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Consider the well-known problem of a cart of mass \(m\) moving along the \(x\) axis attached to a spring (force constant \(k\) ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency \(\omega=\sqrt{k / m} .\) Using the Lagrangian approach, you can find the effect of the spring's mass \(M,\) as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is \(\frac{1}{6} M \dot{x}^{2} .\) (As usual \(x\) is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still \(\frac{1}{2} k x^{2}\).) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency \(\omega=\sqrt{k /(m+M / 3)} ;\) that is, the effect of the spring's mass \(M\) is just to add \(M / 3\) to the mass of the cart.

Short Answer

Expert verified
The spring's mass adds \( \frac{M}{3} \) to the cart's effective mass.

Step by step solution

01

Calculate Kinetic Energy of the Spring

To find the kinetic energy of the spring, assume that it is uniform and stretches uniformly when the cart moves. When the end of the spring connected to the cart moves with velocity \( \dot{x} \), the velocity at a distance \( x' \) from the fixed end of the spring is \( \frac{x'}{x} \dot{x} \). The kinetic energy of an infinitesimal piece of the spring is then \( \frac{1}{2} \left( dx' \right) M \left( \frac{x'}{L} \right) \dot{x}^{2} \), where \( L \) is the equilibrium length of the spring. Integrate this expression from \( 0 \) to \( L \) to obtain the total kinetic energy of the spring: \( \int_{0}^{L} \frac{1}{2} \left( dx' \right) M \left( \frac{x'}{L} \right) \dot{x}^{2} = \frac{1}{6} M \dot{x}^{2} \).
02

Write the Lagrangian for the System

The Lagrangian \( \mathcal{L} \) for the cart and the spring is the difference between their kinetic and potential energies. The kinetic energy of the cart is \( \frac{1}{2} m \dot{x}^{2} \), and the kinetic energy of the spring, as determined in Step 1, is \( \frac{1}{6} M \dot{x}^{2} \). The potential energy is given as \( \frac{1}{2} k x^{2} \). Thus, the Lagrangian is: \( \mathcal{L} = \frac{1}{2} m \dot{x}^{2} + \frac{1}{6} M \dot{x}^{2} - \frac{1}{2} k x^{2} \).
03

Derive Equation of Motion Using Lagrange's Equation

Lagrange's equation is \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) - \frac{\partial \mathcal{L}}{\partial x} = 0 \). First, compute \( \frac{\partial \mathcal{L}}{\partial \dot{x}} = \left( m + \frac{1}{3} M \right) \dot{x} \). Then, \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) = \left( m + \frac{1}{3} M \right) \ddot{x} \). Next, \( \frac{\partial \mathcal{L}}{\partial x} = -kx \). Substitute these into Lagrange's equation: \( \left( m + \frac{1}{3} M \right) \ddot{x} + kx = 0 \).
04

Simplify the Equation of Motion

The equation from Step 3 represents a simple harmonic oscillator, \( \left( m + \frac{1}{3} M \right) \ddot{x} + kx = 0 \), where \( m + \frac{1}{3} M \) is the effective mass of the system. Comparing with the standard form \( m_{\text{eff}} \ddot{x} + kx = 0 \), we identify the effective mass \( m_{\text{eff}} = m + \frac{1}{3} M \). The angular frequency \( \omega \) is then given by \( \omega = \sqrt{\frac{k}{m_{\text{eff}}}} = \sqrt{\frac{k}{m + \frac{M}{3}}} \).

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Key Concepts

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

Simple Harmonic Motion
Simple Harmonic Motion (SHM) is a type of periodic motion where an object moves back and forth through an equilibrium position. In the context of a cart attached to a spring, the motion is controlled by the restoring force of the spring. This force is proportional to the displacement of the cart from its equilibrium position according to Hooke's Law, which is defined as \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement.
  • The system oscillates because the spring force continuously attempts to return the cart to its original position.
  • The motion occurs at an angular frequency \( \omega \), which provides insight into how rapidly the cart oscillates.
Using Lagrangian mechanics, we can determine how the mass of the spring affects the system. Without considering the spring's mass, the angular frequency is \( \omega = \sqrt{\frac{k}{m}} \). However, when the spring's mass \( M \) is considered, the system's behavior changes. It adjusts the effective mass in the system, altering the angular frequency to \( \omega = \sqrt{\frac{k}{m + \frac{M}{3}}} \). This demonstrates that the spring's mass indeed impacts the oscillatory behavior, adding \( \frac{M}{3} \) to the cart's mass.
Kinetic Energy
Kinetic Energy in mechanics refers to the energy that an object possesses due to its motion. For our system of a cart and a spring, both these components have kinetic energy as they move. The kinetic energy \( K \) is represented by the formula \( K = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity.
  • For the cart, its kinetic energy is straightforwardly computed as \( \frac{1}{2} m \dot{x}^2 \), where \( \dot{x} \) is its velocity.
  • The spring, being uniform, has parts that are moving at different speeds. Its kinetic energy is \( \frac{1}{6} M \dot{x}^{2} \) as derived from integration over its length.
This addition of a spring's kinetic energy shows how energy is distributed in a mechanical system. A weighted portion of the spring’s mass contributes to the system's overall kinetic energy, affecting the system's dynamic behavior.
Analytical Mechanics
Analytical Mechanics, particularly Lagrangian Mechanics, offers powerful tools for solving complex mechanical systems. It focuses on the total energy in systems rather than individual forces and allows for the simplification of the equations that govern motion. The Lagrange approach reformulates Newtonian mechanics, making it easier to handle constraints and conserve energy.
The Lagrangian \( \mathcal{L} \) is defined as the difference between kinetic energy \( T \) and potential energy \( V \):
  • \( \mathcal{L} = T - V \)
  • For a cart-spring setup, this becomes \( \mathcal{L} = \frac{1}{2} m \dot{x}^{2} + \frac{1}{6} M \dot{x}^{2} - \frac{1}{2} k x^{2} \).
Using the Lagrange equation \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) - \frac{\partial \mathcal{L}}{\partial x} = 0 \), we derive the equations of motion. This results in the familiar form of the simple harmonic oscillator equation with an adjusted effective mass: \( (m + \frac{1}{3} M) \ddot{x} + kx = 0 \). Through the lens of analytical mechanics, we can see the elegance and efficiency in solving for the dynamics of physical systems.

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Most popular questions from this chapter

Write down the Lagrangian for a one-dimensional particle moving along the \(x\) axis and subject to a force \(F=-k x\) (with \(k\) positive). Find the Lagrange equation of motion and solve it.

[Computer] Consider a massless wheel of radius \(R\) mounted on a frictionless horizontal axis. A point mass \(M\) is glued to the edge, and a massless string is wrapped several times around the perimeter and hangs vertically down with a mass \(m\) suspended from its bottom end. (See Figure 4.28.) Initially I am holding the wheel with \(M\) vertically below the axle. At \(t=0,\) I release the wheel, and \(m\) starts to fall vertically down. (a) Write down the Lagrangian \(\mathcal{L}=T-U\) as a function of the angle \(\phi\) through which the wheel has turned. Find the equation of motion and show that, provided \(m

The "spherical pendulum" is just a simple pendulum that is free to move in any sideways direction. (By contrast a "simple pendulum"- unqualified - is confined to a single vertical plane.) The bob of a spherical pendulum moves on a sphere, centered on the point of support with radius \(r=R\) the length of the pendulum. A convenient choice of coordinates is spherical polars, \(r, \theta, \phi,\) with the origin at the point of support and the polar axis pointing straight down. The two variables \(\theta\) and \(\phi\) make a good choice of generalized coordinates. (a) Find the Lagrangian and the two Lagrange equations. (b) Explain what the \(\phi\) equation tells us about the \(z\) component of angular momentum \(\ell_{z^{*}}\) (c) For the special case that \(\phi=\) const, describe what the \(\theta\) equation tells us. (d) Use the \(\phi\) equation to replace \(\dot{\phi}\) by \(\ell_{z}\) in the \(\theta\) equation and discuss the existence of an angle \(\theta_{\mathrm{o}}\) at which \(\theta\) can remain constant. Why is this motion called a conical pendulum? (e) Show that if \(\theta=\theta_{0}+\epsilon,\) with \(\epsilon\) small, then \(\theta\) oscillates about \(\theta_{\mathrm{o}}\) in harmonic motion. Describe the motion of the pendulum's bob.

A mass \(m\) is suspended from a massless string, the other end of which is wrapped several times around a horizontal cylinder of radius \(R\) and moment of inertia \(I\), which is free to rotate about a fixed horizontal axle. Using a suitable coordinate, set up the Lagrangian and the Lagrange equation of motion, and find the acceleration of the mass \(m\). [The kinetic energy of the rotating cylinder is \(\frac{1}{2} I \omega^{2} .\)]

Noether's theorem asserts a connection between invariance principles and conservation laws. In Section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of \(\mathcal{L}\) implies conservation of total angular momentum. Suppose that the Lagrangian of an \(N\) -particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the \(z\) axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}\right)\) to \(\left(r_{\alpha}, \theta_{\alpha}, \phi_{\alpha}+\epsilon\right)\) (same \(\epsilon\) for all particles). Hence show that $$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}}=0.$$ (b) Use Lagrange's equations to show that this implies that the total angular momentum \(L_{z}\) about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of \(\mathbf{L}\) are conserved.

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