Stokes' law & Terminal Velocity

Stokes' Law and Terminal Velocity

Let's start this article with stock's law.

Stokes' law

When a body moves through a fluid then the fluid exerts a viscous force on the body to oppose its motion. According to Stokes' law, the magnitude of the viscous force depends on the shape and size of the body, its speed and the viscosity of the fluid.

So, for the below figure

If a sphere of radius r moves with velocity v through a fluid of viscosity η

Then using Stokes' law the viscous force (F) opposing the motion of the sphere is given by,

F=6πηrv

Where
η - coefficient viscosity
r-radius
v− velocity

Terminal Velocity

When the spherical body is dropped in a viscous fluid, it is first accelerated and then it's acceleration becomes zero and it attains a constant velocity and this constant velocity is known as terminal velocity.

For a spherical body of radius r is dropped in a viscous fluid, The forces acting on it are shown in the below figure.

So Forces acting on the body are weight and viscous which are given below:

• Weight of Body (W)

W=mg=43πr3ρg

Where ρ→ density of body
Buoyant/ Thrust Force (T of FB )
T=FB=43πr3σg
where σ→ density of fluid

• Viscous force (F)

F=6πηrv

So when the body attains terminal velocity the net force acting on the body is zero.

Apply force balance

FB+F=W→6πηrv+43πr3σg=43πr3ρg→6πηrv=43πr3g(ρ−σ)→vt=29r2(ρ−σ)ηg

Where vT= terminal velocit.

From this formula, we can say that

• Terminal velocity depends on the radius of the sphere/body.
• The greater the density of the solid greater the terminal velocity
• The greater the density and viscosity of the fluid lesser the terminal velocity.
• If ρ > σ then Terminal velocity will be positive.

I.e Spherical body attains constant velocity in a downward direction.

• If ρ < σ then Terminal velocity will be negative.

I.e Spherical body attains constant velocity in an upward direction.

Terminal velocity graph

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Solved Example Based on Stokes' law & Terminal Velocity

Example 1: A ball with density ρ, is falling inside a liquid of density σ, the ball will move downward with terminal velocity
1) Always
2) Only when ρ>σ
3) Only when ρ<σ
4) It will never move with constant velocity

Solution:

If ρ>σ.
Terminal velocity will be positive.
- wherein

The spherical body attains constant velocity in a downward direction.
vT=29(ρ−σ)ηr2g
∴vT is downward only if ρ>σ

Hence, the answer is option (2).

Example 2: Spherical balls of radius R are falling in a viscous fluid of viscosity η with a velocity ν. The retarding viscous force acting on the spherical ball is:

1) directly proportional to $R$ but inversely proportional to $\nu$

2) directly proportional to both radius $R$ and $\nu$

3) inversely proportional to both radius $R$ and velocity $\nu$

4) inversely proportional to $R$ but directly proportional to velocity $\nu$

Solution:

From Stoke's law

F=6πηrv

For
Fαv

Hence, the answer is option (2).

Example 3: In a experiment verify stokes law, a small spherical ball of radius r and density ρ falls under gravity through distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to :(ignore viscosity of air)

1) r4
2) r3
3) r
4) r2

Solution:

As,
v=2gh

So.
ρ×43×π×r3×g=6×π×η×r×2gh

From here
r2=kh
where k= constant
So,
r4∝h

Hence, the answer is option (1).

Example 4: The velocity of a small ball of mass ' m′ and density d1, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is d2, then the viscous force acting on the ball, will be :

1)mg(1−d1d2)

2)mg(1−d2 d1)
3) mg (d1 d2−1)
4)mg(d2 d1−1)

Solution:

B+FV=mgFv=mg−B∵B=d2vg=d2×md1 g∴Fv=mg(1−d2 d1)

Hence, the answer is option (2).

Example 5: A raindrop with radius R=0.2 mm falls from a cloud at a height h=2000 m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is : [Density of water fw=1000 kg m3 and Density of air fa=1.2kgm3,g=10 m/s2 Coefficient of viscosity of air =1.8×10−5Nsm−2 ]

1) 250.6 ms−1
2) 43.56 ms−1
3) 4.94 ms−1
4) 14.4 ms−1

Solution:

F→ Frictional drag face (stokes law)
B→ Buoyancy force B≃0 (neglected)
At terminal velocity,
F=mg6πηRVT=mg=(ρw)×43πR3gVT=ρw×4πR33×6πηRV=45ρw18ηR2g

VT=29×103×(2×10−4)2×101.8×10−5h=8×10−4×1059×1.8=408.1VT=4.94 m/s

Hence, the answer is option (3).

Summary

Stokes' Law defines the motion of objects in fluids, including air and water: such is the force acting upon a small, smooth body moving along through a thick fluid that it depends both upon its size and on the thickness of the fluid it travels at. This force is called drag. When an object falls through a fluid, it accelerates until the force of gravity pulling it down is balanced by the drag force pushing it up. At that time, it starts falling with constant velocity, called terminal velocity. The larger and heavier the objects, the higher their terminal velocity. Actually, this concept explains why the tiny particles in the air settle slowly and why larger objects fall faster.