/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 =A rigid rod of length \(2 a\) h... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 8

=A rigid rod of length \(2 a\) has its lower end in contact with a smooth horizontal floor. Initially the rod is at an angle \(\alpha\) to the upward vertical when it is released from rest. The subsequent motion takes place in a vertical plane. Take as generalised coordinates \(x\), the horizontal displacement of the centre of the rod, and \(\theta\), the angle between the rod and the upward vertical. Obtain Lagrange's equations. Show that \(x\) remains constant in the motion and verify that the \(\theta\)-equation is equivalent to the energy conservation equation.

Short Answer

Expert verified
The subsequent motion of the rod follows Lagrange's equations with \(x\) constant and \(\theta\) following \(\ddot{\theta}+ \frac{g}{a} \sin \theta = 0\). The conservation of energy is also confirmed.

Step by step solution

01

Set Up the Coordinate System

Choose a convenient coordinate system. Here, \(x\) represents the horizontal displacement of the rod center, and \(\theta\) is the angle made by the rod, initially assumed to be \(\alpha\), with the upward vertical.
02

Find the Kinetic and Potential Energies

Find the kinetic (\(T\)) and potential (\(V\)) energy of the system. The kinetic energy of the rod is given by \(T = \frac{1}{2}m \left( \dot{x}^2 + (a\dot{\theta})^2 \right)\), where \(\dot{x}\) and \(\dot{\theta}\) represent time derivatives. The potential energy is \(V = mga \cos\theta\), where \(\theta\) is the angle, \(m\) is the mass of the rod, \(g\) is the acceleration due to gravity, and \(a\) is half the length of the rod.
03

Write the Lagrangian

The Lagrangian \(L\) of the system is the difference of the kinetic and potential energy: \( L = T - V = \frac{1}{2}m \left( \dot{x}^2 + (a\dot{\theta})^2 \right) - mga \cos\theta \).
04

Apply Lagrange's Equations

Lagrange's equations of motion are obtained by applying Euler-Lagrange equation: \(\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i} = 0\), where \(q_i\) are the generalized coordinates \(x\) and \(\theta\). Applying this gives two equations: one for \(x\), which yields \(\ddot{x} = 0\), meaning \(x\) is constant (as per the problem statement). The second equation is for \(\theta\), which gives \(\ddot{\theta}+ \frac{g}{a} \sin \theta = 0\).
05

Verify the Energy Conservation

We can multiply the obtained \(\theta\)-equation with \(\dot{\theta}\) achieving \(\dot{\theta}\ddot{\theta}+ \frac{g}{a} \dot{\theta}\sin \theta = 0\). This equation is the derivative with respect to time of \(\frac{1}{2}\dot{\theta}^2 - \frac{g}{a}\cos\theta = C\), where \(C\) denotes constant of motion. Thus, the conservation of energy is verified.

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Key Concepts

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

Lagrange's equations
Lagrange's equations form a fundamental part of Lagrangian mechanics, allowing us to describe the dynamics of a mechanical system. It's a powerful method to solve motion problems using energy rather than force, focusing on three main components:
  • Kinetic Energy (T) - The energy due to the system鈥檚 motion.
  • Potential Energy (V) - The energy stored due to the system鈥檚 position or configuration.
  • Lagrangian (L) - Defined as the difference between kinetic and potential energy, i.e., \(L = T - V\).
In the exercise, Lagrange's equations are derived from the Euler-Lagrange equation:
\[\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0\]
This equation is used to express the equations of motion for generalized coordinates \(x\) and \(\theta\). By applying it, we confirm the motion consists of constant horizontal positioning (\(x\)) and evolving angle (\(\theta\)). This elegant approach simplifies solving problems that otherwise involve complex force analysis.
Energy Conservation
Energy conservation is a key concept that states total energy in a closed system remains constant over time. In mechanics, this involves balancing kinetic and potential energies.
During motion, energy fluctuations occur鈥攌inetic energy might convert into potential energy and vice versa鈥攂ut the total stays the same. This is the essence of conservation laws.
In the given problem, the energy conservation principle helps verify that the equation for \(\theta\) aligns with this concept. The equation, \(\ddot{\theta} + \frac{g}{a} \sin \theta = 0\), when multiplied by \(\dot{\theta}\) and integrated, proves the conservation principle holds for the rod's motion. Thus, ensuring no energy is mysteriously gained or lost, making it a cornerstone of Lagrangian mechanics.
Kinetic Energy
Kinetic energy refers to the energy associated with the motion of an object. It depends specifically on velocity, and is mathematically expressed as:
\[ T = \frac{1}{2} m v^2 \]
Here, \(m\) is the mass and \(v\) is the velocity of the object.
For the rod in the exercise, its kinetic energy is more complex due to rotational and translational motions. It is given by:
\[ T = \frac{1}{2} m (\dot{x}^2 + (a \dot{\theta})^2) \]
Here, \(\dot{x}\) describes the translational aspect while \(a \dot{\theta}\) accounts for rotational motion around the center of the rod. As you can see, both components play an essential role in the system's dynamics and influence how the total energy changes as the rod moves.
Potential Energy
Potential energy is the stored energy based on an object's position or configuration, and is central in analyzing mechanical systems. It often stems from gravitational, elastic, or electric forces.
In our exercise, potential energy derives from the gravitational pull acting on the rod, calculated as:
\[ V = m g a \cos \theta \]
This formula considers the vertical height of the rod鈥檚 center of mass, which changes with the angle \(\theta\). You can see that as \(\theta\) changes, \(\cos \theta\) changes, altering the potential energy.
This stored gravitational energy converts into kinetic energy as the rod moves, leading to a fascinating interplay visible in systems studied through Lagrangian mechanics. Understanding potential energy鈥檚 role in this transformation helps reinforce the conservation of energy concept.

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Most popular questions from this chapter

A uniform rectangular door of width \(2 a\) can swing freely on its hinges. The door is misaligned and the line of the hinges makes an angle \(\alpha\) with the upward vertical. Take the rotation angle of the door from its equilibrium position as generalised coordinate and obtain Lagrange's equation for the motion. Deduce the period of small oscillations of the door about the equilibrium position.

A uniform disk of mass \(M\) and radius \(a\) can roll along a rough horizontal rail. A particle of mass \(m\) is suspended from the centre \(C\) of the disk by a light inextensible string of length \(b\). The whole system moves in the vertical plane through the rail. Take as generalised coordinates \(x\), the horizontal displacement of \(C\), and \(\theta\), the angle between the string and the downward vertical. Obtain Lagrange's equations. Show that \(x\) is a cyclic coordinate and find the corresponding conserved momentum \(p_{x}\). Is \(p_{x}\) the horizontal linear momentum of the system? Given that \(\theta\) remains small in the motion, find the period of small oscillations of the particle.

A particle of mass \(m\) and charge \(e\) moves in the magnetic field produced by a current \(I\) flowing in an infinite straight wire that lies along the \(z\)-axis. The vector potential \(\boldsymbol{A}\) of the induced magnetic field is given by $$ A_{r}=A_{\theta}=0, \quad A_{z}=-\left(\frac{\mu_{0} I}{2 \pi}\right) \ln r $$ where \(r, \theta, z\) are cylindrical polar coordinates. Find the Lagrangian of the particle. Show that \(\theta\) and \(z\) are cyclic coordinates and find the corresponding conserved momenta.

A particle \(P\) is connected to one end of a light inextensible string which passes through a small hole \(O\) in a smooth horizontal table and extends below the table in a vertical straight line. \(P\) slides on the upper surface of the table while the string is pulled downwards from below in a prescribed manner. (Suppose that the length of the horizontal part of the string is \(R(t)\) at time \(t\).) Take \(\theta\), the angle between \(O P\) and some fixed reference line in the table, as generalised coordinate and obtain Lagrange's equation. Show that \(\theta\) is a cyclic coordinate and find (and identify) the corresponding conserved momentum \(p_{\theta}\). Why is the kinetic energy not conserved? If the constant value of \(p_{\theta}\) is \(m L\), find the tension in the string at time \(t\).

A particle \(P\) is attached to a support \(S\) by a light rigid rod of length \(a\), which is freely pivoted at \(S . P\) moves in a vertical plane through \(S\) and at the same time the support \(S\) is made to oscillate vertically having upward displacement \(Z=\epsilon a \cos p t\) at time \(t\). Take \(\theta\), the angle between \(S P\) and the upward vertical, as generalised coordinate and show that Lagrange's equation is $$ \ddot{\theta}-\left(\Omega^{2}+\epsilon p^{2} \cos p t\right) \sin \theta=0 $$ where \(\Omega^{2}=g / a\). The object is to show that, for suitable choices of the parameters, the pendulum is stable in the vertically upwards position! First write the equation in dimensionless form by introducing the dimensionless time \(\tau=\) \(p t\). Then \(\theta(\tau)\) satisfies $$ \frac{d^{2} \theta}{d \tau^{2}}-\left(\frac{\Omega^{2}}{p^{2}}+\epsilon \cos \tau\right) \sin \theta=0 $$ Solve this equation numerically with initial conditions in which the pendulum starts from rest near the upward vertical. Plot the solution \(\theta(\tau)\) as a function of \(\tau\) for about twenty oscillations of the support. Try \(\epsilon=0.3\) with increasing values of the parameter \(p / \Omega\) in the range \(1 \leq\) \(p / \Omega \leq 10\). You will know that the upside-down pendulum is stable when \(\theta\) remains small in the subsequent motion.
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