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When an object moves through a fluid, the fluid exerts a viscous force \(\overrightarrow{\mathbf{F}}\) on the object that tends to slow it down. For a small sphere of radius \(R\) moving slowly with a speed \(v\) , the magnitude of the viscous force is given by Stokes' law, \(F=6 \pi \eta R v,\) where \(\eta\) is the viscosity of the fluid. (a) What is the viscous force on a sphere of radius \(R=5.0 \times 10^{-4} \mathrm{m}\) that is falling through water \(\left(\eta=1.00 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) when the sphere has a speed of 3.0 \(\mathrm{m} / \mathrm{s} ? \quad\) (b) The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed If the sphere has a mass of \(1.0 \times 10^{-5} \mathrm{kg}\) , what is its terminal speed?

Step by step solution

01

Understand Stokes' Law Formula

Stokes' law for the viscous force is given by \( F = 6 \pi \eta R v \). This formula calculates the viscous force on a small sphere moving through a fluid. We need to find the force for given values of radius \( R \), viscosity \( \eta \), and speed \( v \).

02

Calculate the Viscous Force

Substitute the given values into Stokes' formula: \( R = 5.0 \times 10^{-4} \text{ m} \), \( \eta = 1.00 \times 10^{-3} \text{ Pa} \cdot\text{s} \), and \( v = 3.0 \text{ m/s} \). The formula becomes:\[ F = 6 \pi (1.00 \times 10^{-3}) (5.0 \times 10^{-4}) (3.0) \]Calculate the answer.\[ F \approx 2.8 \times 10^{-5} \text{ N} \]

03

Understand Terminal Speed Condition

The terminal speed is reached when the viscous force \( F \) equals the gravitational force acting on the sphere \( W = mg \), where \( m \) is mass and \( g = 9.8 \text{ m/s}^2 \). The forces balance out when \( F = W \).

04

Calculate Weight of the Sphere

Calculate the weight of the sphere using the formula \( W = mg \):\[ W = (1.0 \times 10^{-5} \text{ kg}) \times 9.8 \text{ m/s}^2 = 9.8 \times 10^{-5} \text{ N} \]

05

Solve for Terminal Speed

Set \( F = W \) using Stokes' law: \[ 6 \pi \eta R v_t = mg \]Substitute known values:\[ 6 \pi (1.00 \times 10^{-3}) (5.0 \times 10^{-4}) v_t = 9.8 \times 10^{-5} \]Solve for \( v_t \):\[ v_t = \frac{9.8 \times 10^{-5}}{6 \pi (1.00 \times 10^{-3}) (5.0 \times 10^{-4})} \approx 10.4 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stokes' Law

When discussing fluid dynamics and the movement of objects in fluids, it's crucial to understand Stokes' Law. This law describes the viscous force experienced by an object as it moves through a fluid. For a sphere, specifically, Stokes' Law is expressed as: \[ F = 6 \pi \eta R v \]where:

• \( F \) is the viscous force,
• \( \eta \) represents the fluid's viscosity,
• \( R \) is the radius of the sphere,
• \( v \) is the velocity of the sphere.

The law essentially quantifies how the fluid resists the movement, which is essential for predicting how different objects behave under various fluid conditions. This resistance is due to the internal friction between the fluid layers as they slide past one another.

Terminal Speed

Terminal speed refers to the constant velocity that an object reaches as it moves through a fluid, where the forces acting on it are balanced. When a small sphere falls through a fluid like water, it initially accelerates due to gravity. However, as its speed increases, so does the viscous force resisting its fall.At a certain speed, called the terminal speed, the gravitational force pulling the sphere down is exactly balanced by the viscous force pushing it up. At this point, the sphere no longer accelerates and continues to fall at this steady speed.The equation to determine the terminal speed \( v_t \) is derived from:\[ 6 \pi \eta R v_t = mg \]where \( m \) is the mass of the sphere and \( g \) is the acceleration due to gravity. This balanced force condition ensures the sphere falls at a uniform rate, highlighting the interaction between gravitational and viscous forces.

Fluid Dynamics

Fluid dynamics is the branch of physics that studies the behavior of fluids (liquids and gases) in motion. It's pivotal in understanding phenomena ranging from weather patterns to hydraulics, and indeed, the movement of objects experiencing viscous forces. In the context of Stokes' Law and terminal speed, fluid dynamics helps us understand:

• The interaction between the fluid molecules and solid surfaces as frictional forces,
• How different shapes and sizes of objects influence the flow pattern and forces experienced,
• How velocity affects the forces; faster velocities generally experience higher viscous resistance.

Fluid dynamics integrates with our understanding of forces to predict how fluid properties like viscosity impact motion, making it indispensable for engineering, meteorology, and even in the design of sports equipment.

Viscosity

Viscosity is a measure of a fluid's resistance to deformation or flow. Imagine how honey flows more slowly compared to water; this is a direct result of higher viscosity in honey. Viscosity is quantified by the internal friction between fluid layers that occurs when they move relative to each other. In chemical or mechanical systems, it's important because:

• High viscosity implies greater resistance to flow, affecting the rate of an object’s motion through the fluid,
• It influences energy dissipation in systems, with more viscous fluids dissipating energy quickly,
• Affects heat transfer properties in fluids, crucial for designing cooling or heating systems.

The unit of viscosity in the SI system is the Pascal-second (Pa·s), representing how forcefully a fluid layer drags another. In problems involving Stokes' Law, viscosity is a core variable that impacts how forces are calculated and understood.