Fnding the Fluid Resistance on a Ball Bearing Falling Through Glycerine
In this experiment I want to investigate the terminal velocity of the
ball bearings that are falling through glycerine and the viscosity of
I predict that the as the balls fall through glycerine, their
acceleration will decrease until they will reach their terminal
velocity, (due to fluid resistance) and after that they will travel
with a constant velocity, because the opposing forces
The bigger the ball, the longer time it needs to reach its terminal
velocity, so it will take less time to fall through the fluid than a
I measured the density of glycerine with a gravity hydrometer and
using Stokes' law I want to find out the viscosity of glycerine.
According to Stokes' law, 'when a sphere is falling
conditions, terminal velocity
has the following formula:
(Ïs - Ï1)g
Vt = 2r2 9Î·
In this formula 'Ïs, Ï1 and Î· are the density of solid, the density of
liquid and viscosity of liquid, respectively'.
Another way to calculate, also shown by Stokes, is to find the
frictional force which equals: F = 6Î aÎ·v, where a (cm) is the radius
of the sphere, Î· is the coefficient of viscosity (poises) and v (cm/s)
is the velocity.
'When the sphere reaches a constant velocity then F balances the
weight of the sphere less the upthrust of the liquid on it.'
mg( 1- Ï ) = 6Î aÎ·v , where Ï is the density of steel and Ïƒ is the
density of glycerine.
Speed is the distance travelled on time taken. Velocity is the speed
of an object in a specific direction.
The viscosity of a fluid is its resistance to flow; the easier the
fluid flows, the lower is its viscosity.
"Archimede's Principle says that the upthrust is always equal to the
weight of fluid displaced. A body in water experience an upthrust
equal to the weight of the water it displaces. If the upthurst equals
the weight of the body, the object does not sink."
The hydrometer uses this principle to measure the density of a liquid,
which is given 'by the level of the liquid on the float', then the
density is read off the scale.
To find out the density of glycerine I used a hydrometer and I wrote
down the value.
I found the density of steel on National Physics Laboratory web site
address; steel density = 8.0 gcm-3
I also base my experiment on Newton's third law of motion: 'to every
action there is an equal and opposite reaction'.
Ball bearings that fall through a fluid accelerate initially due to
the force of gravity, but as the balls fall the fluid resistance
increases until it balances the gravitational force. When the
resultant forces are balanced, the balls fall at their terminal
Terminal velocity is reached when a 'falling object stops accelerating
and continue to fall at a constant speed, due to balance between
gravity on air resistance' (in my experiment fluid resistance). When
this happens the resultant force is equal to zero, because the
opposing forces are balanced.
The terminal velocity depends on the size of the fluid resistance
force, the shape of the object and its mass.
I used ball bearings because due to their shape the fluid resistance
is smaller than it would be with some other shape.
After that I have to calculate the viscosity of glycerine.
- five steel balls of different sizes
- glass cylinder with glycerine
- two strong magnets
- stop watch
- gravity hydrometer
The apparatus is shown in the diagram below.
I set out the apparatus as shown in the diagram. In a glass cylinder
of 1000cm3 I put glycerine from bottom to top. With the ruler I
measured the distance that the balls have to travel through glycerine
and kept it constant for all my measurements.
I used the micrometer to measure the diameter of the balls and the
scale to measure the mass of the balls.
With the thermometer I found out the temperature of the glycerine,
because at different temperature the density may vary, and after that
I used the hydrometer to measure the density of glycerine.
Then I started to let the balls fall through glycerine, one by one,
measuring and recording the time needed to reach the bottom, using a
After every measurement I recovered the ball with the two strong
To make my experiment a fair test I kept constant the distance the
balls travelled through glycerine, the volume of glycerine and its
temperature, so the density should remain the same and the material
the balls were made of and I repeated each experiment six times.
I changed only the size of the balls - diameter and mass.
For my experiment I chose glycerine because it's ser than water, so
the balls will fall slower, due to increased fluid resistance and I
can record the time more accurate.
The table shows the standard values for viscosity of aqueous glycerine
at different temperatures and concentrations.
Viscosity of Aqueous Glycerine Solutions in Centipoises/mPa s
(1) Viscosity of water taken from "Properties of Ordinary
Water-Substance." N.E. Dorsey, p. 184. New York (1940)
The apparatus is relatively safe, the only thing that I have to be
careful about is not to touch my face or eyes while I am working with
glycerine, because I can have an allergic reaction and also to clean
the ball bearings after use.
I recorded my results in the following table:
Mean of the times (s)
Distance = 42 cm
Fluid temp. = 26 ËšC
Density of glycerine = 1.246 kg/l
From the evidence collected I can calculate the average speed the
balls travelled through glycerine.
distance travelled (in cm)
Average speed (in cm/s) = time taken (in seconds)
Ball 1: Average speed = 42 Ã· 1.235 = 34.008 cm/s (to 3dp)
Ball 2: Average speed = 42 Ã· 1.118 = 37.567 cm/s (to 3dp)
Ball 3: Average speed = 42 Ã· 0.678 = 61.946 cm/s (to 3dp)
Ball 4: Average speed = 42 Ã· 0.598 = 70.234 cm/s (to 3dp)
Ball 5: Average speed = 42 Ã· 0.493 = 85.192 cm/s (to 3dp)
Using Stokes formula I can find the viscosity of glycerine.
F = 6Î aÎ·v
mg( 1- Ï ) = 6Î aÎ·v
In my experiment m (mass) and a (radius of the sphere) have 5
different values, because I used 5 different balls, so I will
calculate the value for each ball.
For g (the gravitational acceleration), Ï( the density of steel) and Ïƒ
(the density of glycerine) I have only one value, because these are
constant. Î is also a constant and I'll use the value Î = 3.142
I will name the masses m1, m2â€¦.m5 for different balls; ball 1, ball
2â€¦ball 5, respectively. The same kind of notation I'll use for radius
- radius = diameter Ã· 2
In this case I have the following calculations:
Ï = 8.0 gcm-3 Î = 3.142 Ïƒ = 1.246 kg/l = 1.246 Ã— 103 g/ cm-3 g = 9.8
For ball 1:
m1 = 2.10g v1 = 34.008 cm/s a1 = 3.94 mm = 0.394 cm = 3.94 Ã— 10-1cm
2.10 Ã— 9.8 ( 1- 1.246 Ã—103 Ã· 8.0) = 6 Ã— 3.142 Ã— 3.94 Ã— 10-1 Ã— Î· Ã—
2.10 Ã— 9.8 ( 1 - 1.557 Ã—102) = Î· Ã— 2.526 Ã— 102
2.10 Ã— 9.8 Ã— 1.547 Ã— 102
Î· = 2.526Ã— 102
Î· = 31.837 Ã· 2.526 â†’ Î· = 12.603 centipoise/mPa s (to 3dp)
For ball 2:
m2 = 3g v2 = 37.567 cm/s a2 = 4.425 mm = 4.425 Ã— 10-1cm
3 Ã— 9.8 ( 1- 1.246 Ã—103 Ã· 8.0) = 6 Ã— 3.142 Ã— 4.425 Ã— 10-1Ã— Î· Ã— 37.567
3 Ã— 9.8 Ã— 1.547 Ã— 102
Î· = 3.133 Ã— 102 â†’ Î· = 14.517 centipoise/mPa s (to 3dp)
For ball 3:
m3 = 16.41g v3 = 61.946 cm/s a3 = 7.90 mm = 7.90 Ã— 10-1cm
16.41 Ã— 9.8 ( 1- 1.246 Ã—103 Ã· 8.0) = 6 Ã— 3.142 Ã— 7.9 Ã— 10-1Ã— Î· Ã—
16.41 Ã— 9.8 Ã— 1.547 Ã— 102
Î· = 9.225 Ã— 102 â†’ Î· = 26.968 centipoise/mPa s (to 3dp)
For ball 4:
m4 = 23.79g v4 = 70.234 cm/s a4 = 8.945 mm = 8.945 Ã— 10-1cm
23.79 Ã— 9.8 ( 1- 1.246 Ã—103 Ã· 8.0) = 6 Ã— 3.142 Ã— 8.945 Ã— 10-1Ã— Î· Ã—
23.79 Ã— 9.8 Ã— 1.547 Ã— 102
Î· = 1.184Ã— 103 â†’ Î· = 30.462 centipoise/mPa s (to 3dp)
For ball 5
m5 = 56.41g v5 = 85.192 cm/s a5 = 11.925 mm = 1.192 cm
56.41 Ã— 9.8 ( 1- 1.246 Ã—103 Ã· 8.0) = 6 Ã— 3.142 Ã— 1.192 Ã— Î· Ã— 85.192
56.41 Ã— 9.8 Ã— 1.547 Ã— 102
Î· = 1.914 Ã— 103 â†’ Î· = 44.481 centipoise/mPa s (to 3dp)
As we can see in the graph 1 and 2 the time needed for the smaller
balls to travel the same distance is longer than the time needed for
the bigger ball to travel the distance.
All the values fit to the trend, showing that the smaller the ball,
the larger the time needed to reach the bottom of the cylinder - the
graphs have a descending trend line. So far my prediction is
After I calculated the viscosity for the different balls and calculate
my results with the predicted values for glycerine viscosity I
realised that I should have taken the temperature of the glycerine
after every experiment and I should have asked what is the
concentration of the glycerine solution. Although the initial
temperature of the glycerine was 26ËšC, my values tell me that the
temperature must have risen, considering the fact that the
concentration of the solution remained constant.
From my result and the table of 'Viscosity of Aqueous Glycerine
Solution' I think that the concentration of the solution was about 70%
and the temperature dropped from 26ËšC to about 10ËšC, although I have
no evidence that this happened.
Another mistake I've made in my experiment is that I didn't record the
time needed for the balls to travel a smaller distance to be able to
calculate terminal velocity.
My measurement are reliable and my calculations are also correct, but
I am not sure that I reached the purpose of the experiment because I
neglected some of the factors and didn't realize that I am suppose to
take some extra measurements.
Considering the facts written above I can't draw a firm conclusion
about my experiment. The only thing that I can confirm is that average
speed of a ball falling through a fluid increases directly
proportional with the mass of the ball.
I think that if I had more time to do my experiment, and I was more
enlightened about the implication of the title of the experiment, my
results could be much better.
If I will do my experiment again I will place a change the distance
the balls will travel through glycerine, using two pieces of rubber
tied out to the cylinder, outside - one at the top and one at the
bottom and move the top one 5cm down for each measurement. After that
record the time needed to travel smaller distances. All this will help
me calculate the terminal velocity.
I will also try to find out more information about viscosity and
search for more detailed values of glycerine viscosity at intermediate
temperatures, because the values I have are for every 10ËšC. I will
also record the temperature of the glycerine solution after each
AQA GCSE, Physics -Specification B, 2004
Key Science, Physics - Jim Breithaupt, 1997- Stanley Thornes
Advanced Level, Physics - Nekon & Parker, 2001 Heinemann
Advanced Practical Physics - Leslie Beckett, 1982 John Murray