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Nneka Ezeigwe. Physics coursework.



A series of different sized ball bearings were dropped into a measuring beaker containing glycerine. The time taken to travel between two marks on the beaker was recorded. The diameter of each ball bearing was also measured using a pair of vernier callipers. Each experiment was repeated three times. The diagram and readings are shown below.  







































Note that the average readings have been taken to reduce or eliminate any error present.

I will now plot a graph to make the relationship between the diameter of the ball bearings and the time taken to travel a distance of 162mm.

The graph shows that there is an inverse relationship between the time taken and the diameter of the steel ball. The curve has a decreasing negative slope.

        When a steel ball (or any object for that matter) is dropped into a liquid, it experiences certain forces.  They are:

  • The weight of the object acting downwards
  • The upthrust acting upwards and t
  • The Viscous drag F acting upwards.


Every fluid has a property called viscosity. This is an internal property of the fluid that offers resistance to the movement of particles through it.

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Where η= the viscosity of the fluid.

From this formula, we can see that the viscous drag on the bigger ball is bigger than the viscous drag on the smaller ball. This is because the velocity increases as well as the radius. The weight downward also increases and the graph implies that this does so on a greater scale than the velocity and the radius. Hence, the time taken is less for a bigger ball bearing than for a smaller one.

The resultant force acting on the ball bearing is given by:


The ball bearing continues accelerating downwards until:


At this point, the ball has attained a terminal velocity and is in a state of free fall.

Now, W=4/3πr3ρg

And   U=4/3πr3σg

        Sir George Gabriel Stokes also gave the formula for finding terminal velocity. This is given by:

                       4/3πr3ρg – 4/3πr3σg – 6πηrvt =0

∴ vt = (2r2(ρ-σ)g) / 9η

This can be compared to the formula:

                                  Y  =   M    x    +   C  

Therefore, I will plot a graph of the terminal velocity against the radius squared (r2).

The gradient of this straight-line graph will be 2((ρ-σ) g) / 9η.

Where ρ is the density of the sphere, which is equal to 7930 kg/m3

     and σ is the density of glycerine, which is equal to   1260 kg/m3

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        This can easily be proved with Stokes law. It states that:

                                   vt = (2r2(ρ-σ)g) / 9η

                           Therefore η=(2r2(ρ-σ)g) / 9 vt

        I then chose two random corresponding vales of vt and r2.

I chose the values r2= 2.5*10-7  vt  =2.04*10-3

And the values      r2= 40*10-7 vt  =31.25*10-3

Using the formula above, if indeed, the velocity I calculated is equal to the terminal velocity; I would obtain the same value for η in both cases.

Calculating gives me η=1.78012 pa for the first set and 1.8593 for the second set. Within the limits of experimental error, these are the same and so I can safely assume that the velocity between the two rubber bands is equal to the terminal velocity.


        This experiment was a successful one. The results obtained were within the limits of experimental error. Though errors occurred, this is unavoidable and they were detected and reduced to as bare a minimum as possible. The expected relationship between the velocity and the radius squared was obtained from the graph.

        I can safely conclude that the viscosity of the glycerine used in the experiment is 1.8560775Pa within the limits of experimental error.

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