viscosity of a liquid

Question

The viscosity of a liquid can be determined using a so-called falling ball viscometer. You will also carry out this experiment yourself in the physics practical course. To do this, metal balls with a known mass mball = 205 mg and a known ball radius rball = 1.85 mm are dropped into the liquid whose viscosity η (η: Greek letter eta) is to be determined. One measures the constant rate of fall v, which has set in after some time. According to Newton's principle of inertia, all forces acting on the body cancel each other out when moving in a straight line at constant speed. The gravitational acceleration is g = 9.81 ms2.

a) Calculate the density ρ of the metal sphere!

b) In glycerol with a density ρglycerol = 1.26 g cm3 the ball falls with a velocity of vglycerol = 3.9 cms.

Calculate the viscosity ηglycerol of glycerol!

Answer 1

A metal sphere with mass of 205 milligrams and radius of 1.85 millimeters is dropped into a fluid with viscosity η (to be determined). The sphere slows to a constant velocity of descent or v_sphere.  

Newton's Law Of Inertia has all forces acting on the sphere cancelling or balancing each other to zero difference for the sphere in linear motion at constant velocity v_sphere.

Take acceleration due to Earth's gravity as 9.80665 meters per second every second.

Density of the sphere or ρ_sphere is defined as mass per unit volume and is obtained by way of 0.000205 kilogram divided by (4/3)π(0.00185 Meter)³, which reduces to 7729.47624 kilograms per cubic meter.

Take the density of glycerol or ρ_glycerol as 0.00126 kilograms divided by 0.000001 cubic meter which goes to 1260 kilograms per cubic meter.

Write the descent velocity of the sphere through glycerol as v_glycerol equal to 3.9 centimeters per second or 0.039 meters per second.

Total gravitational force acting on the sinking sphere

(which is the combined effect of sphere weight and buoyant force)

is written as (weight) − (buoyant force) or

ρ_sphereg(4/3)π(r_sphere)³ − ρ_glycerolg(4/3)π(r_sphere)³

or (4/3)π(r_sphere)³g(ρ_sphere − ρ_glycerol).

For the sphere slowed to constant or "terminal" speed,

(4/3)π(r_sphere)³g(ρ_sphere − ρ_glycerol) can be equated to the

upward viscous drag that is given by Stokes' Law as 6πη_glycerolr_spherev_glycerol.

Place values to gain (4/3)π(0.00185 meter)³(9.80665 meters-per-second-squared)(7729.47624 − 1260) kilograms per cubic meter equals 6πη_glycerol(0.00185 meter)(0.039 meter per second).

This last isolates the viscosity (or resistance to flow) of glycerol as 1.237246215 Pascal-dot-seconds (or Poiseuilles).

{Note that viscosity of glycerol at 20°C is given online as 1.412 Pascal-dot-seconds (or Poiseuilles).}