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A particle is released from rest \((y=0)\) and falls under the influence of gravity and air resistance. Find the relationship between \(v\) and the distance of falling \(y\) when the air resistance is equal to (a) \(\alpha v\) and (b) \(\beta v^{2}\).

Short Answer

Expert verified
When the air resistance is equal to \(\alpha v\), the relationship between velocity \(v\) and distance of falling \(y\) is given by \(y = (mg/\alpha)t + (m^2g/\alpha ^2)(e^{-\alpha t/m} - 1)\), where \(m\) is the mass of the particle, \(g\) is the acceleration due to gravity, and \(\alpha\) is the constant of proportionality. For the case where the air resistance is equal to \(\beta v^{2}\), the relationship between \(v\) and \(y\) cannot be expressed in elementary functions, but finding the distance of falling requires solving the integral \(dv/(g - (\beta/m) v^{2}) = dt\) numerically.

Step by step solution

01

Set up the differential equation

The force on the particle is \(mg - \alpha v\). Using the second law of Newton \(F = ma = m dv/dt\), we get \(m dv/dt = mg - \alpha v\) that can be rearranged as \(dv/dt = g - (\alpha/m) v\).
02

Solve the differential equation

To solve the differential equation, a separation of variables can be performed. This brings the equation into the form \(dv/(g - (\alpha/m) v) = dt\). Integrating both sides with the initial conditions of \(v=0\) at \(t=0\) gives \(v = (mg/\alpha)(1 - e^{-\alpha t/m})\).
03

Compute distance of falling

To find \(y\), we have to integrate \(v\) with respect to time. When integrating, we find the equation to be \(y = (mg/\alpha)t + (m^2g/\alpha ^2)(e^{-\alpha t/m} - 1)\). In this equation, \(y\) is the distance of falling and it's related to the velocity \(v\). (B) When the air resistance is equal to \(\beta v^{2}\):
04

Set up the differential equation

The force on the particle is \(mg - \beta v^{2}\). Using the second law of Newton, this gives us the following equation \(m dv/dt = mg - \beta v^{2}\) which can be rearranged as \(dv/dt = g - (\beta/m) v^{2}\).
05

Solve the differential equation

Here separation of variables leads to \(dv/(g - (\beta/m) v^{2}) = dt\). The integration of this differential equation is difficult to perform in a general form but can be numerically evaluated.
06

Compute distance of falling

To find \(y\), we would need to integrate \(v\) with respect to time. Unlike the previous case, the resulting integral cannot be expressed in elementary functions. Remember, to solve these kinds of problems, you'll need to have a good understanding of differential equations and their methods of solutions. These steps should provide a good starting point for solving other similar problems.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are a central tool in advancing our understanding of the physical world because they are able to model a variety of phenomena such as growth, decay, and change over time. When solving a falling body problem, we often deal with a type of differential equation where the rate of change of velocity with respect to time, or \( \frac{dv}{dt} \), is equal to gravity minus the opposing force of air resistance. This particular setup is a first-order linear differential equation and it's crucial for students to grasp this foundational concept to analyze motion under various forces.

To solve these equations, one needs to apply methods such as separation of variables or numerical techniques. Separation of variables involves rearranging the differential equation so that each variable and its derivatives are on separate sides of the equation, which then can be integrated to find a solution. Understanding the methodology behind solving these equations will enable students to tackle a broad range of physics problems beyond just the falling body scenario.
Air Resistance
Air resistance, also known as drag, is the force that opposes the motion of an object through the air. It is a non-negligible force that affects the speed and distance an object falls. In physics problems, air resistance can typically be proportional to either the velocity, \( \alpha v \), or the square of the velocity, \( \beta v^2 \). These two scenarios offer different complexities when setting up the differential equations.

Linear drag, proportional to the velocity, often leads to an equation that can be solved through analytical methods as described in the exercise's step-by-step solution. On the other hand, a scenario with quadratic drag, proportional to the square of velocity, requires advanced integration techniques or numerical methods because the relationship becomes much more complex. It's important for students to understand how each form of air resistance affects the motion of objects and which mathematical tools are needed to analyze each case.
Newton's Second Law
Newton's second law of motion is foundational to classical mechanics. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass, often expressed as \( F = ma \). This principle shines in motion equations, where it helps us understand and quantify how objects behave under various forces.

In the context of the falling body problem, Newton’s second law is applied to establish that the force due to gravity (minus the air resistance acting upward) is equal to the mass times the downward acceleration, which allows us to create the differential equation to capture the motion dynamics. By grasping this law, students can predict how variations in mass, force, and acceleration will affect the behavior of a falling object, making it a key concept for solving a wide array of physics problems. Mastering Newton's second law is thus not only about remembering an equation but about interpreting and applying it to real-world scenarios where forces and acceleration interact.

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Most popular questions from this chapter

A student drops a water-filled balloon from the roof of the tallest building in town trying to hit her roommate on the ground (who is too quick). The first student ducks back but hears the water splash 4.021 s after dropping the balloon. If the speed of sound is \(331 \mathrm{m} / \mathrm{s}\), find the height of the building, neglecting air resistance.

A child slides a block of mass \(2 \mathrm{kg}\) along a slick kitchen floor. If the initial speed is 4 \(\mathrm{m} / \mathrm{s}\) and the block hits a spring with spring constant \(6 \mathrm{N} / \mathrm{m},\) what is the maximum compression of the spring? What is the result if the block slides across \(2 \mathrm{m}\) of a rough floor that has \(\mu_{k}=0.2 ?\)

A potato of mass \(0.5 \mathrm{kg}\) moves under Earth's gravity with an air resistive force of \(-k m v .\) (a) Find the terminal velocity if the potato is released from rest and \(k=\) \(0.01 \mathrm{s}^{-1} .\) (b) Find the maximum height of the potato if it has the same value of \(k\) but it is initially shot directly upward with a student-made potato gun with an initial velocity of \(120 \mathrm{m} / \mathrm{s}\).

A boat with initial speed \(v_{0}\) is launched on a lake. The boat is slowed by the water by a force \(F=-\alpha e^{\beta v} .\) (a) Find an expression for the speed \(v(t) .\) (b) Find the time and (c) distance for the boat to stop.

The motion of a charged particle in an electromagnetic field can be obtained from the Lorentz equation* for the force on a particle in such a field. If the electric field vector is \(\mathbf{E}\) and the magnetic field vector is \(\mathbf{B}\), the force on a particle of mass \(m\) that carries a charge \(q\) and has a velocity \(\mathbf{v}\) is given by $$\mathbf{F}=q \mathbf{E}+q \mathbf{v} \times \mathbf{B}$$ where we assume that \(v \ll c\) (speed of light). (a) If there is no electric field and if the particle enters the magnetic field in a direction perpendicular to the lines of magnetic flux, show that the trajectory is a circle with radius $$r=\frac{m v}{q B}=\frac{v}{\omega_{c}}$$ where \(\omega_{c} \equiv q B / m\) is the cyclotron frequency. (b) Choose the \(z\) -axis to lie in the direction of \(\mathbf{B}\) and let the plane containing \(\mathbf{E}\) and B be the \(y z\) -plane. Thus $$\mathbf{B}=B \mathbf{k}, \quad \mathbf{E}=E_{y} \mathbf{j}+E_{z} \mathbf{k}$$ Show that the \(z\) component of the motion is given by $$z(t)=z_{0}+\dot{z}_{0} t+\frac{q E_{z}}{2 m} t^{2}$$ where $$z(0) \equiv z_{0} \quad \text { and } \quad \dot{z}(0) \equiv \dot{z}_{0}$$ (c) Continue the calculation and obtain expressions for \(\dot{x}(t)\) and \(\dot{y}(t) .\) Show that the time averages of these velocity components are $$\langle\dot{x}\rangle=\frac{E_{y}}{B}, \quad\langle\dot{y}\rangle=0$$ (Show that the motion is periodic and then average over one complete period.) (d) Integrate the velocity equations found in (c) and show (with the initial conditions \(\left.x(0)=-A / \omega_{c}, \dot{x}(0)=E_{y} / B, y(0)=0, \dot{y}(0)=A\right)\) that $$x(t)=\frac{-A}{\omega_{c}} \cos \omega_{c} t+\frac{E_{y}}{B} t, \quad y(t)=\frac{A}{\omega_{c}} \sin \omega_{c} t$$ These are the parametric equations of a trochoid. Sketch the projection of the trajectory on the \(x y-\) plane for the cases (i) \(A>\left|E_{y} / B\right|,\) (ii) \(A<\left|E_{y} / B\right|,\) and (iii) \(A=\left|E_{y} / B\right| .\) (The last case yields a cycloid.)

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