Calculating viscosity of a liquid with a falling object

In summary, the problem involves a small steel-bearing falling in two different liquids, glycerol and castor oil, and the goal is to calculate the viscosity of the castor oil. Using the equation for viscosity, and taking into account the density and velocity, a value for the viscosity of castor oil can be obtained. However, the equation does not account for the density of the steel-bearing, so further consideration may be needed to accurately calculate the viscosity. The use of Stokes' Law may be helpful, but it requires knowledge of the radius of the steel-bearing, which is not given in the problem.
  • #1
Beyar
7
0

Homework Statement


[/B]
A small steel-bearing falls 25.0 cm in glycerol in 23.8 s and the same distance in castor oil in 15.1 s. The densities are for glycerol 1260 kg m−3 , for castor oil 961 kg m−3 , and for steel 7830 kg m−3 . The viscosity for glycerol is 1.490 Pa s. Calculate the viscosity for castor oil. All values are valid for 20 ◦C.

Homework Equations


I guess Viscosity=Density*Velocity where the velocity is equal to Distance/time.

The Attempt at a Solution


Thought I'd put the equation for the distance equal to each other and then rewrite it to get the viscosity of the castor oil, but I get the wrong value. It should be 0,988 Pas. I get the final equation to:
Viscosity of Glycerol= (Density of Castor Oil*Viscosity of Castor*Time the steel bearing fell in the Glycerol)/(Density of Glycerol * time steel bearing fell in Castor oil)

The equation has not regarded the denisty of the steel-bearing though, so that might be the problem but I don't see how I would get around to fit it in.
It is very frustrating.
 
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  • #2
Beyar said:
The equation has not regarded the denisty of the steel-bearing though
Well then, could it be that the net driving force for falling is a density difference ?
 
  • #5
Beyar said:
But stokes relation has accounted the radius of the spherical particle, I don't have a radius to utilize.
Call it ##R##, cross your fingers and hope it divides out in the answer :smile:
 

1. How do you measure the viscosity of a liquid with a falling object?

The viscosity of a liquid can be measured by using a falling object and measuring its terminal velocity. The terminal velocity is directly related to the viscosity of the liquid and can be calculated using the Stoke's Law equation.

2. What is the Stoke's Law equation?

The Stoke's Law equation is given by F = 6πηrv, where F is the drag force, η is the viscosity of the liquid, r is the radius of the falling object, and v is the terminal velocity of the object.

3. What is the relationship between terminal velocity and viscosity?

The terminal velocity of a falling object is directly proportional to the viscosity of the liquid. This means that as the viscosity of the liquid increases, the terminal velocity decreases.

4. What are some common units for measuring viscosity?

The most common units for measuring viscosity are poise (P) and centipoise (cP).

5. Can the Stoke's Law equation be used for all types of liquids?

The Stoke's Law equation is most accurate for Newtonian fluids, which have a constant viscosity regardless of the applied force. It can also be used for non-Newtonian fluids, but the results may not be as accurate due to the varying viscosity of these fluids.

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