mass = density x volume
volume of a sphere =
We have the diameter of the low mass ball.
radius =
We now have the volume of the low mass ball, to calculate its mass we need to find an average density of the steel used.
density =
The density for each ball is calculated excluding the unknown mass and the average is calculated. We can now use the first equation to give an estimate of the unknown mass.
From my calculations, the unknown mass is 0.0161kg
This result seems reasonable, but I cannot ensure that it is reliable since the density of steel may be different from the other balls.
Liquid Friction
If an object moves through a fluid it will experience friction due the object trying to move pass the fluid. The friction drag is caused by the shear between the liquid and the object. In this case the ball bearing moving downwards through the glycerol.
This shearing force acts on a body parallel to the direction of flow, and is caused by the viscosity of the liquid. The ball bearings move downwards through the liquid, in this case glycerol because of the gravitational pull of the earth, which in this experiment I will consider to remain constant at 9.81Nkg-1
Viscosity is not only the resistance of an object through the fluid, but also the internal friction between the liquid. It can be thought as the ‘stickiness’ of the material.
The coefficient of viscosity is the ratio of the shearing stress to the velocity gradient. In a liquid, viscosity is determined by the strength of the bonds between its molecules, something which temperature will effect. As temperature increases a liquid will become less viscous as its bonds become weaker and start to break more easily.
As long as laminar or streamline flow exists then, the stress exerted on the liquid is directly proportional to the velocity gradient that causes it. Viscosity can be expressed in (known as the Poiseuille (PI) or Poise (P)).
Different shapes moving through a liquid will distort the measurement of viscosity unless they are spherical this is because the surface area exposed to the liquid must remain constant. Also different shapes may encourage turbulence.
Viscosity is the primary factor resisting motion in this experiment, but when velocity exceeds a critical point at which turbulence starts to occur then pressure differences start to become more important that the viscosity.
This turbulence effect is illustrated on the diagram to the right. The movement of the ball flowing through the liquid creates an area of low pressure; the liquid then flows into this creating turbulence. This area of low pressure creates something called liquid pressure drag. The inertial effects of this turbulence however can be accounted for.
Forces
The above diagram is an illustration of a ball bearing, showing forces acting upon it.
The forces do not balance,
Force (downwards) > Force (frictional drag)
This means that the ball is accelerating downwards, if the two forces above become equal then acceleration will be 0 and the ball will be moving at a constant velocity which will be its terminal velocity where the forces are in equilibrium. We are unable to determine that the balls reach terminal velocity, but I am going to assume that they do not.
Force (downwards) = Force (gravity) – Force (buoyancy)
Force (buoyancy) is the upthrust on the ball, created by the weight of liquid that it has displaced.
- radius of ball bearing
- density of liquid
- density of ball bearing
- volume
- weight
We can calculate the weight of liquid that has been displaced by the ball by finding the balls volume and multiplying it by the density of the liquid.
= x
=
=
We can therefore equate this. This gives us the force in Newton’s.
Force (buoyancy) =
We can also similarly work out the force of gravity acting on the ball, this force will be mg, but to calculate the mass we need to find the volume of the ball and multiply it by the density of the steel.
Force (gravity) =
= x
=
We can therefore equate this. This gives us the force in Newton’s.
Force (gravity) =
This allows us to make a statement, which then can be simplified.
Force (downwards) = -
Force (downwards) =
The force acting downwards can also be expressed using stokes law.
- radius of ball bearing
- velocity of ball (at terminal velocity)
- coefficient of viscosity
Force (downwards) =
This is an expression for stokes law, and requires that the ball is travelling at its terminal velocity. We can use stokes law, and the equation for Force (downwards) found earlier to allow us to find the coefficient of viscosity of the liquid that the ball is moving through.
=
=
I used this approach to find the coefficient of friction of the liquid as shown in the table below.
Velocity was calculated by the equation below.
Speed =
The coefficient of friction that the experiment data has produced seems reasonable, and follows a trend. As the radius of a sphere increase so does the coefficient of friction. This is a positive relationship. However, the coefficient of friction should be the same for all the different ball bearings used.
The actual frictional drag will increase with radius, but viscosity is a ratio between shearing stress and the velocity gradient, hence it should remain constant as long as other factors such as temperature are not changed.
There is a positive relationship between radius2 and velocity. This is to be expected, as the gradient represents the coefficient of viscosity.
Y = where Y is the line shown in graph 1.
Graph 1 shows this positive relationship with the upwards-sloping line, it also shows that the balls velocity increases with radius. The error bars in the graph show where the points could actually be, and give an idea of inaccuracy.
Initial Evaluation
Possible reasons why the coefficient of viscosity did not remain constant are:
- Terminal velocity was not reached, stokes law uses the velocity over an infinite distance, which should be the terminal velocity of the object.
- Effects from the bottom of the cylinder, and the sides can cause the viscosity to be artificially inflated.
- Inertial effects and turbulence
I am going to analyse these effects in more detail, and introduce mathematical corrections to help normalise the results I initially have got.
Corrections
I have identified possible problems in the initial evaluation, however these can be partially be overcome using mathematical corrections.
Stokes law applies to a fluid of an infinite extent, but in this investigation we are using a cylinder. The first correction accounts for the boundaries at the walls and the second correction accounts for the bottom of the cylinder.
- radius of cylinder
- radius of ball bearing
- height of liquid
- corrected velocity
- initial velocity
- density of liquid
- density of ball bearing
Ladenburg correction for wall boundaries
Ladenburg correction for bottom of cylinder
As you can see, these corrections alter the velocity to a value, which should be nearer the terminal velocity. The effect of the walls is going to be much greater than the effect of the bottom of the cylinder when dealing with a long narrow cylinder as we are.
The other correction, which I can apply, accounts for the inertial effects
Corrections Analysis and Conclusions
As expected the Ladenburg correction for wall boundaries affect the coefficient of viscosity much more than the Ladenburg correction for the bottom of the cylinder.
The correction for the bottom of the cylinder only had a 0.95% change from the average, whilst the correction for wall boundaries had a change of 14.7%. Both these percentage changes show how much smaller the coefficient of viscosity was.
The determining factor identified in this investigation is clearly the velocity of the ball. The velocity recorded by the experiment seems to be smaller than the actual terminal velocity of the ball.
For low Reynolds numbers, the inertia term in the equation is smaller than the viscous term and therefore can be ignored.
The equation for viscosity as shown below can only be used when the Reynolds number is low.
The Reynolds number (Re) is calculated by this formula.
D = characteristic Length
G = mass velocity
= Coefficient of viscosity
This equation is only accurate for Re < 0.1 however the Goldstein correction allows us to calculate the viscosity still. The Goldstein correction sees a 9.02% change in viscosity, which shows that the terminal velocity had a greater affect that the inertial effects. However the inertial effects are still significant.
This table shows the Reynolds No for each different radius of sphere.
I can combine the Goldstein and Ladenburg corrections for walls to create a final result for viscosity.
How each correction affects the viscosity can be seen on graph 2.
The textbook answer for viscosity of glycerol was 1.2 , my result of 1.1±0.1 seems reasonable, and the average of 1.1676 is close to the textbook value. This is not conclusive since the glycerol water content is not known, but certainly is reassuring.
Evaluation
Anomalous results
There may have been some anomalous results, either balls of radius 3.00E-03 m or 3.18E-03 m have inaccuracies, I would expect the former to have a slightly higher viscosity in trend with the other results, but this is not so.
Theoretically the viscosity should remain the same, but as expressed before it increases with radius – this suggests that the terminal velocity is not reached.
I am not sure why the results do not follow this trend, but one possible reason is that one of the balls hit the side of the cylinder causing more friction than normal.
A possible solution would be to widen the cylinder, and ensure the ball was dropped from the middle. This solution would also help reduce the effect the wall boundaries have on the viscosity, as I have shown using the Ladenburg corrections – these have quite a significant effect. By widening the cylinder, the effect would be reduced.
Terminal Velocity
I already identified the problem of that the balls do not reach terminal velocity in my initial evaluation, but this problem was addressed by a correction. This does not give a definitive answer but certainly a more appropriate one. The reason why terminal velocity is important is that stokes law requires a fluid of infinite length to be used, this would mean that the object would be moving at its maximum speed, would have constant acceleration and hence would be at terminal velocity.
I cannot conclusively prove that the ball bearings did not reach terminal velocity, but since the coefficient of viscosity was not constant – this must be the case.
Steel density
The text book value for the density of steel varies, but tends to be between 7800 and 7900 kg/m3. The average density that I used to calculate the low mass was 7820 kg/m3 which shows that it is fairly accurate and should not have a significant effect on the experiment.
Measurement
The way the ruler was used with the cylinder wasn’t really satisfactory, and could have been the cause of inaccurate results. I was not satisfied that enough care was taken when measuring the 0.7m.
But the ruler was held by a clamp stand, so any effect that it would have had would have been universal and all my results would have been effected in the same way. This would not prevent me from coming to suitable conclusions.
The top balance may have created problems too, but as before these too would be universal they also would not effect the experiment.
Human error
This is likely to contribute the biggest problem to reliability and accuracy to the investigation; calculations are based on data collected from a human.
Humans can only judge things to an accuracy of ±0.1s. The human reaction time coupled with the accuracy of the basic stop watched used could see a variation of almost ±0.2s.
This means that the velocity calculated based on the time it takes the ball to travel the 0.7m distance can be inaccurate because of the human error involved when using a stopwatch.
The problems of the stopwatch and the use of are probably the biggest factor sin causing uncertainties and inaccurate results. But this did not prevent me from making suitable analysis.
Temperature
Temperature can have a large effect on the viscosity of a fluid, since it gives the molecules more energy and the bonds will become weaker, meaning that the viscosity will be decreased. But the experiment was all done on the same day, in the same room so any change in temperature will be negligible.
Improvements
To solve the problem of human error, a photodiode arrangement could be setup to electronically calculate the speed of the ball bearing. Also, a more instantaneous result could be attained if the distance was shorter, and closer to the bottom of the cylinder.
There are problems with the photodiode setup, and would require the ball bearing to be accurately dropped from a specific position.
An extension to the investigation could be the measurement of velocities at more than one intervals, perhaps every 0.1m, this would allow me to work out if the ball had reached terminal velocity or not.
Also, more care should be taken when releasing the ball bearing. The position from which the ball is dropped should be measured, and kept the same each repeat.
End of Investigation
pages type
17 total
12 normal
02 graphs
03 data