Viscosity Troubles: Solving a Physics Dilemma

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SUMMARY

This discussion focuses on calculating the coefficient of viscosity of motor oil by measuring the terminal velocity of a sphere using Stokes' Law. The rearranged equation for viscosity is n = F/6πRV, where F is the frictional force. The frictional force can be determined by balancing it with buoyancy and gravitational forces acting on the sphere. The final formula for viscosity is η = (ρs - ρf)Vsg / (6πrV), highlighting the importance of consistent viscosity values across various test spheres.

PREREQUISITES
  • Understanding of Stokes' Law and its application in fluid dynamics.
  • Knowledge of buoyancy and gravitational forces in a fluid medium.
  • Familiarity with the concepts of terminal velocity and drag force.
  • Basic principles of hydrostatic pressure distribution in fluids.
NEXT STEPS
  • Research the application of Stokes' Law in different fluid types and conditions.
  • Learn about the effects of temperature on the viscosity of fluids, specifically motor oil.
  • Explore methods for measuring terminal velocity in various experimental setups.
  • Investigate the impact of velocity-induced changes in pressure profiles on viscosity measurements.
USEFUL FOR

Physics students, researchers in fluid dynamics, and anyone conducting experiments related to viscosity and fluid behavior will benefit from this discussion.

Mystery_Man
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Viscosity Woes :(

Hi,

I'm conducting a Physics Investigation where I'm changing the temperature of Motor Oil and seeing how the change in temperature affects viscosity of the oil. I am measuring the terminal velocity of the sphere in the oil (V). I'm Using Stokes' Law Equation: F=6piRNV. I've re-arranged this equation to: n = F/6piRV. I want to calculate the coefficient of viscosity directly, however I can't understand how to calculate the frictional force (F). Can anybody shed some light as to how I calculate this?
 
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i have never had to work out a problem as such, but in general for drag force you have

F = -Cv or F = -Dv^2

where you use the first for very small speeds and the second for large speeds. C and D are constants that depend on the shape of the object and the viscosity of the material
 
Well, one way of doing this is the following:
If the sphere is falling through the oil at a uniform rate, then the OTHER forces acting upon the sphere must balance the force of friction.

Assuming hydrostatic pressure distribution (which, at the very least, ought to require that the dimensions of the falling sphere is a lot less than the fluid volume), we ,may calculate the buoyancy force [itex]F_{b}=\rho_{fluid}V_{sphere}g[/itex]acting upon the sphere. In addition, you'll have gravity working [tex]F_{g}=\rho_{sphere}V_{sphere}g[/tex].

where the indiced V is the volume of the sphere, and the rho's the densities
Thus, the force of friction will need to balance these to forces, which
means that your viscosity should be calculable from:
[tex]\eta=\frac{\rho_{s}-\rho_{f})V_{s}g}{6\pi{r}V}=\frac{2}{9}\frac{(\rho_{s}-\rho_{f})r^{2}}{V}[/tex]

Note that the assumption that we have hydrostatic pressure can only be held if the given equation gives CONSISTENT values for [itex]\eta[/itex] for a large variety of test spheres.

If we do not get consistent [itex]\eta[/itex]-values, the most likely explanation is that we cannot neglect the velocity-induced changes in the pressure profile of the fluid.

Thus, if you do this experiment, you might find that using a falling sphere through a viscous fluid is not a particularly good way to determine the viscosity of the fluid..
 
Last edited:

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