Calculating viscosity of a liquid with a falling object

Click For Summary
The discussion revolves around calculating the viscosity of castor oil using a steel-bearing's fall time in glycerol and castor oil. The user attempts to derive the viscosity using the relationship between density, velocity, and time but encounters issues with their calculations. They express frustration over not incorporating the density of the steel-bearing, which they suspect might be affecting their results. Suggestions include considering the density difference as a driving force and referencing Stokes' law, although the user lacks a radius for the spherical particle. The conversation highlights the complexities of viscosity calculations in fluid dynamics.
Beyar
Messages
7
Reaction score
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.
 
Last edited by a moderator:
Physics news on Phys.org
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 ?
 
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:
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
33K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
9
Views
4K
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
10K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
7K