Motion of Hexagonal Pencil on an inclined Plane ∗
Amin Rezaeezadeh
Physics College , Sharif University of Technology , Iran, Tehran
Abstract This article is describing the motion of a hexagonal shape- briefly, a hexagonal pencil- which is rolling down on
an inclined plane. This attempt is done by introducing a simple model, without focusing on interactive forces
acting on the pencil due to the plane. Experiment results are going to be compared with what we discover from
theory of the problem. The analysis is a simplified sample of motion of a cylinder, which has a plane of contact
with its subplane, instead of a line.
∗
amin67r@gmail.com
I.
Introduction
Motion of a shape of hexagonal cross section, on an
inclined plane is a demonstration of inelastic collisions
between faces of the shape and the plane. The best
thing to work on is, hexagonal pencil that can be found
easily. It is observed that pencil reaches a constant
terminal speed. This is due to the fact that, each
collision between faces of the pencil and the plane,
causes some decrease in mechanical energy of the
pencil. This dissipation of kinetic energy is a function
of coefficient of restitution of the inclined plane, that is,
value of the terminal velocity (vt), depends on elastic
properties of the inclined plane. At section II, we will
see, how the terminal speed, depends on the angle of
inclination. Sections III and IV, report the experiment
observations and then, compare the data with predicted
results.
inclined plane to the horizontal and μ is the coefficient
of
a
FIG 1
Next section, I try to explain the motion, more clear and
name it as a “rolling” motion.
II.
Theory
Let us accept some assumptions for solving the
problem. Consider the pencil to be a hexagonal prism
of uniform density. It would be an adequate
approximation, if the graphite lead is thin and of
density not very different from that of the wood. I
assume the pencil remains in contact with the plane and
does not slip or slide. During one complete revolution
of the pencil, each of its six edges serves in turn, as the
instantaneous axis of rotation. Thus, the pencil rotates
by one-sixth of a turn while a given edge serves as the
axis of rotation. The main axis remains horizontal.
For a sharp- edged pencil, the radius of curvature of
any edge is very small, compared with width of a face
of the pencil –section V, shows the effect of radius of
curvature.
Rolling friction should be negligible. Pure rolling can
occur only for 1 μ > tan θ, where θ is the angle of the
1
In practice a mixture of rolling and sliding can be
observed for inclinations larger than the bound of μ>tanθ.
:
Cross-section of the discussed pencil. limiting static friction of the pencil on the plane.
In the remainder of this article, I assume that pure
rolling takes place and that the pencil does not bounce
after any of its collisions with the plane.
A. Motion Analysis Assuming a Coefficient of
Restitution
If θ < π / 6 the center of mass of the pencil rises
during the first part of any 1/6 turn. Hence, the pencil
will not roll down the plane from rest unless θ > π / 6 .
In practice it is found that spontaneous rolling may
occur for θ≈ 20º because of the corners of the
hexagonal shape are usually a little rounded.
The energy dissipation in the collision of a face of the
pencil with the inclined plane can be described in terms
of a coefficient of restitution. This approach introduces
an unknown parameter, but it appears to the author to
be the most appropriate simple description of a
situation in which the elastic properties of materials
cannot be ignored.
I try to use energy analysis to estimate the terminal
linear and angular speeds.
During each one-sixth of a turn, the center of mass of
the pencil falls through a height a sin θ , where "a" is
the width of each of the six faces of the pencil. (see
FIG2)
On continuing this process, we find that the kinetic
energy at the beginning of the nth 1/6 turn – after the
(n-1)th collision– is
E n = εΔE (1 + ε + ε 2 + ... + ε n − 2 ) = εΔE
1 − ε n −1
.
1−ε
(2)
For any value of ε less than one, the energy of the
pencil at the beginning of a 1/6 turn approaches the
asymptotic value
E∞ =
ε
ΔE
1− ε
(3)
When the pencil has tipped through an angle α after the
beginning of the nth 1/6 turn (see FIG3), its kinetic
energy E (α) is then
FIG 2
:
One revolution of the pencil.
E(α) = En + 2mgasin(π / 12 − θ / 2 + α / 2) sin(α / 2 + θ / 2 − π / 12)
Thus, gravity adds energy
ΔE = mga sin θ
(1)
to the pencil at each, one-sixth of a turn.
At the end of each one-sixth of a turn, the pencil
collides with the surface of the inclined plane. Let
ε ( 0 ≤ ε ≤ 1 ) be the coefficient of restitution for this
collision, defined so that ε is the ratio of the kinetic
energy immediately after a collision to the kinetic
energy immediately before it. We shall adopt the
notation that E n is the kinetic energy at the beginning
of the nth 1/6 turn – that is, immediately after the(n-1
1)th collision. If the pencil starts from rest, its initial
kinetic energy, at the beginning of the first 1/6 turn, is
E1 = 0 , and the kinetic energy immediately before the
first collision (after the first 1/6 turn) is ΔE. The kinetic
energy immediately after the first collision – i.e. at the
beginning of the second 1/6 turn– is E 2 = εΔE. The
kinetic energy immediately before the second collision
will be (1 + ε)ΔE , and immediately after it – i.e. at the
beginning of the third 1/6 turn– is, E3 = ε(1 + ε ) ΔE .
FIG 3.
The average kinetic energy over this 1/6 turn is
θ +π / 6
⟨ E⟩ = θ −
∫π [ E
n
+ 2 mga (sin 2 (α / 2) − sin 2 (π / 12 − θ / 2))]dα
/6
π /3
That is,
⟨ E ⟩ = mga (
−3
π
+
mga sin θ
1
3
) cos θ + E n +
= I ⟨ω n ⟩ 2
2
2
2
(4)
Where I is the moment of inertia of a pencil about an
edge and ⟨ω⟩ is the average angular velocity of the
pencil. For the terminal velocity (as n → ∞ ) the
asymptotic linear velocity , ⟨v ∞ ⟩ ,of the pencil is
⟨v ∞ ⟩ =
3a
⟨ω∞ ⟩
π
(5)
Noting that the pencil advances through a distance 6a
during one full turn whose period is 2 π / ⟨ ω∞ ⟩ , and by
FIG 4
using equations(1),(3)and (4) together, we obtain an
estimated value for the terminal velocity
⟨v ∞ ⟩ =
ε
3a
3 2 ga ⎛⎜ − 3 cos θ
3
sin θ ⎞⎟
⟨ ω⟩ =
+
cos θ +
sin θ +
π
π
π
k ⎜⎝
2
1− ε
2 ⎟⎠
(6)
Here k is the coefficient in the formula I = kma 2 for
the moment of inertia. For a uniform hexagonal prism,
about an edge, k has the value 17
.
12
For a positive real solution, the expression under the
radical in eq(6) should be positive. This requires that
ε >
3
π
−
3
π
−
3
2
3
2
−
1
2
tan θ
+
1
2
tan θ
(7)
:
Pencil needs initial energy to rise.
The pencil will not continue to roll through a second
1/6 turn unless sufficient energy remains after its
collision with the plane. That is, we need the energy at
the beginning of the second 1/6 turn to satisfy
E 2 = ε( E1 + ΔE ) > mga (1 − cos( π / 6 − θ)).
Similarly, the energy at the beginning of the nth 1/6
turn must satisfy
E n = ε n −1 E1 + εΔE
1 − ε n −1
> mga (1 − cos( π / 6 − θ)).
1− ε
Recalling the argument that led to eq(2), we find that
the asymptotic condition is
εΔE
> mga(1 − cos(π / 6 − θ)),
1− ε
B. Calculation of ε and the Condition of
Continuous Rolling
or
If θ < π / 6 the pencil will not roll unless it is given an
initial kinetic energy
ε >
E1 > mga (1 − cos( π / 6 − θ))
For small θ we must have
such that the center of mass of the pencil can rise to the
vertical during the first 1/6 turn as shown in FIG4.
ε >
1 − cos(π / 6 − θ)
1 − cos(π / 6 − θ) + sin θ
1
θ
1+
1− 3 / 2
≈
1
1 + 7.5θ
(8)
The value of ε can be determined from eq(8)2, taken as
an equality for the smallest angle of inclination θ at
which the pencil continues to roll after being given an
initial velocity.
ε =
1 − cos(π / 6 − θ min )
1 − cos(π / 6 − θ min ) + sin θ min
(9)
It is possible to show that the ε that is calculated by eq.
(9) always satisfies the inequality (7). In another word,
equation (8) guarantees condition (7). Thus, there is
always a real answer for the terminal speed.
C. When does the pencil lose contact with the
plane?
component of the weight also occurs right before the
contact point changes, and is equal to g cos(θ + π / 6).
Thus, the condition of remaining in contact with the
plane, would be
g cos(θ + π / 6) ≥ ω2 a.
The angular velocity of the pencil, ω, is determined,
using energy conservation at the end of each turn.
mga sin θ
1 2
Iω = E∞ + ΔE =
.
2
1− ε
Substituting I, it gives that
2 g sin θ
ka(1 − ε)
The center of mass of the pencil moves on a circular
arc around the axis of rotation. When the pencil has
angular velocity ω, its center of mass must be subject to
ω2 =
However, the
a centrifugal force equal to mω 2 a .
centrifugal force cannot be greater than the component
of the weight of the pencil along the radius vector from
the center of mass to the point of contact. The maximal
centripetal acceleration occurs right before the contact
point changes (because ω is maximum). The minimal
Thus, eq(10) and eq(11) lead to
cos(θ + π / 6) ≥
2 sin θ
k (1 − ε)
3
1+
21
It might be misconceived that it is possible to measure ε by
However, there is no reason that the mentioned method
would give the same result as eq(9). This is because, when
the pencil hits the plane (in free fall) the force acting on it is
distributed uniformly on the face of contact, whereas in case
of rolling , the distribution of the force is not uniform, and
hence the dissipation of energy is different for both cases and
then the value of ε would be different. This is borne out by
experiment, where I have found that the value of ε in the
case of falling is somewhere between 0.1 and 0.3, while for
rolling condition, it is larger than 0.6.
(11)
(12)
On simplifying, we obtain
tan θ ≤
a very simple experiment, Namely to drop the pencil from a
height h0 on to a horizontal plane with the same material as
the inclined plane. Then measure the height to which the
pencil rises after hitting the plane. The ratio of these heights
gives the value of ε .
(10)
4
k (1 − ε)
(13)
Maximum available angle of inclination (θmax), for
k =
17
12
and ε = 0.8 is almost 7 degrees.
At the best conditions, in which ε → 0 –pure rolling–
the maximum value of θ is
θ max ≈ 24.7°
That is, the pencil cannot start pure rolling, from rest.
Pure rolling requires a small initial kick.
It is possible to determine ε experimentally and using
eqs(13) and (9).
III.
Experiment 1
FIG 5 shows the matter, I used in the experiment 1. It is
a uniform solid wooden hexagonal prism, in the shape
of a pencil –but with no graphite lead. Although it
needs a small impulse to start the first rotation, I
neglect it to be consider in the equations of motion.
values
Wooden plane
Plane covered by
Cloth
sin θ min
0.032
0.060
0.79
0.68
0.127m/s
0.090m/s
0.122 m/s
0.084 m/s
0.039
0.065
ε
v ex
v th
v th − v ex
v ex
Table 1 - v ex and v th are the terminal speeds determined
from experiment and theory, respectively.
FIG 5
:
The data are for the case sin θ = 0.076 and a = 0.005m.
The matter used in Exp. 1
By filming the motion, it would be possible to display
the distance covered in terms of time, on a graph. The
experiment is done on two inclined planes, with the
same θ, but made of different surface materials. Both
data is displayed in Graph1. One surface is soft, the
other one is some harder.
In graphs (2) and (3) ,the terminal speeds are calculated
separately.
0.6
L(m)
y = 0.127x - 0.0336
2
R = 0.9984
0.5
0.4
0.3
The terminal speed would be different for both cases,
since ε is different for the two materials. According to
table 1, the softer the plane, the smaller the coefficient
of restitution ε .
In table 1 , the experimentally
measured terminal speed(vth), is compared with the
theoretical value.
0.2
0.1
0
0
1
Graph 2 :
2
3
4 Time(sec)5
Time-position graph for wooden surface.
L(m)
0.55
0.45
0.6
0.35
0.5
0.25
wood
0.3
0.05
-0.05 0
y = 0.0901x - 0.0473
R2 = 0.9999
0.4
cloth
0.15
L(m)
0.2
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
0.1
Time(sec)
Time(sec)
0
0
Graph 1 :
Time-position graph for cloth and wooden
surfaces.
Graph 3 :
1
2
3
4
5
6
Time-position graph for cloth surface.
Cloth
surface
L(m)
0
0.125
0.25
0.417
0.625
0.75
0.875
1.125
1.375
1.917
2.125
2.5
2.79
3.125
3.75
4.08
4.29
4.75
5.08
5.417
5.625
6.08
0
0.01
0.02
0.03
0.04
0.05
0.06
0.08
0.1
0.14
0.16
0.19
0.21
0.24
0.29
0.32
0.34
0.38
0.41
0.44
0.46
0.5
Wooden
surface
L(m)
0
0.25
0.42
0.54
0.63
0.75
0.79
1.08
1.25
1.42
1.58
1.75
1.96
2.33
2.5
2.71
2.92
3.04
3.29
3.42
3.58
3.79
4.04
4.38
4.58
0
0.01
0.02
0.03
0.04
0.05
0.06
0.09
0.11
0.13
0.15
0.17
0.2
0.25
0.27
0.3
0.33
0.35
0.38
0.4
0.42
0.45
0.48
0.52
0.55
Table 2 :
The data for graphs (2) and (3). L is the
distance traversed, measured in meter.
0.45
L(m)
0.4
y = 0.0845x - 0.0165
0.35
0.3
0.25
0.2
0.15
0.1
0.05
Time(sec)
0
0
1
2
3
4
5
Graph 4 :
Position versus time for a real pencil in
case of sin θ = 0.046.
A. Sound Wave Analyzing
When the pencil is rolling down the plane, every
collision, makes a weak sound. This hitting sound could
help us to find the terminal velocity. The sound is
analyzed by a simple program vbWavAnalyzer3.
Analyzing the sound of collisions of Graph 5, is shown
in FIG 6.
0.8
IV. Experiment 2
L(m)
y = 0.141x - 0.0724
0.7
0.6
This experiment is done with a real pencil (see FIG1).
Values of the parameters are gathered in Table 3. The
data are displayed in Graph 4 and the terminal speed
has been calculated there.
0.5
0.4
0.3
0.2
0.1
values
mpencil
The real pencil and
wooden plane
9.8 m/s
ρ graphite
1.85 g/cm3
k
1.38
4º.6
ε
a
0.86 ± 0.02
0.003m
0.076 m/s
θ max
sin θ =
v th
0.046
v ex
rgraphite
Table 3 :
0.085 m/s
0.001 m
Parameters of Exp2.
Time(sec)
0
0
0.85 ± 0.01 g
g
6
1
2
3
4
5
6
7
Graph 5 :
Position versus time for a real pencil in
case of sin θ = 0.066.
FIG 6 :
Sonogram of the pencil hitting the plane
for sinθ = 0.066. The vertical lines indicate the collision of
pencil with the inclined plane.
3
You could find it here: http://amin67r.googlepages.com v n is calculated from energy conservation.
FIG 7
:
En =
Zooming FIG 6.
According to FIG 6, the time interval for a complete
revolution –i.e. six collisions – is almost 0.147 s. The
distance, in which, the pencil covers during this period
can be measured by its perimeter. Note that, this value
is more than 6a, because of the finite radius of
curvature of each edge. We obtain for the measured
terminal speed
v∞ =
p 0.021m
=
≈ 0.143 m/s,
T 0.147 s
(14)
where p is the perimeter of the pencil. As expected,
calculated value for the terminal velocity is very close
to what we saw in Graph 5.
V. In search of the equation of motion
In this section, I want to estimate an equation of
motion, by using a simple method. Consider the
moment of the beginning of nth turn (see FIG 8),
where, the speed of the center of mass is v n .
1 2 1 v 2n 1
Iω = I 2 = kmv 2 n
2
2 a
2
And from eqs (1) and (2) ,
En = εmga sin θ
⇒vn =
1 − ε n −1
1− ε
2εga sin θ . 1 − ε n −1
k
1− ε
A similar result is obtained for v n +1 .
Thus, the increase in “parallel to the plane component”
of velocity, during each turn, is
Δv = (v n +1 − v n ) cos( π / 6).
The time-average of a complete turn is determined
by ⟨ ω n ⟩ . Therefore,
⟨ ω n ⟩ Δt = π / 3
So, the acceleration during a turn is
3 3 (v n +1 − v n )
Δv
⟨ ωn ⟩
=
Δt
2π
N-th turn
(15)
(16)
I expand it to the first order of ε n (note that the
expansion is just to ε n , and 0 ≤ ε ≤ 1 )
vn
⎡3 3g sinθ 1 − ε ⎛ 3 3 ⎞
⎤
⎛
ε ⎞
Δv
⎜
⎟⎟sinθ⎥.εn
− ⎟ cosθ + ⎜⎜ 0.5 +
≈ ⎢
ε ⎜⎝ 2
π ⎟⎠
Δt
1− ε⎠
⎢ 2kπ
⎥
⎝
⎣
⎦
vn +1
(17)
θ
FIG 8
:
The pencil, during its n-th rotation.
The number of collisions, n, is determined by the
distance “s” through which the pencil passes along the
inclined plane. That is,
n=
s
+1
a
(18)
Note that, beginning of the second turn, happens at s =
a.
By differentiating from eq(22), we came up with a
statement for velocity.
Since the time of each turn is small enough compared
to the total time of motion, Therefore we can say
Δv
dv
=
= &s&(t ),
Δt
dt
(19)
⎛ − ρ ln ε
− 2aρ
tanh⎜⎜ t
ln ε
2a
⎝
s&(t ) =
Graph 7 shows
⎞
⎟ (23) ⎟
⎠
versus time and for different values
of coefficient of restitution.
This leads to
&s&(t ) = ρε s / a = ρe ln(ε ) s / a , ρ > 0,
(20)
where
ρ=
3 3 g sin θ
ε(1 − ε)
2kπ
⎛ 3
⎛
3⎞
ε ⎞
⎜
⎟ sin θ .
− ⎟ cos θ + ⎜⎜ 0.5 +
⎜ 2
⎟
π
1
−
ε ⎟⎠
⎝
⎝
⎠
(21)
Equation (20) is readily integrable to
s(t ) =
2a ⎡ ⎛ − ρ ln ε
ln ⎢sec⎜ t
ln ε ⎢⎣ ⎜⎝
2a
⎞⎤
⎟⎥
⎟
⎠⎦⎥
(22)
Graph 6, shows the function s(t) for different values of
(noting that, ρ is a function of ε).
For small ε, the pencil quickly reaches a constant
speed.
Graph 7 :
of ε.
as a function of t and different values
A. Terminal Velocity Determination
Another way of calculating terminal velocity is using
eq(23), taking t→∞, it simplifies to v∞ =
− 2aρ
ln ε
(24)
Therefore, Using data of Table 3, we have,
ρ=0.20
And then ,
v ∞ = 0.089 m/s
While, what we have obtained for
v ∞ , from the
Experiment 2 (see Table 3) was 0.085m/s. That is, less
than 5% is our error.
Graph 6 :
Function s(t) for different values of ε.
fit = NonlinearRegressAdata,
B. Finding Effective “a”
From the fact that, real pencil (used in Exp2) is not
sharp edged, Thus, radius of curvature of the edges
could affect the motion. I mean, the effective width of a
face, can be found from the real data.
i
0.03 % y
j
z
j
$%%%%%%%%%%%
j
z
EEz
j2 a ∗ LogASechAx
z, x, 8a<,
2
a
k
{
RegressionReport → 8BestFit, SummaryReport<E
1
Log@0.86D
:BestFit→ −0.0376186Log@Sech@2.29946xDD,
BestFitParameters→ 8a→ 0.00283688<,ParameterCITable→
Estimate
Asymptotic SE
CI
,
a
0.00283688
0.0000284937
80.00277614,0.00289761<
Using data of Graph 4 in this section and finding the
best curve of s(t) to fit the data, is a way of determining
the effective “a”. I get help from Mathematica to do
this work – finding the best fitting curve among
different values of “a”. Graph 8 shows the observed
points fitted to the theoretical curve for s(t).
EstimatedVariance→ 0.0000174782,ANOVATable→
Model
Error
Uncorrected Total
Corrected Total
DF
1
15
16
15
SumOfSq
0.846838
0.000262174
0.8471
0.219044
MeanSq
0.846838
0.0000174782,
AsymptoticCorrelationMatrix→ H1.L,
sHmL
0.4
Max Intrinsic
FitCurvatureTable→
Max Parameter−Effects
95. % Confidence Region
R = 0.998
2
Curvature
0.000587892
>
0.00605878
0.469164
0.3
0.2
VI. Conclusion
0.1
1
2
3
4
5
TimeHsL
observed points fitted to the theoretical
Graph 8 :
curve for s(t)
In fitting points to the curve, it is very important to use
points in which ε n << 1 since, it was our condition to
find the equation of motion. This means, for instance
n>10 or ε n <0.2 is a good region, and this happens for
s>4cm.
Calculating “a” from fitted curve,
a= 0.00284
± 0.00003 m
Which is less than what we had measured (3mm).
By means of a series of experiments and theoretical
calculations, It is determined that a pencil rolling down
an inclined plane quickly reaches a terminal speed. If
we examine other shapes with a different number of
faces, we would come to the same result. One may
question whether or not this is to be characterized as a
kind of rolling friction. For the motion of cylinder on
an inclined plane, usually, we describe the problem in
terms of "rolling friction"4. Rolling friction is caused,
when the surface of contact is not a point or a line, but
also it is a plane of contact. Thus, the normal force
acting on the cylinder ( or a sphere) has a toque about
center of the mass. This effect is somewhat complex,
but it can be solved by considering that the deviation
from center of mass is small enough compared to the
dimensions of the matter. If this approximation does
not work, then, there would be an approach to the
4
A. Domenech, T. Domenech and J. Cebrian, “Introduction
to the study of rolling friction” , Am. J. Phys. 55, 231-235
(1987).
D. Shaw, “Frictional force on rolling objects”, Am. J. Phys. 47,
887-887 (1979).
problem that says, the cylinder would reach to a
terminal velocity, as we discussed in the pencil
problem. In the case of a rolling polygonal cylinder,
the reason for the constant terminal speed is that energy
is dissipated at the collision of each face, and the same
must apply to a circular cylinder, which is merely the
limiting case of a polygonal cylinder. We know that,
when a circular cylinder is placed on an inclined plane,
it makes a small contact surface with the plane, and so
it can be described as a pencil with a large number of
faces. Hence, according to the motion of a pencil, it
finally rolls down the inclined plane at a constant
speed, too.
Acknowledgement
The author is indebted to Dr. Jeremy Tatum for
drawing his attention to the problem and for supplying
the relevant literature. Also, I would like to thank Prof.
Kirk. T. McDonald for his helpful comments and
suggestions.
References
[1]
Boston Area Undergraduate Physics Competition,1995,Problem6
http://hutrough.harvard.edu/penanen/competition/images/
exam95.pdf
http://hutrough.harvard.edu/penanen/competition/images/
solut95.pdf