
( sin )cos
E m M X ML
ML X
MgL

(17)
Below we present two alternative equations of
motion, of first and second order respectively,
which have to be numerically solved.
3.1 First-order differential equation
The energy conservation Eq. (14) is directly solved
in 󰇗 thus we receive:
0
()EC
B

, (18)
Equation (18) is a first-order ordinary differential
equation of 󰇛󰇜 in , thus can be easily solved in a
numerical way, for example implementing the
Runge-Kutta algorithm. The only difficulty with Eq.
(18) is the sign which is related to whether the
bob keeps going in a certain direction. In more
detail, considering the definition of the azimuthal
angle according to Fig. 1a, when the bob rotates in
the anti-clockwise direction (thus increasing the
function 󰇛󰇜) the proper sign is () whereas when
rotates in the clockwise direction (thus decreasing
the function 󰇛󰇜) the suitable sign is ().
To control the sign in the above mentioned
procedure, it is useful to determine the two positions
at which the condition 󰇗 is met, called “turn
points” (extreme oscillation points), and correspond
to the maximum possible angle  (amplitude
). Therefore, setting in Eq. (18) the
condition 󰇗 , it simplifies to , and then
we obtain:
2
10 0 0 0
2
max
( )( sin )
cos ( sin )
m M X A E
MgL

(19a)
Then, the maximum value (half-amplitude) is:
22
10
max cos
cos 1
2 sin ( )
LM
g m M








(20a)
Therefore, in the first half-period (angle interval
󰇟 󰇠) the sign is (+), in the second half-
period the sign is (–), in the third half-period the
sign is (+), and so on.
3.2 Second-order differential equation
Taking the first derivative in time of both parts in
Eq. (14), we derive 󰇗󰇗 󰇗󰇘 󰇗 , whence
we can solve in 󰇘:
22
BC
BB
(21)
After substitution of Eq. (15) and Eq. (16) in
Eq.(21), and then replacing the constant according
to Eq. (6), we eventually obtain the desired 2nd-order
ODE:
22
22
sin ( )sin cos cos
( cos sin )
g M m LM
L M M m





(22)
The advantage of Eq. (22) compared to Eq. (18) is
that it requires no book-keeping on the sign in front
of the square root, while the disadvantage is that its
numerical solution does not ensure energy
conservation. Of course, since Eq. (22) was derived
from the energy conservation after differentiation,
the initial energy is lost and is merely substituted by
the initial conditions (󰇗) that ensure ,
according to Eq. (17). In any case, the degree in
which the numerical solution ensures the energy
conservation is a safe criterion to measure and judge
its quality.
4. Estimation of the time period
4.1 A primitive model
One may observe that the equation of motion, Eq.
(22), differs from the following standard nonlinear
equation that characterizes large-angle pendulum
oscillations:
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sin ( ) 0
gt
L


, (23)
As a very primitive approximation, we may set the
term  in the denominator of Eq. (22) equal to
unity thus the  in the numerator will be equal to
zero. Then, transferring all terms in the left-hand side
we get the approximation:
2
( )sin sin 0
( cos )
Mm
gL M M m





(24)
Comparing the very approximate Eq. (24) with the
golden standard Eq. (23), one may observe that the
model of this paper is related to an effective length
given by:
2
( cos )
( )sin
eff
LM M m
LMm
(25)
Having obtained the above efficient (equivalent)
length , according to the state-of-the-art the
small angle oscillation formula  is
replaced by one of the following alternative
formulas, (see, [8], [9]):
2
max
1
1
2
max
2
3
8
max
3
max
21
16
2 cos 2
sin
2
eff
eff
eff
T L g
T L g
T L g













(26)
4.2 A more accurate approximate model
Starting from the definition of the instantaneous
angular velocity, , we can derive 
, after integrating in the angle interval
󰇟 󰇠 we can derive the half-period ,
thus for the entire period we get
max
max
2()
d
T

(27)
Working for the first half-period in which the sign of
Eq. (18) is positive, thus considering that
󰇛 󰇜 󰇛󰇜, the angular velocity can be
written in terms of four constants as follows:
2
cos
() sin
ab
cd

, (28)
with
22
0
2
( cos )
2 ( )sin
()
cos
a L m M M
b g m M
c L m M
d LM




(29)
The infinite integral of Eq. (27) in conjunction with
Eq. (28) cannot be analytically found, however
instead one can easily apply either the Simpson’s
trapezoidal rule or Gauss numerical integration.
Alternatively, we can resort to the substitution of
the involved trigonometric functions by:
2 4 6
3 5 7
cos 1 2 24 720
sin ,
6 120 5040

(30)
but then only a lengthy expression in the form of the
Taylor series may be derived using symbol
manipulation software such as MATHEMATICA®.
Whatever method is applied to determine the
period , having determined the half-amplitude
 using Eq. (20a), and knowing that at the initial
time corresponds to a position that is a little
after the initial turn point, a closed-form analytical
solution could be:
max
( ) cos( )tt
, (31)
with
2T

denoting the average angular
frequency.
Applying Eq. (31) for , we receive
0 max cos

, thus we can derive the unknown
phase difference from the formula:
10
max
cos





(32)
Combining (31) and (32) we obtain the following
approximation:
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10
max
max
( ) cos costt








(33)
5. Numerical application
Based on the experimental prototype, [7], the
parameters of the device under consideration are:
= 0.445 kg (bob mass)
= 2.500 kg (chassis, i.e. cart without bob)
=0.203 radians (11.65 degrees)
= 0.225 m (shaft length)
= 9.81 m/s2, gravitational acceleration
The initial conditions for the two variables at
are:
(initial position of the cart)
󰇗 (initial velocity of the cart)
(90 degrees), initial azimuthal angle
󰇗  (rad/s), initial azimuthal angular
velocity.
5.1 Time response
It is noted that, according to Eq. (8b), the maximum
possible initial angular velocity is 󰇛󰇗󰇜
. Therefore, the selected value 󰇗
 is a rather small one.
For the particular conditions, ( , 󰇗 ,
as well as and 󰇗 ), Eq. (19a) provides:
max
cos 0.999999881725856
(19b)
thus:
max 1.571282689119088radians,
90.027866508490348degrees
(20b)
One may observe that Eq. (20b) provides two
angles, one close to the initial position  and
another close to the final position 󰇛󰇜 of the
half oscillation. From the computational point of
view, these two values are valuable to control the
sign in Eq. (18) when the first-order formulation is
used, so that when they are met the sign
immediately changes into its opposite value than the
previous one. In contrast, the second-order
formulation given by Eq. (22) recognizes the
reversal of motion (turn points) automatically,
because the unknown variable 󰇗 is a part of the
solution.
For  , the obtained results are
shown from Fig. 2, Fig. 2, Fig. 3, Fig. 4, Fig. 5, and
Fig. 6. One may observe that the overall behavior of
the results is very similar to other inertial drives, i.e.,
it constitutes an oscillating motion around a linearly
increased (with respect to time) displacement
(according to Eq. (5a), (5b)). In other words, the
cart, of mass , is progressively displaced
to the -direction.
The absolute velocity of the cart is according to
the derivative of 󰇛󰇜 described by Eq. (5b).
Therefore, for the particular initial conditions, (
, 󰇗 , and ), it is given as 󰇗
󰇛󰇗 󰇗󰇜, thus the absolute velocity of
the cart oscillates in accordance to a repeated
pattern, as shown in Fig. 3: .
Fig. 2: Cart displacement.
Fig. 3: Cart velocity.
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Next, the calculated “azimuthal” (longitudinal)
angle 󰇛󰇜 is harmonic as shown in Fig. 4, and
again, the two formulations (first- and second-order
ODEs) visually coincide with one another.
Also, the ‘azimuthal’ angular velocity 󰇗󰇛󰇜
is harmonic, as shown in Fig. 5. Interestingly, in
contrast to the well known contra-rotating drives
which are based on an almost constant rotation ,
[10], in the case of this paper the function of the
instantaneous cyclic frequency 󰇛󰇜 is harmonic.
Moreover, the altitude 󰇛󰇜 of the bob mass is
shown in Fig. 6, and depicts that the bob mass is
always below the horizontal plane.
Fig. 4: “Azimuthal” angle 󰇛󰇜.
Fig. 5: “Azimuthal” angular velocity 󰇗󰇛󰇜.
Fig. 6: Altitude of bob mass (-coordinate).
5.2 Energy Breakdown
In the first formulation (first-order ODE) the energy
conservation is ensured per se since the governing
equation, Eq. (18), is coming from it.
In the second formulation (second-order ODE,
i.e., Eq. (22)) the conservation of total energy works
as a criterion to judge the quality of the numerical
solution.
Actually, while in both formulations the total
energy is practically preserved at the level of
  Joule, Fig. 7 shows that in the
second formulation it slightly decreases.
Regarding the energy breakdown in kinetic and
potential energy, Fig. 8 shows that the kinetic
energy of the cart is very small compared to the
kinetic energy and potential energy of the bob. Note
that, although due to the scale the total energy seems
close to zero, actually it is equal to  
Joule, as also mentioned above.
Fig. 7: Energy conservation.
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Fig. 8: Energy breakdown.
Now, we present results regarding the analytical
estimation of the time period of the oscillation
using Eq. (26), as well as the possible
approximation of the azimuthal angle function 󰇛󰇜.
Considering the time period  
coming from the peak-to-peak graph in the Runge-
Kutta solution (in our case we have:
 󰇗 ), the comparison of the
latter with the approximate values (Eq. (26)) is
adequately satisfactory, as shown in Table 1.
Table 1. Ratio of the estimated period over the accurate
 (Runge-Kutta) value
Normalized period
Eq. (26)
Value

0.9517

0.9806

0.9767
Fig. 9: Approximation of the angle 󰇛󰇜.
It is noted that when using the first two terms of
Eq. (30) and then calculate the integral of Eq. (27)
through a series expansion (powers of
) using MATHEMATICA®, the
calculated value is not of adequate accuracy
( ), that is worse than all those
shown in Table 1.
Then, in Fig. 9 we present the estimation of the
azimuthal angle 󰇛󰇜, in two ways. The former is
based on the actual period   [in the
legend of the graph is labeled as ‘Eq. (33)’], while
the latter is based on the best estimation in Table 1
[in the legend of the graph is labeled as ‘Eqs.
(26)+(33)’]. One may observe that the former
visually coincides with the Runge-Kutta solution
whether in the progress of time the latter (with
 ) appears a small shift to the left.
In any case, having a closed-form analytical
solution for 󰇛󰇜 given by Eq. (33), we can
immediately apply Eq. (5a) and thus determine the
cart position 󰇛󰇜. The conservation of Linear
Momentumn is presented in Fig. 10.
Fig. 10: Conservation of Linear Momentum.
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6. Conservation of linear and angular
momentum
6.1 Linear momentum
Regarding the conservation of the linear momentum
in the system, Fig. 1 shows that it is preserved at the
level of  .
From the theoretical point of view, the total
momentum equals that of the center of mass which
possesses a constant velocity (the initial one)
because no external force is exerted in the -
direction. Therefore, by virtue of the first equality in
Eqs. (3) we have:
cmbtotal xP mX Mx m M
. (34)
And since the initial velocity of the cart is 󰇗
whereas that of the bob is 󰇛󰇗󰇜 󰇗
󰇗, the initial linear momentum of
the system will be:
0 0 0 0 0
0 0 0
sin cos
( ) ( cos ) sin .
P mX M X L
m M X ML
(35)
Therefore, for the particular initial conditions
adopted in Section V (i.e.,  and 󰇗
), we receive:
0
-3
(0.445 0.225 cos0.203) 0.1 ( 1)
9.8069 10 kgm/s.
P

Then we shall show the ‘mechanism’ according to
which the momentum of the bob mass is transferred
to the cart. Actually, when the pendulum reaches the
most forward position (almost at the end of
the first quarter of the oscillation period), the first
component of the velocity (i.e., 󰇗 󰇗
󰇗) of the bob mass becomes 󰇛󰇗󰇜
󰇗, that means it shares the same velocity with the
cart, thus the total momentum is 󰇛 󰇜󰇗,
whence 󰇗 󰇛 󰇜 
and
this point corresponds to the intersection of the two
branches in the center of the eight-shaped curve
shown in Fig. 11.
Fig. 11: Cart velocity versus azimuthal angle .
As the bob mass keeps moving toward the
position about , it loses momentum (it
becomes negative) which is gained by the cart thus
increasing the travel length . The whole procedure
is shown in Fig. 12, with respect to both time and
angle parameters.
6.2 Angular momentum
Regarding the angular momentum, first of all it is
interesting to note that the rigid rod  is always
perpendicular to the unit vector
󰇍
that is normal to
the slant plane, and the same happens with the
relative velocities of the bob and cart mass with
respect to the center of mass of the system. In other
words, these two masses rotate about an axis of
constant direction (i.e., parallel to the
󰇍
-vector)
given by Eq. (1), which (axis) always passes
through the moving center of mass.
The position of the center of mass is given by
the distances and from the cart and the bob
mass, respectively, given by:
1M
LL
mM
, and
2m
LL
mM
(36)
Therefore the total second moment of inertia is
2 2 2
12 .
()
mM
I mL ML L
mM
(37)
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Fig. 12: Exchange of linear momentum between
pendulum and cart.
With respect to the inclined
󰇍
-axis of rotation
(passing through the center of mass), the equation of
motion (Newton’s second law for rotation) is
written as follows (see, [11], [12]):
2
2
dd
ddext
I
tt

Gτ
(38)
with
󰇍
and
ext extn
τ
.
The key factor is the careful consideration of the
external forces based on Newton’s second law for
translation, which for the two point masses are
according to Table 2.
To determine the algebraic value of the external
torque, , we need to find the sum of torques due
to external forces with respect to the center of mass,
and then to perpendicularly project it on the
󰇍
-axis.
Therefore, we find:
1 1 2 2
()
ext r F r F n
. (39)
Obviously, the fact that the system is constrained
(supported on the ground) is the cause that the total
angular momentum
󰇍
󰇍
is not preserved, but instead
Eq. (38) shows that its time derivative (d
󰇍
󰇍

󰇘) equals to the sum of the nonzero external
torques. In other words, the fact that the total
angular momentum is not preserved is fully
consistent with Newton’s second law for rotation.
Table 2. External forces () and position vectors ()
Cart mass ():
Mass No.1
Bob mass ():
Mass No.2


 󰇘

 󰇛󰇘 󰇜
 
󰇟󰇛 󰇜󰇛 󰇜󰇛
󰇜󰇠
󰇟󰇛 󰇜󰇛
󰇜󰇛 󰇜󰇠
Actually, using a computer program the
interested reader may see that the two parts of Eq.
(38) are the same, as clearly is shown in Fig. 13.
The last important issue is to start from Eq. (38)
and eventually to derive Eq. (22), without passing
through the condition of energy conservation (on
which Eq. (22) was based). This is accomplished as
follows. The left part of Eq. (38) remains as is, i.e.,
󰇘 (although one could forget the first paragraph of
this subsection regarding the second moment of
inertia and, alternatively, could blindly derive it
simply by differentiating the general expression
󰇍
󰇍

󰇍

󰇍
). Moreover, the first
term in the right part of Eq. (39), i.e.,
, is replaced according to Table 2. Then, the
involved variables ( ) are replaced by Eq.
(2), while the required second derivatives (󰇘 󰇘) by
differentiating Eq. (10) thus receiving:
2
( cos sin ),
sin ( sin ).cos
b
b
yL
zL


(40)
Also, the coordinates ( ) of the center of
mass are given by Eq. (3).
Substituting Eq. (2), (3) and (40) into
the right-hand side of Eq. (39) using also Eq. (1), we
derive a certain expression which includes 󰇘 and
󰇘 thus depends on the parameter 󰇘 (see, Eq.
(40)). Therefore, 󰇘 appears in both parts of
Eq. (38) and after extensive manipulation and
rearrangement, when solving in 󰇘, Eq. (22) is
eventually obtained again.
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Fig. 13: Coincidence between the left and right parts of
Eq. (38).
7. Discussion
In this study, the friction has been entirely neglected
on purpose, just to test the conservation of energy
and momentum as well as the overall feasibility of
the prototype mechanism and the model. Actually,
while the conservation of energy and linear
momentum are fulfilled, the angular momentum is
not preserved but its change in time equals to the
external torque.
In other words, the system ‘cart + pendulum’ is
an open system with the characteristic that it
undertakes no external forces in the -direction of
motion (thus linear momentum is preserved) and has
no energy loses (thus energy is preserved).
Regarding the abovementioned angular momentum,
the three-dimensional motion of the bob mass
causes inertial forces in all the three directions
(  ) but only those in the ( ) directions are
transferred to the wheels as support forces (see,
Table 2) thus causing external torque in addition to
the dead weights. We recall that the absence of
support force in the -direction is due to the lack of
friction.
A weakness is that the model of this study has
not considered the possibility of the cart either to
move toward the -direction or to overturn, because
both facts would be inconsistent with the
experiment, [7], in which the friction with the
ground plays a significant role. From a different
point of view, we could additionally assume that the
wheels are forced to roll on rails toward the -
direction and this resolves the supposed weakness of
our model.
Regarding the angular momentum, for instructive
purposes it is worthy to point out that while a
similar finding (i.e., the change of angular
momentum equals the external torque) is valid for a
fixed pivot spinning top where the torque is taken
with respect to the fixed point, in the present paper
we had to implement Newton’s second law (for
rotation) with respect to the center of mass of the
mechanical system.
It was clearly shown that an inclined large-angle
pendulum attached on a cart in the form of a
‘winding’ clock (where the winding is replaced by
the initial potential energy), may feed it with linear
momentum thus it can travel at an infinite distance,
provided the friction between the wheels of the cart
and the ground as well as the friction at the pivot 
is negligible, otherwise cart’s motion progressively
ceases. The advantage of the pendulum over other
inertia drives is that it actually requires much time
to decay due to friction at the pivoting point . In
addition, we only need to offer an initial potential
energy to the bob mass by elevating the pendulum
setting it at best parallel to the desired direction of
motion, and then leaving it free to oscillate.
In simple words, the key factor for the
continuous motion of the cart is the conservation of
the linear momentum in the -direction. Actually,
the corresponding velocity of the bob mass [see, Eq.
(10)] is given by 󰇗 󰇗 󰇗, which
means that with respect to an inertial observer at the
pivot point , the product 󰇗 plays an
important role. The quantity
󰇗 󰇗
󰇛 󰇜󰇗󰇗 remains constant,
which means that at the minimum value of the
product 󰇗 we have the maximum value of
󰇗 and vice versa. Moreover, regarding the absolute
position of the cart given by 󰇛󰇜 󰇛
󰇗󰇜, for the particular initial conditions of our test
case, it is a simple superposition of a constant
velocity 󰇛󰇗󰇜 and a parasitic cosine term. The
whole procedure was explained by details in Section
6. Interestingly, while the amplitude in the
oscillation of the linear momentum is relatively high
(see, Fig. 12), the magnitude of the constant total
linear momentum is rather small.
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In Section 5, results were presented for a
relatively low initial angular velocity 󰇗
, thus consequently the calculated
displacement of the cart was rather small. If the
aforementioned 󰇗 increases up to its maximum
allowable value given by Eq. (8b), that is close to
the value 󰇗 , in the same time
interval, , the displacement
increases a lot, as shown in Fig. 14. One may
observe that, the higher the initial angular velocity
the longer the travelled length in the same time
interval.
It is obvious that in any large-angle oscillation,
the angular velocity vanishes at the extreme points.
For example, if a small initial velocity such as 󰇗
 had been given at an initial angle of
 degrees, the zero value of the angular velocity
would happen earlier at  degrees, while
the other extreme point would happen later at
 degrees. Similarly, now that in the
present study we have chosen the initial value be
󰇗  at  degrees, it has been
shown that the half-amplitude  of the
‘azimuthal’ angle is slightly larger than  
(i.e.,  degrees in absolute value).
According to Eq. (20a), the inverse cosine of 
is proportional to the term 󰇗
thus the higher the
initial angular velocity the higher the half-amplitude
() is.
Interestingly, although the system has two
degrees of freedom, the azimuthal angle is the
primary variable and even it satisfies either a first-
order (due to energy itself) or a second-order (due to
the differentiation of the total energy in time)
differential equation. For didactic purposes, the
same second-order equation was derived following a
much more difficult way, by considering the change
of the angular momentum. A fourth straightforward
formulation is to implement Lagrange’s equations,
which include the difference between the kinetic
and potential energies (see Appendix B).
Fig. 14: Cart’s displacement for various initial angular
velocities 󰇗   .
APPENDIX A: Derivatives in time t
(cos ) sin
g
(A-1)
(sin ) cos

g
(A-2)
2
(cos ) sin cos
gg
(A-3)
2
(sin ) cos sin
gg
(A-4)
APPENDIX B: Lagrange’s equations
Lagrange equations are given by
12
0, ,
ii
d L L q X q
dt q q





(B-1)
and
kin pot
L E E
(B-2)
Setting the kinetic energy  as the sum of Eq. (9)
and Eq. (11b), while the potential energy  is
according to Eq. (12), the Lagrangian in (B-2)
becomes:
2
2 2 2
1
2
1( 2 sin cos )
2
cos sin
L mX
M X L LX
MgL

(B-3)
The first Lagrange equation (with ) leads to:
( ) ( cos )(cos ) 0m M X ML

gg
, (B-4)
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which coincides with Eq. (4). The second equation
(with ) leads to:
2( cos ) sin
sin sin 0
ML ML X
MgL


(B-5)
Eliminating the quantity 󰇘 between Eq. (B-4) and
Eq. (B-5), we eventually obtain Eq. (22).
Conflict of interest: "The authors have no conflicts
to disclose."
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https://en.wikipedia.org/wiki/Eric_Laithwaite
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[5] Anonymous:
https://en.wikipedia.org/wiki/Dean_drive
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[9] Kidd R.B, and Fogg S.L., A simple formula for the
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[10] Vartholomeos P., and Papadopoulos E., Analysis
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Contribution of Individual Authors to the
Creation of a Scientific Article (Ghostwriting
Policy)
The author(s) contributed in the present research,
at all stages from the formulation of the problem
to the final findings and solution.
Sources of Funding for Research Presented in a
Scientific Article or Scientific Article Itself
No funding was received for conducting this study.
Conflict of Interest
The author(s) declare no potential conflicts of
interest concerning the research, authorship, or
publication of this article.
Creative Commons Attribution
License 4.0 (Attribution 4.0
International, CC BY 4.0)
This article is published under the terms of
the Creative Commons Attribution License
4.0
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