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An object moving in a liquid experiences a linear drag force: \(\vec{D}=(b v, \text { direction opposite the motion), where } b\) is a constant called the \(d r a g\) coefficient. For a sphere of radius \(R,\) the drag constant can be computed as \(b=6 \pi \eta R,\) where \(\eta\) is the viscosity of the liquid. a. Find an algebraic expression for the terminal speed \(v_{\mathrm{tem}}\) of a spherical particle of radius \(R\) and mass \(m\) falling through a liquid of viscosity \(\eta\) b. Water at \(20^{\circ} \mathrm{C}\) has viscosity \(\eta=1.0 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} .\) Sand grains have density \(2400 \mathrm{kg} / \mathrm{m}^{3} .\) Suppose a 1.0 -mm-diameter sand grain is dropped into a 50 -m-decp lake whose water is a constant \(20^{\circ} \mathrm{C}\). If the sand grain reaches terminal speed almost instantly (a quite good approximation), how long will it take the sand grain to settle to the bottom of the lake?

Short Answer

Expert verified
The terminal speed of the particle is about 0.17 m/s, and it will take approximately 294 seconds for the sand grain to settle to the bottom of the lake.

Step by step solution

01

Derive the expression for terminal speed

The net force on the sphere when it reaches terminal speed is zero. This means the gravitational force downwards is equal to the drag force upwards. Therefore, we have \(mg = bv_{tem}\). Given \(b = 6\pi \eta R\), we substitute it into the previous equation to get \(mg = 6\pi \eta R v_{tem}\). Solving for \(v_{tem}\) we get \(v_{tem} = \frac{mg}{6\pi \eta R}\)
02

Apply the equation to given data

Next, for a spherical particle, the mass \(m\) can be expressed as \(\frac{4}{3}\pi R^3 \rho\) where \(\rho\) is the density of the particle. Substituting \(m\) into the equation for \(v_{tem}\), we have \(v_{tem} = \frac{4R^2 \rho g}{9\eta}\). Now, we can plug in the given values and calculate \(v_{tem}\). The radius \(R = 0.5\) mm, density \(\rho= 2400\) kg/m^3, gravitational acceleration \(g = 9.81\) m/s^2 and viscosity \(\eta =1.0 \times 10^{-3}\) Ns/m^2. After doing the math, we get \(v_{tem} \approx 0.17\) m/s.
03

Determine the time it takes for the sand grain to settle

Once we know the terminal speed, we can calculate the time it takes for the sand grain to settle to the bottom of the lake. We use the formula \(t = \frac{d}{v_{tem}}\), where \(d = 50\) m is the depth of the lake. Substituting the known values into the formula, we get \(t \approx 294\) seconds.

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Key Concepts

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

Drag Force
When an object moves through a fluid, it experiences a resistance known as drag force. This force acts in the direction opposite to the object's motion and its magnitude is influenced by several factors such as the object’s shape, speed, and the fluid's properties.

For a spherical object, the drag force, \( \vec{D} \), can be mathematically described as \( \vec{D} = (bv, \) direction opposite to the motion), where \( b \) is the drag coefficient. It's crucial to understand that the drag force is directly proportional to the velocity of the object in the fluid, denoted here as \( v \).

In the case of a sphere moving through a liquid, the drag coefficient \( b \) has a special form: \( b = 6\pi\eta R \), where \( R \) is the radius of the sphere, and \( \eta \) is the viscosity of the liquid. The drag force plays a significant role in determining the terminal speed of an object, which is the constant speed reached when the drag force equals the force due to gravity.
Viscosity
Viscosity is a measure of a fluid's resistance to flow and deformation by external forces. It is often referred to as the 'thickness' or 'stickiness' of a fluid. The higher the viscosity, the greater the resistance to flow.

In scientific terms, viscosity is represented by the Greek letter \( \eta \). This property comes into play when calculating the drag force experienced by objects moving through a fluid. As the viscosity of the liquid increases, so does the drag force. For example, moving through honey (which has a high viscosity) would create a greater drag force on an object than if it were moving through water (which has a lower viscosity).

Understanding viscosity is crucial because it directly affects how an object's speed will change over time as it moves through a fluid. In the terminal speed example, we can see that \( \eta \) is used in the calculation of \( b \) the drag coefficient.
Spherical Particle Dynamics
The behavior of spherical particles in fluids is determined by several forces, primarily gravitational forces pulling them down and the drag force pushing against them as they move. When a spherical particle falls through a liquid, these two forces gradually balance each other, leading the particle to eventually move at a constant speed, known as its terminal speed.

To analyze the dynamics of a spherical particle, we apply Newton's second law, considering both the gravitational force \( (mg) \) and the drag force provided by the fluid resistance. The effective dynamic behavior of the particle can be expressed by the drag coefficient, taking into account attributes such as the particle's radius and the fluid's viscosity.

For spherical particles, the symmetry of their shape simplifies analysis and calculations. In particular, their uniform geometry makes it possible to derive a specific mathematical expression for the drag force, which depends on the particle's radius and the fluid's viscosity.
Settling Velocity
Settling velocity, also known as terminal velocity in the context of sinking objects, is the steady speed attained by an object as it falls through a viscous medium like a liquid, where the upward drag force equals the downward gravitational force.

For a spherical particle of radius \( R \) and mass \( m \) in a fluid with viscosity \( \eta \), we can use the derived expression \( v_{tem} = \frac{mg}{6\pi\eta R} \) to calculate this terminal speed. In the case of our example with the sand grain in water, once the terminal speed is found, we can determine the time it will take for the particle to settle at the bottom of a container—such as a lake—by dividing the depth of the lake by the settling velocity.

Terminal velocity is an important factor not just in physics, but also in various applications including environmental science and engineering, as it helps predict how long particles will remain suspended in a fluid or when they might settle.

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Most popular questions from this chapter

A horizontal rope is tied to a \(50 \mathrm{kg}\) box on frictionless ice. What is the tension in the rope if: a. The box is at rest? b. The box moves at a steady \(5.0 \mathrm{m} / \mathrm{s} ?\) c. The box has \(v_{x}=5.0 \mathrm{m} / \mathrm{s}\) and \(a_{x}=5.0 \mathrm{m} / \mathrm{s}^{2} ?\)

An object moving in a liquid experiences a linear drag force: \(\vec{D}=(b v, \text { direction opposite the motion }),\) where \(b\) is a constant called the drag coefficient. For a sphere of radius \(R\), the drag constant can be computed as \(b=6 \pi \eta R,\) where \(\eta\) is the viscosity of the liquid. a. Find an algebraic expression for \(v_{x}(t),\) the \(x\) -component of velocity as a function of time, for a spherical particle of radius \(R\) and mass \(m\) that is shot horizontally with initial speed \(v_{0}\) through a liquid of viscosity \(\eta\) b. Water at \(20^{\circ} \mathrm{C}\) has viscosity \(\eta=1.0 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} .\) Suppose a \(4.0-\mathrm{cm}\) -diameter, \(33 \mathrm{g}\) ball is shot horizontally into a tank of \(20^{\circ} \mathrm{C}\) water. How long will it take for the horizontal speed to decrease to \(50 \%\) of its initial value?

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You've entered a "slow ski race" where the winner is the skier who takes the longest time to go down a \(15^{\circ}\) slope without ever stopping. You need to choose the best wax to apply to your skis. Red wax has a coefficient of kinetic friction \(0.25,\) yellow is 0.20 green is \(0.15,\) and blue is \(0.10 .\) Having just finished taking physics, you realize that a wax too slippery will cause you to accelerate down the slope and lose the race. But a wax that's too sticky will cause you to stop and be disqualified. You know that a strong headwind will apply a \(50 \mathrm{N}\) horizontal force against you as you ski, and you know that your mass is 75 kg. Which wax do you choose?

A \(50 \mathrm{kg}\) box hangs from a rope. What is the tension in the rope if: a. The box is at rest? b. The box moves up at a steady \(5.0 \mathrm{m} / \mathrm{s} ?\) c. The box has \(v_{y}=5.0 \mathrm{m} / \mathrm{s}\) and is speeding up at \(5.0 \mathrm{m} / \mathrm{s}^{2} ?\) d. The box has \(v_{y}=5.0 \mathrm{m} / \mathrm{s}\) and is slowing down at \(5.0 \mathrm{m} / \mathrm{s}^{2} ?\)

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