An investigation into the terminal velocity of steel ball bearings as they pass through glycerol.

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An investigation into the terminal velocity of steel ball bearings as they pass through glycerol

Hypothesis

According to stokes law, spheres falling through a fluid exhibit the following relationship.

4/3?r3 (?-?)g = 6?nrvt

vt = 2r2 (?-?)g

9?

where ? is viscosity, ? is fluid density, ? is sphere density and vt is terminal velocity

Therefore

Vt r2

Variables

In this investigation there are many variables that will affect the terminal velocity of ball bearings as they pass through glycerol. These have been considered and the appropriate precautions taken so that their influence can be observed and analysed and the correct conclusions drawn.

Viscosity - any object moving through a viscous fluid is acted on by friction due to the fluid. A higher viscosity will increase this friction that opposes its motion. In order to keep the viscosity of the glycerol constant in this investigation, the same glycerol of concentration 1moldm-3 will be used throughout the experiments.

Sphere radius

Temperature - An increase in temperature decreases the viscosity of a fluid and therefore decreases the friction opposing the motion of the ball bearings. In this investigation the volume of glycerol is great enough to ensure that fluctuations in temperature during experiments will be very small. However, since the investigation takes place over a number of days, the temperature was monitored closely.

Gravitational field - The ball bearings fall through the glycerol due to the attraction created by the gravitational field from the Earth. In this investigation, this can be considered constant at 9.8ms-2.

Friction - Whilst friction created by the viscosity of the fluid has already been discussed, friction due to the sides of the tube may also play a part if the vertical tube is narrow. To minimize this, in the investigation, ball bearings were dropped from the centre of the tube so that they were less likely to experience friction due to the sides of the tube.

Medium - the fluid that the glycerol passes through must be kept constant if the viscous drag experienced by the ball bearings is to be kept constant and any variation in terminal velocity related to the ball's radius.

Distance - In order to make accurate measurements on the terminal velocity of the ball bearings the distance over which their speed is measured must be kept constant and must be selected so that accurate times can be generated................???

Material - the surface of the ball bearings must be constant for each radius since this will affect the area exposed to the friction of the fluid opposing the motion. Ball bearings with a smooth shiny surface were selected for this investigation as these were less likely to have a worn surface that would increase the friction independent of radius.

Likely outcomes

Released form rest - there is no frictional force F and resultant force = w

Velocity increases while F is less than w, resultant force = w-f downwards

Terminal velocity 0 reached when F=w, resultant force is zero: w-F=0 so there is not acceleration

The frictional forces oppose the motion. The magnitude of the frictional force increases from zero with the speed of the falling object

When an object is released and u=0 there is no frictional force and the resultant force is w. So at the moment of release the acceleration for the object is given by

A = resultant force = w = mg = g

mass m m

i.e. the full acceleration due to gravity of g = 9.8ms-2

When the object has gained velocity a frictional force F opposes its weight w. Now the downwards resultant force acting on the object is w-f and its acceleration is given by:

a = resultant force = w-F

Mass m

So the acceleration of the falling object is now reduced to less than 9.8ms-s

As the velocity of the falling object increases so the magnitude of the frictional force increases until it reaches a value equal and opposite to w;

So the resultant force = w - F = 0

And the acceleration is also zero.

So the object continues to fall at a constant speed known as terminal velocity, which is the maximum downwards speed possible for a particular object falling through a particular fluid. This illustrates Newton's first law of motion.
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According to stokes law

4/3?r3 (?-?)g = 6?nrvt

vt = 2r2 (?-?)g

9?

where ? is viscosity, ? is fluid density, ? is sphere density and vt is terminal velocity

Therefore

Vt r2

Apparatus

The apparatus selected for this investigation were chosen in order to generate accurate and reliable results.

Ball bearings of various radii

Glycerol (1moldm-3)

Large wide tube

Thermometer

Retort stand

Boss and Clamps

Stop watch

Micrometer

Metre ruler

Marker pen

Magnet

Procedures

The set ...

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