International Baccalaureate Programme
Session: 2015
HL Mathematics
Internal Assessment
Balancing a Can of Coke On Its Bottom Edge
by
Gede Sumantara Adi Pranata
Introduction
Friday night there was a Gwynedd house bonding where the students did not go
to dinner but instead ordering food and chilling in the dayroom. Most of the
students ordered a can of coke. By the time it arrived, student did not directly eat
the food but drink the coke first. Gwynedd students had a chat and jumping
around. At some point my fellow housemate knocked my coke that I had drunk, it
was not a really hard knock but a gentle unintentional push. Everyone was
screaming but the coke did not fall as everyone thought. The can of coke was
balance and stood on its bottom
edge. Everyone try to balance their
coke so it could stand on its bottom
edge, later we found out not many
people succeed to balance the can of
coke. I had previous knowledge of
centre of mass from Physics courses,
I was wondering why we had to
drink the coke and left particular amount of coke so it could balance. This
brought me to choose this idea as my Maths IA.
Research Question
What is the range of the volume of the coke so that the can of coke can stand on
its bottom edge?
Problem Analysis
The can of coke edge was curvy but for simplification it will be straight metal.
The diagram below shows the can of coke dimension in 2D with full of liquid
inside. The dimension is really important data through out the analysis of the
problem.
5.5 cm
1.5 cm
11.5 cm
9.5 cm
0.5 cm
6.5 cm
First, I will investigate whether the can full of liquid standing on its bottom edge
will balance. The centre of mass of the can full with liquid in 2D is the
average/arithmetic mean position of all the points in the shape of the can. The
centroid (centre of mass of a can full of liquid could be determined by geometric
decomposition, by dividing it into a finite number of simpler shapes. This
assumes that the can of coke has negligible mass; liquid inside the can of coke is
the only one that contribute to the centre of mass.
Cx =
Σ Cix 𝐴𝑖
Cy =
Σ 𝐴𝑖
Σ Ciy 𝐴𝑖
Σ 𝐴𝑖
Where:
Cix
: X-axis centroid of each of the plane
Ciy
: Y-axis centroid of each of the plane
Ai
: Area of each plane
The can of coke can be divided into 3 parts; 2 trapezoids and 1 rectangle, to
determine its centre of mass.
I
II
III
The trapezoid will be split into 3 parts; 2 triangles and a rectangle.
y
x
The triangle centroid can be determine from diagram below
C
P, Q, and R are the midpoints of AC, BC, and AB
respectively. CR, PB, and AQ are the medians of
the triangle. A median of a triangle is the
Q
P
G
segment from a vertex to the midpoint of the
opposite side. The centroid (G) of a triangle is
A
B
R
the common intersection of three medians. PQ
is called the midsegment of the triangle
because it is the segment that connects the midpoints of two sides of triangle.
Since it is the midsegment, it is parallel to the third side AB and half the length of
AB. Since PQ is parallel to AB therefore the PGQ is congruent to AGB, hence
the  PQG is similar to  ABG. So,
𝑃𝑄
1
𝑄𝐺
= =
𝐴𝐵
2
𝐺𝐴
GA = 2 QG
1
QG = QA
3
2
GA = 3 QA
The triangle centroid proof is not considering the fact that it is a right triangle.
Vector will be used to determine the coordinate of G from A.
1
⃗⃗⃗⃗⃗
𝑅𝐶
3
x2 − x1
x1 + x2
−
𝑥
1
⃗⃗⃗⃗⃗ = ( 2
𝐴𝑅
)= ( 2 )
𝑦1 − 𝑦1
0
C (x1, y2)
⃗⃗⃗⃗⃗ = 𝐴𝑅
⃗⃗⃗⃗⃗ +
𝐴𝐺
2
P
1
Q
G 1
2
A
(x1, y1)
1
R
, y1)
x1 + x2
(
2
2
⃗⃗⃗⃗⃗
𝑅𝐶 = (
B
(x2, y1)
x + x
x1 − x2
𝑥1 − 1 2 2
)= ( 2 )
𝑦2 − 𝑦1
𝑦2 − 𝑦1
x2 − x1
x −x
1 12 2
2
⃗⃗⃗⃗⃗ = (
𝐴𝐺
)+ (
)
3 𝑦2 − 𝑦1
0
1
3 (x2 − x1 )
⃗⃗⃗⃗⃗
𝐴𝐺 = (1
)
(y
−
y
)
1
3 2
Can be concluded from ⃗⃗⃗⃗⃗
𝐴𝐺 , in general, a right triangle with base X and height Y as
shown below, the G position is
1
Gx = 3 X
y
Y
1
Gy = 3 Y
G
0
x
X
The trapezoid (I) centroid is
y
5.5 cm
0.5 cm
x
0
6.5 cm
The Cix every shape counts from the origin will be count from origin.
Cx
=
2
3
1
2
0.5
1
)𝑐𝑚 ×( ×0.5×0.5)𝑐𝑚2
3
2
( ×0.5)𝑐𝑚 ×( ×0.5×0.5)𝑐𝑚2 +3.25 𝑐𝑚 ×(5.5×0.5)𝑐𝑚2 +(0.5+5.5+
1
2
2×( ×0.5×0.5)𝑐𝑚2 +(5.5×0.5)𝑐𝑚2
= 3.25 cm
Cy
=
1
3
1
2
1
3
1
2
( ×0.5)𝑐𝑚 ×( ×0.5×0.5)𝑐𝑚2 +0.25 𝑐𝑚 ×(5.5×0.5)𝑐𝑚2 +( ×0.5 )𝑐𝑚 ×( ×0.5×0.5)𝑐𝑚2
= 0.243
1
2
2×( ×0.5×0.5)𝑐𝑚2 +(5.5×0.5)𝑐𝑚2
The trapezoid (III) centroid is
y
5.5 cm
1.5 cm
x
O
6.5 cm
Cx
=
2
3
1
2
0.5
1
)𝑐𝑚 ×( ×1.5×0.5)𝑐𝑚2
3
2
( ×0.5)𝑐𝑚 ×( ×1.5×0.5)𝑐𝑚2 +3.25 𝑐𝑚 ×(5.5×1.5)𝑐𝑚2 +(0.5+5.5+
1
2
2×( ×1.5×0.5)𝑐𝑚2 +(5.5×1.5)𝑐𝑚2
= 3.25 cm
Cy
=
1
3
1
2
1.5
1
1
𝑐𝑚 ×(5.5×1.5)𝑐𝑚2 +( ×1.5 )𝑐𝑚 ×( ×1.5×0.5)𝑐𝑚2
2
3
2
1
2×( ×1.5×0.5)𝑐𝑚2 +(5.5×1.5)𝑐𝑚2
2
( ×1.5)𝑐𝑚 ×( ×1.5×0.5)𝑐𝑚2 +
= 0.729 cm
The centroid of the rectangular part of the can of coke from symmetry is
y
9.5 cm
Cx
=
9.5
Cy
=
6.5
2
2
= 4.75
= 3.25
6.5 cm
x
O
From the centroids of trapezoids and the rectangle, the can of coke centroid with
a full liquid and mass of metal assume to be negligible can be determine.
y
I
O
II
III
x
Cx
=
𝐶𝐼𝑥 × 𝐴𝐼 +𝐶𝐼𝐼𝑥 × 𝐴𝐼𝐼 +𝐶𝐼𝐼𝐼𝑥 × 𝐴𝐼𝐼𝐼
Cy
=
𝐶𝐼𝑦 × 𝐴𝐼 +𝐶𝐼𝐼𝑦 × 𝐴𝐼𝐼 +𝐶𝐼𝐼𝐼𝑦 × 𝐴𝐼𝐼𝐼
𝐴𝐼 +𝐴𝐼𝐼 +𝐴𝐼𝐼𝐼
𝐴𝐼 +𝐴𝐼𝐼 +𝐴𝐼𝐼𝐼
Cx
=
(0.5−0.243)𝑐𝑚×(0.5×
(5.5+6.5)
(5.5+6.5)
)𝑐𝑚2 +(0.5+4.75)𝑐𝑚×(9.5×6.5)𝑐𝑚2 +(10+0.729)𝑐𝑚×(1.5×
)𝑐𝑚2
2
2
(5.5+6.5)
(5.5+6.5)
(0.5×
) 𝑐𝑚2 +(9.5×6.5) 𝑐𝑚2 +(1.5×
) 𝑐𝑚2
2
2
= 5.72 cm
Cy
=
(5.5+6.5)
(5.5+6.5)
)𝑐𝑚2 +3.25 𝑐𝑚×(9.5×6.5)𝑐𝑚2 +3.25 𝑐𝑚×(1.5×
)𝑐𝑚2
2
2
(5.5+6.5)
(5.5+6.5)
(0.5×
) 𝑐𝑚2 +(9.5×6.5) 𝑐𝑚2 +(1.5×
) 𝑐𝑚2
2
2
3.25 𝑐𝑚×(0.5×
= 3.25 cm
Cy is 3.25 cm, which we know by symmetry of the can of coke shape. The can will
be balanced when the centroid is over the base. When the can is full, the centroid
is not right over the base so it will not balance.
Given that amount of water in 2D with
height h is in the can when the can balance
on its bottom edge as shown in the diagram
below. The area occupy by the liquid is
trapezoid to simplify the model and limit the
h
θ
value of h; h is a variable. To find the angle
between the can and the table, it can be
found by using can symmetry.
The next diagram will be focus on lower part of the can in order to find the angle
of the can and the table made.
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
0.5
tan α = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 = 0.5
0.5 cm
β
0.5 cm
α x
α
= 45° hence β = 45°
x
= √0.52 + 0.52 = √0.5
The angle between the bottom edge
and the can base is 135° thus angle
between the red dash line with the
can base is 135°-90° = 45°. This
made the angle between the can and
the table when its balance on its
bottom edge is 180°-135° = 45°. The
same method applies for the angle
between the side of the can and the
45°
45°
table.
There are 2 triangles and 1 rectangle from the
trapezoid form by the water in 2D occupies the
can of coke. The height of the trapezoid is h that
is will be height of the triangle as well. The base
of the triangle can be determine by
h
45°
tan 45° =
45°
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
ℎ
Opposite = h
To be balance the centroid of the trapezoid has to be ℎ + √0.5 > 𝐶𝑥 > ℎ
Therefore the centroid of the trapezoid is
y
√0.5
h
h
h
x
O
Cx
=
2
ℎ
3
1
2
× ( ℎ2 ) + (ℎ+
3
=
1
1
√0.5
) × (√0.5ℎ) + (√0.5+ℎ+ ℎ) × ( ℎ2 )
2
3
2
ℎ2 + √0.5 ℎ
0.5
ℎ3 + √0.5ℎ2 + ℎ
2
2
ℎ2 + √0.5 ℎ
=
(ℎ2 + √0.5ℎ)(ℎ+
√0.5
)
2
(ℎ2 + √0.5ℎ)
=ℎ+
√0.5
2
This is satisfy inequalities given, which mean ℎ + √0.5 > ℎ +
√0.5
2
>ℎ
Cy
=
=
2
ℎ
3
1
2
1
2
2
3
1
2
× ( ℎ2 ) + ( ℎ) × (√0.5ℎ) + ( ℎ) × ( ℎ2 )
ℎ2 + √0.5 ℎ
4 3 √0.5 2
ℎ +
ℎ
3
2
2
ℎ + √0.5 ℎ
The mass of the can will be taken into
consideration to get the centre of mass of the
liquid and the can. The distance of the average
centre of mass of the can and the liquid in it
q
from the original line can be calculated with
t
formula
t=
h
0 × 𝑤𝑙𝑖𝑞𝑢𝑖𝑑 + 𝑞 × 𝑤𝑐𝑎𝑛
𝑤𝑙𝑖𝑞𝑢𝑖𝑑 + 𝑤𝑐𝑎𝑛
I weigh the empty can using balance; wcan = 13.07 g. The mass of the liquid is
calculated by 𝜌 × 𝑉. The density of the liquid (𝜌) was using water density,
which is 1 g cm-3. The volume of the liquid will be determined using a solid
revolution of the line from the trapezoid in the y-axis. The line in the trapezoid
that will be used is in red.
y
(h +
√0.5
, h)
2
The function of the line is necessary in
order to get the volume. The gradient of
the line is
m
√0.5
, 0)
(
2
𝑦 −𝑦
= 𝑥2−𝑥1
2
=
ℎ−0
√0.5 √0.5
h+
−
2
2
=1
𝑦 − 0 = 1 (𝑥 −
√0.5
)
2
𝑦
=𝑥−
√0.5
2
𝑥
=𝑦+
√0.5
2
1
x
ℎ
𝑉 = π ∫ 𝑥 2 𝑑𝑦
0
2
ℎ
√0.5
= π ∫ (𝑦 +
) 𝑑𝑦
2
0
ℎ
= π ∫ 𝑦 2 + √0.5𝑦 +
0
0.5
𝑑𝑦
4
ℎ
𝑦 3 √0.5𝑦 2 0.5𝑦
= π ([ +
+
] )
3
2
4 0
ℎ3 √0.5ℎ2 0.5ℎ
= 𝜋( +
+
)
3
2
4
Alternative method to find the volume of the liquid is a volume of cone method.
The volume of the water is the volume of a cone, which is the tip, is taken.
(h +
(0, −
√0.5
, h)
2
√0.5
)
2
The volume of the trapezoid = big cone – small cone
1
2
√0.5
)
(ℎ
2
1
3√0.5ℎ2
= 3 𝜋 (ℎ +
= 3 𝜋 (ℎ3 +
ℎ3
=𝜋(3 +
2
√0.5ℎ2
2
+
2
+
+
1.5ℎ
0.5ℎ
4
1
√0.5
√0.5
)
−
𝜋
(
)
2
3
2
4
+
0.5√0.5
8
1
(
√0.5
)
2
)− 3𝜋(
0.5√0.5
8
)
)
It shows both methods will produce the same volume. The mass of the liquid is
ℎ3
the density ×the volume of the liquid = 𝜋 ( 3 +
√0.5ℎ2
2
+
0.5ℎ
4
)𝑔
The distance q can be determined using the can geometry.
𝑞−
x = 5.72 – 0.5
𝛽
y = 3.25
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan θ = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 =
z
√0.5
2
3.25
(5.72−0.5)
𝜃
, θ = 31.91°. We know β that is equal to 45°.
= √𝑥 2 + 𝑦 2
= 6.15
sin (45° − 31.91°) =
𝑞−
sin 13.09°
𝑞−
𝑞−
=
√0.5
2
√0.5
2
𝑧
√0.5
2
6.15
= sin 13.09° ×6.15
q
= 1.747
The average centre of mass distance from origin has to be over the base to be
balance, hence
√0.5
2
> 𝑡. 𝑡 can be calculated since the mass of the liquid inside the
can and the distance of the centre of mass of the can to the original line (q)
already solved.
√0.5
>
2
√0.5
>
2
ℎ3 √0.5ℎ2 0.5ℎ
0 × 𝜋 ( 3 + 2 + 4 ) 𝑔 + 1.747 𝑐𝑚 × 13.07 𝑔
ℎ3 √0.5ℎ2 0.5ℎ
𝜋 ( 3 + 2 + 4 ) 𝑔 + 13.07 𝑔
22.83 𝑔 𝑐𝑚
ℎ3 √0.5ℎ2 0.5ℎ
𝜋 ( 3 + 2 + 4 ) 𝑔 + 13.07𝑔
By plotting the following equations in TI-84 calculator, the h value can be
determined from the intersection of the following functions.
Y1
=
Y2
=
22.83
𝑥3 √0.5𝑥2 0.5𝑥
)+13.07
𝜋( +
+
3
2
4
√0.5
2
The intersection of two functions is x = 3.31, hence h > 3.31 cm.
Conclusion
The can of coke can be balance if it had a certain amount of water inside. There is
a limitation determining the range of the h as h get higher than certain level, the
shape occupy by the water in 2D is no longer trapezoid. Different shape of the
water in 2D would change the balance of the
liquid and the can. The solid revolution has a
biggest weakness because I assume that the
trapezoid in a perfect circle when rotated but in
facts the water is eclipse as shown in the picture.
The volume was the key of the average centre of
mass distance from the original line that I
picked. The can of coke was simplified, which
would be possibly different when the model of
the can of coke used was as it was where the
metal is not uniform all the way across.
However the biggest I have ever learnt is you
can trust what you have seen. The way I
perceive something now different when see an object get knocked, before it was
I really sure it would be fall but in some case it did not happen.
Bibliography
Accessed November 21, 2014.
http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx.
Jim Wilson's Home Page. "Centroid of a triangle." Accessed November 15, 2014.
http://jwilson.coe.uga.edu/EMAT6680Fa2012/Szatkowski/SzatkowskiWU
4/ASwriteup4.html.
Breathe internet. "Finding the centroid of a triangle." Accessed
November 19, 2014. http://www.netcomuk.co.uk/~jenolive/centroid.html.
Wolfram MathWorld: The Web's Most Extensive Mathematics Resource.
"Geometric Centroid -- from Wolfram MathWorld." Accessed
November 10, 2014.
http://mathworld.wolfram.com/GeometricCentroid.html.