Results
Data Analysis
Stokes Law states that spheres falling through a fluid exhibit the following relationship.
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.
Calculating Viscosity
The force needed to separate molecules of the fluid according to Stokes is
Eq. (1) -> F = 6(pi)Rnvc,
Where R is the radius of the sphere, n is the viscosity of the fluid, and vc is the velocity of the sphere through the infinite fluid. This force can be set equal to the gravitational force modified to account for the buoyant effect as follows
Eq. (2) -> 6 (pi) R n vc = 4/3 (pi) R3 (pS-pL) g,
where pS is equal to the density of the sphere, pL is equal to the density of the liquid, and g is the acceleration due to gravity. Constant velocity can be set equal to L/t (distance/time) and the equation solved for n as
Eq. (3) -> n = [2 g R2 (pS-pL) t] / 9L.
The velocity must be modified for the compression of the fluid by the cylinder walls (known as edge effects) by
Eq. (4) -> vc = v (1 + 2.4x),
Where x is the ratio of sphere diameter to cylinder diameter. The velocity must also be modified for the finite falling distance by
Eq. (5) -> vc = v (1 + 1.65y),
Where y is the ratio of sphere diameter to total liquid height.
Continuous velocity can now be described as
Eq. (6) -> vc = v (1 + 2.4x) (1 + 1.65y).
The new values for vc can now be substituted into equation (3) to yield
Eq. (7) -> n = [2 g R2 (pS-pL) t] / [9L (1 + 2.4x) (1 + 1.65y)].
Viscosity of glycerol
2 x 9.8 x 0.0022 x (8.02 – 1.26) x 14.07 = 2.628 poise
9 x 0.515
Resultant Force = W – Fr – u
W = mg
W = 4/3πr3 x density of steel x g
u = 4/3πr3σg
At terminal velocity:
W – Fr – u = 0
Because there is therefore no acceleration
Conclusion
From my results I concluded that as the diameter of the ball increases the speed of the ball increases, but so does the drag. This means that the force pulling the ball the ball is greater than the frictional force pushing the ball upwards. So as the ball’s diameter increases the drag increased but still the speed at which it falls through the liquid increases.
Bibliography
