State stokes law and derive an expression for it
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state Stokes law derive an expression for the terminal velocity of a sphere falling through a viscous fluid
Asked by admin @ in Physics viewed by 427 People
state Stokes law derive an expression for the terminal velocity of a sphere falling through a viscous fluid
Answered by admin @
According to Stokes’ law, the backward viscous force acting on a small spherical body of radius r moving with uniform velocity v through fluid of viscosity η is given by
F=6 πη rv
Where, r = Radius of the spherical body
v = velocity of the spherical body
It gives the relationship between retarding force and velocity. When viscous force plus buoyant force becomes equal to force due to gravity, the net force becomes zero. The sphere then descends with a constant terminal velocity (vt).
Now,
6 π η rv =4/3 πr^3( p-σ )
Where, ρ = Density of the liquid
σ = Density of the spherical body
Terminal velocity of a spherical body of density ρ and radius r moving through a liquid of density ρ’ is
where,
ρ = density of body,
sigma = density of liquid,
η = coefficient of viscosity of liquid and,
g = acceleration due to gravity
1) If ρ > ρ0, the body falls downwards.
2) If ρ < ρ0, the body moves upwards with the constant velocity.
3) If po << ρ, v = (2r2ρg/9η)
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