Stokes’ Law (9.5.1) - MECHANICAL PROPERTIES OF FLUIDS - CBSE 11 Physics Part 2
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Stokes’ Law

Stokes’ Law

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Interactive Audio Lesson

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Introduction to Viscous Drag

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Teacher
Teacher Instructor

Today, we'll dive into Stokes' Law, which gives us a fascinating insight into how a body moves through a viscous fluid. Can anyone share what they think happens when an object falls through water or oil?

Student 1
Student 1

I think it slows down because the fluid pushes against it.

Teacher
Teacher Instructor

Exactly! This retarding force occurs due to the viscosity of the fluid, which resists the motion. This leads us to understand the viscous drag force.

Student 2
Student 2

So the faster it falls, the more drag force it experiences?

Teacher
Teacher Instructor

Correct! The viscous force is indeed proportional to the velocity of the body and acts in the opposite direction. Remembered as 'V-FED' for Velocity-Factor-Effect-Drag. This is a helpful acronym!

Understanding Stokes’ Law

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Teacher
Teacher Instructor

Stokes' Law expresses the viscous drag force mathematically. It states, F equals π over six times the fluid's viscosity times the velocity. Does anyone remember the equation?

Student 3
Student 3

Is it F = (π/6)ηv?

Teacher
Teacher Instructor

Spot on! This equation helps us calculate the drag force on spheres moving in a viscous medium. Now, what factors influence this drag force?

Student 4
Student 4

The viscosity and the velocity?

Teacher
Teacher Instructor

Yes, and also the radius of the sphere! Larger spheres will experience more drag. Does anyone have thoughts on how this impacts raindrops?

Terminal Velocity Explained

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Teacher
Teacher Instructor

Now, let’s explore terminal velocity, which occurs when gravitational force equals drag force. Can someone summarize how that condition is reached?

Student 1
Student 1

It means the object stops accelerating and falls at a constant speed.

Teacher
Teacher Instructor

Exactly! The formula for terminal velocity relates the properties of the sphere and fluid. It’s given by: vt = (2a²(ρ-σ)g) / (9η). What can we infer about the factors affecting terminal velocity?

Student 2
Student 2

If the radius increases, terminal velocity increases too, right?

Teacher
Teacher Instructor

Yes! It increases with the square of the radius but decreases with higher viscosity. Always remember the riddle - 'Big drops versus sticky liquids!'

Applying Stokes’ Law: Example Problem

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Teacher
Teacher Instructor

Let’s put this knowledge into practice. We have a copper ball of radius 2mm falling through oil at 20°C with a terminal velocity of 6.5 cm/s. Can anyone recall how we might calculate the viscosity of the oil?

Student 3
Student 3

We would use the terminal velocity equation!

Teacher
Teacher Instructor

That’s right! Substituting the known values into the formula will help find η. Can anyone do the math?

Student 4
Student 4

I got η = 9.9 × 10⁻¹ kg m⁻¹ s⁻¹!

Teacher
Teacher Instructor

Excellent work! Understanding how to apply the law makes the concept much more tangible. Remember - practice makes perfect!

Introduction & Overview

Read summaries of the section's main ideas at different levels of detail.

Quick Overview

Stokes' Law describes the viscous drag force experienced by a falling body in a fluid, which is proportional to its velocity and dependent on the fluid's viscosity.

Standard

Stokes' Law, articulated by Sir George G. Stokes, explains the relationship between the viscous drag force acting on a falling sphere and its velocity, the viscosity of the fluid, and the sphere's radius. The law leads to the concept of terminal velocity, where a falling raindrop eventually reaches a constant speed.

Detailed

Stokes’ Law

Stokes’ Law illustrates the dynamics of a body falling through a viscous fluid, outlining how the viscous drag force (F) affects its motion. As the body falls, it drags the fluid, creating relative motion among fluid layers and resulting in a retarding force opposite to the motion. This force is proportional to the object's velocity and can be expressed mathematically as:

$$F = \frac{\pi}{6} \eta v$$

where \( \eta \) is the viscosity of the fluid and \( v \) is the velocity. The law leads to important implications regarding terminal velocity (the constant speed a falling body reaches when the forces acting on it are balanced).

For a sphere falling in a fluid, the terminal velocity \( v_t \) is derived from the balance of forces, given by:

$$v_t = \frac{2a^2 (\rho - \sigma) g}{9\eta}$$

Where:
- \( a \) is the radius of the sphere,
- \( \rho \) is the density of the sphere,
- \( \sigma \) is the density of the fluid,
- \( g \) is the acceleration due to gravity.

This relationship indicates that the terminal velocity depends on the square of the radius and inversely on the fluid's viscosity, highlighting the importance of fluid characteristics on falling objects. The example of a copper ball falling through oil is employed to illustrate these principles effectively.

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Audio Book

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Viscous Drag Force

Chapter 1 of 5

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Chapter Content

When a body falls through a fluid it drags the layer of the fluid in contact with it. A relative motion between the different layers of the fluid is set and, as a result, the body experiences a retarding force. Falling of a raindrop and swinging of a pendulum bob are some common examples of such motion.

Detailed Explanation

When an object, like a raindrop, falls through air, it doesn't just fall freely. It interacts with the air in a way that the air pushes back against it, creating a force that opposes the fall. This is similar to how you feel resistance when you try to move your hand through water. As the raindrop falls, it causes the air around it to move and creates layers of air with different velocities due to friction and drag. This interaction leads to a retarding force on the falling object.

Examples & Analogies

Imagine trying to push your hand through thick syrup compared to pushing it through air. Your hand meets much more resistance in syrup than in air. Similarly, as a raindrop falls, it meets resistance from the air, which significantly affects its speed and motion.

Proportionality of Viscous Force

Chapter 2 of 5

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Chapter Content

It is seen that the viscous force is proportional to the velocity of the object and is opposite to the direction of motion. The other quantities on which the force F depends are viscosity η of the fluid and radius a of the sphere.

Detailed Explanation

Stokes’ law states that the force resisting the fall of a sphere in a viscous fluid is directly related to the sphere's velocity and the fluid's viscosity. The higher the viscosity of the fluid, the greater the retarding force on the sphere. This means, for instance, a small ball falling through honey (a viscous fluid) will experience more resistance than one falling through water. Mathematically, this is expressed as: F = 6πηav, where F is the viscous force, η is the fluid's viscosity, a is the radius of the sphere, and v is its velocity.

Examples & Analogies

Think about how easy it is to roll a marble on a flat surface versus rolling it in a pot of thick honey. The marble moves slowly in honey because the viscous force acting against it is significant compared to when it's on a flat surface with little resistance.

Stokes' Law Formula

Chapter 3 of 5

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Chapter Content

Sir George G. Stokes (1819–1903), an English scientist enunciated clearly the viscous drag force F as 6F = avηπ.

Detailed Explanation

Stokes formulated a mathematical equation to describe this resistance force, known as Stokes’ Law. The equation shows that the force due to viscosity increases with the radius of the sphere and the speed at which it is moving through the fluid. This establishes a clear relationship between the size of an object moving through a fluid, its speed, and how sticky or thick the fluid is.

Examples & Analogies

Consider different sizes of balls dropping through a liquid, such as oil. A large ball would experience a greater viscous drag than a smaller one at the same speed due to its larger surface area interacting with the fluid. This principle helps in designing objects like submarines that minimize drag to move efficiently through water.

Terminal Velocity

Chapter 4 of 5

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Chapter Content

The sphere (raindrop) then descends with a constant velocity. Thus, in equilibrium, this terminal velocity vt is given by 6πηavt = (4π/3) a3 (ρ-σ)g where ρ and σ are mass densities of sphere and the fluid, respectively.

Detailed Explanation

As the raindrop falls, it accelerates until the upward forces (viscous drag and buoyancy) balance the downward gravitational force. At this point, the raindrop stops accelerating and descends at a constant speed, known as terminal velocity. The equation also shows that larger drops will fall faster due to the factor of a² in the equation, meaning their size greatly influences how fast they can fall through the fluid.

Examples & Analogies

When you drop a small ball and a large ball simultaneously in a viscous fluid, like oil, you'll notice that the larger ball reaches the bottom first. This is because it has a different terminal velocity, dictated by its size, allowing it to overcome the resistance of the fluid more effectively.

Dependence of Terminal Velocity

Chapter 5 of 5

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Chapter Content

Thus, the terminal velocity vt depends on the square of the radius of the sphere and inversely on the viscosity of the medium.

Detailed Explanation

The relationship found in Stokes’ Law indicates that not only does the size of the sphere affect how fast it can fall through a fluid, but also, the thickness of the fluid matters. A more viscous fluid will slow down the descent of the raindrop or object significantly, leading to a lower terminal velocity. Conversely, a larger ball in a less viscous fluid will sink faster.

Examples & Analogies

Think about how easy or difficult it is to run through water versus running through a thick mud. In thick mud (high viscosity), you slow down considerably, while in water you can move more freely. This illustrates how viscosity directly impacts movement speed and terminal velocity in fluids.

Key Concepts

  • Viscous Drag Force: The force opposing the motion of a body in a fluid.

  • Terminal Velocity: The constant speed when forces are balanced.

  • Viscosity: A fluid's resistance to flow, influencing how bodies fall.

  • Stokes' Law: A formula relating viscous drag force to velocity, viscosity, and radius.

Examples & Applications

A raindrop reaching a constant speed while falling through the air due to viscous drag.

A copper ball falling through oil and reaching a steady velocity, illustrating Stokes' Law.

Memory Aids

Interactive tools to help you remember key concepts

🎵

Rhymes

In the goo or in the oil, big drops fall slower, less recoil.

📖

Stories

Imagine a tiny raindrop trying to escape a thick mud puddle. As it falls, it finds itself slowed by the sticky, viscous mud, allowing it to reach a constant speed - a challenge every raindrop faces!

🧠

Memory Tools

Remember 'V-FED' for Viscosity, Force, Effect, Drag to understand the drag force relationship.

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Acronyms

Use 'TGV' for Terminal = Gravity - Viscosity (where Gravity pulls down and Viscosity pulls up).

Flash Cards

Glossary

Viscous Drag Force

The resistive force experienced by an object moving through a fluid, proportional to its velocity.

Terminal Velocity

The constant speed achieved by an object when the force of gravity is balanced by the resistive forces.

Viscosity (η)

A measure of a fluid's resistance to flow and deformation.

Radius (a)

The distance from the center to the edge of a sphere.

Density (ρ, σ)

The mass per unit volume of a substance; ρ denotes sphere density, σ denotes fluid density.

Reference links

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