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A thin uniform rod of length \(2 L\) and mass \(M\) is suspended from a massless string of length \(l\) tied to a nail. Initially the rod hangs vertically. A weak horizontal force \(F\) is applied to the rod's free end. (a) Write the Lagrangian for this system. (b) For very short times such that all angles are small, determine the angles that string and the rod make with the vertical. Start from rest at \(t=0 .\) (c) Draw a diagram to illustrate the initial motion of the rod.

Short Answer

Expert verified
The angles are \( \theta(t) = \frac{F}{Mgl}t^2 \) and \( \phi(t) = \frac{3F}{3MgL}t^2 \).

Step by step solution

01

Understand the Problem Setup

We have a thin uniform rod of mass \( M \) and length \( 2L \) suspended by a string of length \( l \). A force \( F \) is applied horizontally at the rod’s free end. We need a Lagrangian for the system to determine small angle motions.
02

Define Coordinates and Variables

Choose the angle \( \theta \) for the string and \( \phi \) for the rod with respect to the vertical. Assume both angles are initially small. The horizontal force \( F \) on the rod's end causes small displacements, allowing small-angle approximations.
03

Write the Lagrangian for the System

The kinetic energy \( T \) of the system involves the rotational kinetic energies of the rod and the point mass at the string's end. Using the standard rotational kinetic energy formula, we have:\[T = \frac{1}{2} \left(I_{string} \dot{\theta}^2 + I_{rod} \dot{\phi}^2 \right)\]For small angles, the potential energy \( V \) is given by potential due to heights which is:\[ V = Mgl(1 - \cos \theta) + \frac{Mg(2L)}{2}(1 - \cos \phi) \approx \frac{1}{2}Mg(l\theta^2 + L\phi^2) \]Thus, the Lagrangian \( L \) is:\[ L = T - V \approx \frac{1}{2}(ml^2\dot{\theta}^2 + \frac{1}{3}M(2L)^2\dot{\phi}^2) - \frac{1}{2}Mg(l\theta^2 + L\phi^2) \]
04

Use Small Angle Approximation for Dynamics

Apply the Lagrange equations \( \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0 \) for each coordinate, assuming small angles \( \theta, \phi \). The initial conditions at \( t = 0 \) are \( \theta(0) = 0 \) and \( \phi(0) = 0 \).
05

Solve Equations for Small Angles

For small angles, solve the differential equations for both the string and rod:\[ ml^2\ddot{\theta} + Mgl\theta = F \text{ and } \frac{1}{3}M(2L)^2\ddot{\phi} + \frac{1}{2}MgL\phi = F.\]Solving these, we find that\[ \theta(t) = \frac{F}{Mgl}t^2\] and \[ \phi(t) = \frac{3F}{3MgL}t^2.\]
06

Illustrate Initial Motion

Draw a diagram showing:1. The string and rod both hanging vertically initially at \( t = 0 \).2. The forces acting: gravity downward, \( F \) horizontally at the rod's free end.3. Indicate small angles \( \theta \) and \( \phi \) being formed as the force \( F \) acts.

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Key Concepts

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

Small Angle Approximation
When dealing with pendulum-like systems, the small angle approximation is a vital tool. It simplifies trigonometric functions like sine and cosine when the angles involved are tiny, usually less than about 10 degrees. By relying on this approximation, complex equations become much more manageable.
For small angles, expressed in radians:
  • We can approximate \ \ \( \sin \theta \approx \theta \).
  • Similarly, \ \( \cos \theta \approx 1 - \frac{\theta^2}{2} \).
This simplification is crucial in mechanics because it allows us to linearize equations that would otherwise involve higher-order terms, which are more difficult to solve.
In the given problem, both the angle of the rod (\( \phi \)) and the string (\( \theta \)) with the vertical are considered small. With these approximations, the resulting Lagrangian equations become linear, making them solvable analytically. This approximation is applied to both kinetic and potential energy expressions to streamline the calculations.
Rotational Kinetic Energy
Rotational kinetic energy is an essential part of analyzing the motion of rotating bodies. For a rotating object, it is given by the formula:\[ T = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
For the rod in this exercise, we consider each component's contribution to the system's kinetic energy:
  • The rod itself has a rotational kinetic energy about its center: \[ \frac{1}{2} I_{rod} \dot{\phi}^2 \]
  • For the string, although massless, it influences the kinetic energy through its length and the mass suspended from it: \[ \frac{1}{2} I_{string} \dot{\theta}^2 \]
The moment of inertia for the rod suspended from a point other than its center is adjusted using the parallel axis theorem. This provides the correct inertia values and ensures that the angular velocities \( \dot{\phi} \) and \( \dot{\theta} \) contribute correctly to the kinetic energy. Understanding these contributions allows us to calculate the Lagrangian and analyze how this energy drives the motion of the rod and string when disturbed.
Lagrangian Equations
Lagrangian mechanics offers a powerful way to derive the equations of motion for a dynamic system. It's based on the principle of least action, using the difference between kinetic energy \( T \) and potential energy \( V \), both encapsulated in the Lagrangian function \( L \):
\[ L = T - V \]
In this exercise, once the Lagrangian is constructed, the next step is to apply the Lagrange equations. These equations are:
  • \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0 \]
The variables \( q \) correspond to the generalized coordinates — angles \( \theta \) and \( \phi \) in this case. These coordinates describe the configuration of the system at any given time. Solving these equations lets us find how these angles evolve over time.
In particular, with the assumptions of small angles, the equations simplify considerably into a system of linear differential equations. Solving these gives the time evolution of \( \theta(t) \) and \( \phi(t) \), showcasing the effectiveness of Lagrangian mechanics in handling dynamics efficiently.

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Most popular questions from this chapter

The classical mechanies exam induces Jacob to try his hand at bungee jumping. Assume Jacob's mass \(m\) is suspended in a gravitational field by the bungee of unstretched length \(b\) and spring constant \(k\). Besides the longitudinal oscillations due to the bungee jump, Jacob also swings with plane pendulum motion in a vertical plane. Use polar coordinates \(r, \phi\), neglect air drag, and assume that the bungee always is under tension. (a) Derive the Lagrangian (b) Determine Lagrange's equation of motion for angular motion and identify by name the forces contributing to the angular motion. (c) Determine Lagrange's equation of motion for radial oscillation and identify by name the forces contributing to the tension in the spring. (d) Derive the generalized momenta (e) Determine the Hamiltonian and give all of Hamilton's equations of motion.

A block of mass \(m\) rests on an inclined plane making an angle \(\theta\) with the horizontal. The inclined plane (a triangular block of mass \(M\) ) is free to slide horizontally without friction. The block of mass \(m\) is also free to slide on the larger block of mass \(M\) without friction. (a) Construct the Lagrangian function. (b) Derive the equations of motion for this system. (c) Calculate the canonical momenta. (d) Construct the Hamiltonian function. (e) Find which of the two momenta found in part (c) is a constant of motion and discuss why it is so. If the two blocks start from rest, what is the value of this constant of motion?

A uniform ladder of mass \(M\) and length \(2 L\) is leaning against a frictionless vertical wall with its feet on a frictionless horizontal floor, Initially the stationary ladder is released at an angle \(\theta_{0}=60^{\circ}\) to the floor. Assume that gravitation field \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) acts vertically downward and that the moment of inertia of the ladder about its midpoint is \(I=\frac{1}{3} M L^{2}\). (a) What is the number of generalized coordinates necessary to describe the problem? (b) Derive the Lagrangian (c) Derive the Hamiltonian (d) Explain if the Hamiltonian is conserved and/or if it equals the total energy (e) Use the Lagrangian to derive the equations of motion (f) Derive the angle \(\theta\) at which the ladder loses contact with the vertical wall?

It can be shown that if \(L(q, \dot{q}, t)\) is the Lagrangian of a particle moving in one dimension, then \(L=L^{\prime}\) where \(L^{\prime}(q, \dot{q}, t)=L(q, \dot{q}, t)+\frac{d f}{d t}\) and \(f(q, t)\) is an arbitrary function. This problem explores the consequences of this on the Hamiltonian formalism. (a) Relate the new canonical momentum \(p^{\prime}\), for \(L^{\prime}\), to the old canonical momentum \(p\), for \(L\). (b) Express the new Hamiltonian \(H^{\prime}\left(q^{\prime}, p^{\prime}, t\right)\) for \(L^{\prime}\) in terms of the old Hamiltonian \(H(q, p, t)\) and \(f\). (c) Explicitly show that the new Hamilton's equations for \(H^{\prime}\) are equivalent to the old Hamilton's equations for \(H\).

A rigid straight, frictionless, massless, rod rotates about the \(z\) axis at an angular velocity \(\dot{\theta}\), A mass \(m\) slides along the frictionless rod and is attached to the rod by a massless spring of spring constant \(\kappa\). (a) Derive the Lagrangian and the Hamiltonian (b) Derive the equations of motion in the stationary frame using Hamiltonian mechanics. (c) What are the constants of motion? (d) If the rotation is constrained to have a constant angular velocity \(\dot{\theta}=\omega\) then is the non-cyclic Routhian \(R_{\text {noncyclic }}=H-p_{\theta} \dot{\theta}\) a constant of motion, and does it equal the total energy? (e) Use the non-cyclic Routhian \(R_{\text {noneyclie to derive the radial equation of motion in the rotating frame of }}\) reference for the cranked system with \(\dot{\theta}=\omega\).

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