η = 2r2 (ρ steel - ρ honey) g/9ν.
Where ‘η’ is the viscosity of the honey.
Viscous drag force is described by the relationship known as Stoke’s Law: F = 6πrην
Which can be rearranged to: η = F/6πrν.
To work out the density of the steel I will use the equation ‘density = mass / volume’.
I will do two experiments; one with the honey at room temperature using different sized ball bearings, and the other with the honey at 30°C using different sized ball bearings. When I have my results I will put them into a graph of velocity against radius squared. I will use a magnet to remove the steel ball bearings from the cylinder filled with honey.
This experiment is safe; there is no danger if the experiment is carried out correctly.
Results
I have gathered the following the results:
First experiment conducted at room temperature.
The distance of the region of terminal velocity is 10cm.
The terminal velocities of the ball bearings are as follows:
Speed of 0.0082 m ball bearing = 0.015082956 ms-1
Speed of 0.00252 m ball bearing = 0.008368200837 ms-1
Speed of 0.0022 m ball bearing = 0.006844626968 ms-1
Speed of 0.00152 m ball bearing = 0.002815315315 ms-1
Speed of 0.0012 m ball bearing = 0.001361470388 ms-1
Speed of 0.000752 m ball bearing = 0.0008729050279 ms-1
Second experiment conducted at 30° C.
The distance of the region of terminal velocity is 10cm.
The terminal velocities of the ball bearings are as follows:
Speed of 0.0082 m ball bearing = 0.040816326 ms-1
Speed of 0.00252 m ball bearing = 0.022675736 ms-1
Speed of 0.0022 m ball bearing = 0.011025358 ms-1
Speed of 0.00152 m ball bearing = 0.009560229446 ms-1
Speed of 0.0012 m ball bearing = 0.005561735261 ms-1
Speed of 0.000752 m ball bearing = 0.003158559697 ms-1
The mass of the cylinder = 0.11597 kg.
The mass of the cylinder + honey = 0.2355 kg.
The mass of honey = 0.2355 – 0.11597 = 0.11953 kg.
The volume of the honey = 0.00008499999582 m3.
The density of honey = 0.11953/0.00008499999582 = 1406.235363 kg/m3.
The mass of 0.0082 m steel ball bearing = 0.01671 kg.
The volume of 0.0082 m steel ball bearing = 4/3 π r3 = 4/3 π x 0.0083
= 0.000002144660585 m3.
The density of steel = mass/volume = 0.01671/0.000002144660585
= 7791.442673 kg/m3.
Using these results I have plotted a scatter graph of velocity against radius2.
This scatter graph shows the results from the experiment conducted under room temperature.
This scatter graph shows the results from the experiment conducted at 30°C.
As you can see both of the graphs of the two experiments have an anomalous result each, the anomalous results are the results of the largest (0.0082 m2) ball bearings. This may have been caused be the ‘edge effect’ which is when the ball bearing is too close to the sides of the measuring cylinder and as the ball tries to displace the honey, the honey pushes back because of the limited space. This is why the graph is a curve instead of a straight line, which it should have been; a straight-line graph suggests that the viscosity is constant under the same conditions whereas a curved graph suggests that the viscosity is not a constant under the same conditions.
I will work out the viscosity of the honey using the information from my graphs, I will have to find the gradient of the graph and use the equation:
η = 2g (ρ steel - ρ honey) / 9 x Gradient
To work out the viscosity I will need to find out the gradient of the graph, the gradient should be of the straight part of the graph and to find the gradient I will make another graph of each experiment, without the result of the large ball bearings (the anomalous results) because the other results are accurate enough to calculate the gradient with, by drawing a line of best fit. I will use the following equation to work out the gradient of the graph:
Gradient = (y2 – y1) / (x2 – x1)
The following two graphs are the graphs that I am going to use to work out the gradient with.
For the graph of the experiment conducted at room temperature.
On the graph I have drawn a line of best fit, and marked y1, y2, x1, x2, now using these values I can work out the gradient of this graph.
y1 = 0.0032 y2 = 0.008 x1 = 0.0000022 x2 = 0.0000056
Gradient = (y2 - y1) / (x2 - x1)
= (0.008 – 0.0032) / (0.0000056 – 0.0000022) = 1411.764706
The gradient of the room temperature graph is 1411.764706, and now I can work out the viscosity of the honey at room temperature using:
η = 2g (ρ steel - ρ honey) / 9 x Gradient
η = 2(9.8) (7791.442673 - 1406.235363) / (9 x 1411.764706)
= 19.6 (6385.20731) / 12705.88235 = 9.849773503
The viscosity of the honey is 9.849773503 Nsm-1.
For the graph of the experiment conducted at 30°C.
I have also drawn a line of best fit on this graph, and marked y1, y2, x1, x2, now using these values I can work out the gradient of this graph.
y1 = 0.011 y2 = 0.019 x1 = 0.000003 x2 = 0.0000056
Gradient = (y2 - y1) / (x2 - x1)
= (0.019 – 0.011) / (0.0000056 – 0.000003) = 3076.923077
The gradient of the 30°C graph is 3076.923077, and now I can work out the viscosity of the honey at 30°C using:
η = 2g (ρ steel - ρ honey) / 9 x Gradient
η = 2(9.8) (7791.442673 - 1406.235363) / (9 x 3076.923077)
= 19.6 (6385.20731) / 27692.30769 = 4.519307842
The viscosity of honey at 30°C is 4.519307842 Nsm-1.
One thing my results show is that the viscosity of the honey is lower at 30°C than it is at room temperature, suggesting that the viscosity of the honey decreases with a rise in temperature and increases with a fall in temperature.
The reason why I got inaccurate measurements is due to human error, the time it took me to start and stop the stop clock, one way to improve my measurements is to use a computer to measure the time it took for the ball bearing to pass through the region of terminal velocity. This and other similar errors were inevitable.
