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.
Apparatus
- five steel balls of different sizes
- glass cylinder with glycerine
- ruler
- two strong magnets
- stop watch
- micrometer
- scale
- thermometer
- gravity hydrometer
The apparatus is shown in the diagram below.
Method
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 stop watch.
After every measurement I recovered the ball with the two strong magnets.
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 denser 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.
Risk assessment
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.
Obtaining evidence
I recorded my results in the following table:
Distance = 42 cm
Fluid temp. = 26 ˚C
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 ÷ 1.235 = 34.008 cm/s (to 3dp)
Ball 2: Average speed = 42 ÷ 1.118 = 37.567 cm/s (to 3dp)
Ball 3: Average speed = 42 ÷ 0.678 = 61.946 cm/s (to 3dp)
Ball 4: Average speed = 42 ÷ 0.598 = 70.234 cm/s (to 3dp)
Ball 5: Average speed = 42 ÷ 0.493 = 85.192 cm/s (to 3dp)
Using Stokes formula I can find the viscosity of glycerine.
F = 6Πaηv
σ
mg( 1- ρ ) = 6Πaηv
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), ρ( the density of steel) and σ (the density of glycerine) I have only one value, because these are constant. Π is also a constant and I’ll use the value Π= 3.142
I will name the masses m1, m2….m5 for different balls; ball 1, ball 2…ball 5, respectively. The same kind of notation I’ll use for radius – radius = diameter ÷ 2
In this case I have the following calculations:
ρ = 8.0 gcm-3 Π= 3.142 σ = 1.246 kg/l = 1.246 × 103 g/ cm-3 g = 9.8 m/s2
For ball 1:
m1 = 2.10g v1 = 34.008 cm/s a1 = 3.94 mm = 0.394 cm = 3.94 × 10-1cm
2.10 × 9.8 ( 1- 1.246 ×103 ÷ 8.0) = 6 × 3.142 × 3.94 × 10-1 × η × 34.008 →
2.10 × 9.8 ( 1 – 1.557 ×102) = η × 2.526 × 102
2.10 × 9.8 × 1.547 × 102
η = 2.526 × 102
η = 31.837 ÷ 2.526 → η = 12.603 centipoise/mPa s (to 3dp)
For ball 2:
m2 = 3g v2 = 37.567 cm/s a2 = 4.425 mm = 4.425 × 10-1cm
3 × 9.8 ( 1- 1.246 ×103 ÷ 8.0) = 6 × 3.142 × 4.425 × 10-1× η × 37.567
3 × 9.8 × 1.547 × 102
η = 3.133 × 102 → η = 14.517 centipoise/mPa s (to 3dp)
For ball 3:
m3 = 16.41g v3 = 61.946 cm/s a3 = 7.90 mm = 7.90 × 10-1cm
16.41 × 9.8 ( 1- 1.246 ×103 ÷ 8.0) = 6 × 3.142 × 7.9 × 10-1× η × 61.946
16.41 × 9.8 × 1.547 × 102
η = 9.225 × 102 → η = 26.968 centipoise/mPa s (to 3dp)
For ball 4:
m4 = 23.79g v4 = 70.234 cm/s a4 = 8.945 mm = 8.945 × 10-1cm
23.79 × 9.8 ( 1- 1.246 ×103 ÷ 8.0) = 6 × 3.142 × 8.945 × 10-1× η × 70.234
23.79 × 9.8 × 1.547 × 102
η = 1.184× 103 → η = 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 × 9.8 ( 1- 1.246 ×103 ÷ 8.0) = 6 × 3.142 × 1.192 × η × 85.192
56.41 × 9.8 × 1.547 × 102
η = 1.914 × 103 → η = 44.481 centipoise/mPa s (to 3dp)
Analysis
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 confirmed.
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 26˚C, 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 26˚C to about 10˚C, 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.
Evaluation
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 10˚C. I will also record the temperature of the glycerine solution after each experiment.
References:
AQA GCSE, Physics –Specification B, 2004
Key Science, Physics – Jim Breithaupt, 1997- Stanley Thornes (Publishers) Ltd
Advanced Level, Physics – Nekon & Parker, 2001 Heinemann
Advanced Practical Physics – Leslie Beckett, 1982 John Murray (Publishers) Ltd
http://www.neolytica.co.uk/glycerine/resources/table18.htm
http://www.npl.co.uk/mass/research.html