VISCOSITY COEFFICIENT OF GLYCERIN
Çisel Özbütün
120520042
Introduction
When we drop any object from a certain height, there is always some friction acting on that object
during the movement. Just as there is air friction when we drop that object from air, there is a
friction caused by a liquid when we drop the object trough a liquid. The friction that the liquid
creates on the object is a result of the liquids viscosity.
Viscosity of liquids: Viscosity is the internal resistance flow, commonly known as the 'thickness' of
a fluid. A higher viscosity means that a fluid will have higher resistance to change of form, and
cause more friction on an object flowing trough it. Viscosity is temperature dependent, as in at
higher temperatures, viscosity is lesser.
Theory
As we drop (Vo=0) an object into a viscous fluid;
The flow speed of said object, moving vertically trough a viscous fluid, is dependent on the
following factors,
• Viscosity
• Size of the of the object
• Gravity
• Ascending force of fluid
Stokes Law:
Defining the friction force created by viscosity as F(v), as we drop a sphere with no initial speed
(Vo=0), into a viscous and homogeneous liquid, Stokes law formulates F(v) as:
Where
F(v)= Friction force
r = sphere radius
η = viscosity coefficient
V= velocity of sphere
F(v) acts as an opposite force on the movement V(created by gravity), slowing down the vertical
movement. F(a) (ascending force) will be another force that acts as opposite.
As an object falls from a height, the object will gain speed (V=gt), thus an accelerated movement
occurs. For the sphere falling trough a viscous fluid, as the speed of sphere increases, the friction
force acting on the globe will also increase proportionally. And will get to a point where Friction
Force F(v) will equal to F(g)-F(a), which was initially causing the sphere to accelerate. Thus
causing the net force to become 0, consequently acceleration as well (F=ma). Then the sphere will
continue it’s movement trough the viscous fluid at a constant speed, which is called the terminal
speed.
To find the terminal speed of the sphere:
Experiment
In this experiement we examine an objects movement trough a viscous fluid, and understand how
the viscosity coefficient correlates to the process.
We will use tiny steel globes of different radius’s, and use glycerin as our viscous liquid. We vill use
a graduated cylinder filled with glycerin.
We will use a stopwatch and a ruler to examine the terminal speed of our spheres. Calculate the
speed by using the data we collected from the experiment using x=Vt, and use that data calculate
the viscosity coefficient of glycerin.
Experiement apparatus:
• 3 steel spheres in different radius
• Thermometer
• Glycerin
• Caliper
• Stopwatch
• Ruler
• Magnet
• Graduated cylinder
Experimental procedure:
1. We filled our graduated cylinder with glycerin
2. Measured the temperature of the glycerin.
3. Measured the diameter of 3 different sized steel spheres using a caliper.
4. To avoid spheres to have initial speed before they hit the liquid, we used a magnet as a tool
to hold the steel spheres from outside the graduated cylinder, and dropped them from almost
no distance into the glycerin.
5. We marked the points at which the sphere hit terminal speed.
6. We used the stopwatch to measure the falling time from since the sphere hit terminal speed,
to when they hit the surface of the cylinder.
7. We used a ruler to measure the height of the marked points at which the spheres hit terminal
speed, wrote them down under “x” at the Data Table.
8. Repeated each 3 times, for 3 different spheres and wrote down the data for time under “t” at
Data Table.
Data And Calculations
Data Table:
Temperature of glycerin: T = 25,2 °C
Falling time, t (s) Height, x Steel Spheres Terminal Viscosity
Speed Coefficients
t(1) t(2) t(3) t(avr) cm m Diameter Radius (cm) V (m/s) η (N/s)
d (cm) r (cm)
0.68 0.58 0.62 0.62 12 0.12 1 0.5 0.25 0.19 2.00
1.82 1.94 2.0 1.92 18.5 0.185 0.5 0.25 0.062 0.10 0.949
2.21 2.24 2.14 2.19 20.5 0.205 0.25 0.125 0.016 0.10 0.237
η (avr) = 1.062
Slope of V-r^2 graph = 2.95
Possible causes of error: That the graph was drawn in r^2 (cm) while experimental value was in r^2
meters.Or an error made by me while determining scales for x and y values of the graph, or while
drawing the graph.
