Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

[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

Short Answer

Expert verified
Set up the Lagrangian, derive the motion equation, find equilibrium, then solve numerically.

Step by step solution

01

Analyze the System

The system is composed of a massless wheel with radius \( R \) and two point masses: \( M \) at the edge of the wheel, and \( m \) hanging from a string wrapped around the wheel, both subject to gravity \( g \). Initially, \( M \) is directly below the axis.
02

Define the Lagrangian

The kinetic energy \( T \) is due to the mass \( m \) falling and the rotation of the mass \( M \) around the wheel. The potential energy \( U \) arises from the heights of both masses. The Lagrangian is given by \( \mathcal{L} = T - U \).- Kinetic energy: \( T = \frac{1}{2}MR^2\dot{\phi}^2 + \frac{1}{2}m(R\dot{\phi})^2 = \frac{1}{2}(M+m)R^2\dot{\phi}^2 \).- Potential energy: \( U = MgR(1 - \cos\phi) + mgR\phi \).So, \( \mathcal{L} = \frac{1}{2}(M+m)R^2\dot{\phi}^2 - \left( MgR(1 - \cos\phi) + mgR\phi \right) \).
03

Find the Equation of Motion

Using the Euler-Lagrange equation \( \frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{\phi}}) - \frac{\partial \mathcal{L}}{\partial \phi} = 0 \), we derive the equation of motion.Compute \( \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = (M+m)R^2\dot{\phi} \).Then, \( \frac{d}{dt}(\frac{\partial \mathcal{L}}{\partial \dot{\phi}}) = (M+m)R^2\ddot{\phi} \).Compute \( \frac{\partial \mathcal{L}}{\partial \phi} = MgR\sin\phi - mgR \).Substitute into Euler-Lagrange: \[ (M+m)R^2\ddot{\phi} = MgR\sin\phi - mgR. \]Simplifying gives:\[ \ddot{\phi} = \frac{g}{R} \left( \frac{M\sin\phi - m}{M+m} \right). \]
04

Determine Equilibrium Positions

For equilibrium, \( \ddot{\phi} = 0 \), implying \( M\sin\phi = m \). Therefore, equilibrium positions occur at \( \phi = \sin^{-1}(\frac{m}{M}) \). For stability, require \( \phi \) to be a minimum or maximum. Checking second derivative, stable equilibrium occurs if \( m < M \).
05

Sketch Potential Energy for Part (b) and Interpret

Sketch \( U(\phi) = MgR(1-\cos\phi) + mgR\phi \) for \( -\pi \leq \phi \leq 4\pi \). The potential energy has periodic minima and maxima which represent equilibrium points. When \( m < M \), there is one potential energy minimum corresponding to \( \phi = \sin^{-1}(\frac{m}{M}) \).
06

Predict and Solve Motion Numerically for Part (c)

Set \( M = g = R = 1 \) and \( m = 0.7 \). Plot \( U(\phi) \) against \( \phi \). From rest, initially at \( \phi = 0 \), expect periodic oscillation due to the potential minima assuming small oscillations around this position. Verify this prediction by solving the derived equation of motion numerically using a suitable method (e.g., Runge-Kutta) for \( 0 \leq t \leq 20 \).
07

Repeat Calculation for Part (d) with Modified Parameters

Repeat numerical simulations with \( m = 0.8 \). The shift in mass will alter the equilibrium positions, slightly increasing potential oscillation around the new stable position.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Stable Equilibrium
In Lagrangian mechanics, equilibrium positions represent states where the system experiences no net force, meaning it is at rest. An equilibrium is stable if, when slightly perturbed, the system tends to return to the equilibrium position. This happens because the potential energy is at a minimum.

For a massless wheel setup, as described in the exercise, the equilibrium points are found by setting the angular acceleration to zero, i.e., where \( M \sin\phi = m \). Hence, the position \( \phi = \sin^{-1}(\frac{m}{M}) \) is an equilibrium point.

When \( m < M \), the effect of gravity pulls the mass \( M \) more strongly downward, creating a restoring force.
  • This results in potential energy being minimized at these equilibrium points, indicating stability.
Any deviation from this point results in the wheel naturally trying to return to this equilibrium state, making it a stable equilibrium.
Potential Energy
Potential energy in this context arises from the gravitational forces acting on the masses attached to the wheel. It defines the energy storage due in part to their positions relative to a datum point, often taken as the lowest point in the gravitational field.

For the wheel system, the potential energy \( U(\phi) \) changes with the angle \( \phi \). It is calculated as:
  • \( U = MgR(1-\cos\phi) + mgR\phi \)
The term \( MgR(1-\cos\phi) \) comes from the mass \( M \) moving with the wheel, while \( mgR\phi \) accounts for the vertical motion of mass \( m \).

Potential energy graphs can provide insights into motion. Minima on such graphs indicate stable equilibrium positions. The graph undulates with maxima and minima, representing various positions of high and low potential energy. When plotted for \( -\pi \leq \phi \leq 4\pi \), the periodic nature of the graph clearly shows the recurring points of stable equilibrium.”
Equation of Motion
The equation of motion arises from the Euler-Lagrange equation, which employs the Lagrangian to derive the dynamic behavior of the system.

The relevant equation for the wheel system:
  • \[ \ddot{\phi} = \frac{g}{R} \left( \frac{M\sin\phi - m}{M+m} \right) \]
This second-order differential equation links the angular acceleration \( \ddot{\phi} \) to the system's parameters.

When \( M > m \), oscillatory behavior emerges due to this equation's nature. As it cannot be straightforwardly solved using elementary functions:
  • Numerical methods, like Runge-Kutta, are employed to simulate and analyze the actual motion.
By tuning the mass values \( M \) and \( m \), one predicts how any disturbance from the equilibrium will evolve over time. Thus, numerical solutions verify predictions about the wheel's oscillations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Using the usual angle \(\phi\) as generalized coordinate, write down the Lagrangian for a simple pendulum of length \(l\) suspended from the ceiling of an elevator that is accelerating upward with constant acceleration \(a\). (Be careful when writing \(T\); it is probably safest to write the bob's velocity in component form.) Find the Lagrange equation of motion and show that it is the same as that for a normal, nonaccelerating pendulum, except that \(g\) has been replaced by \(g+a\). In particular, the angular frequency of small oscillations is \(\sqrt{(g+a) / l}\).

Consider a mass \(m\) moving in a frictionless plane that slopes at an angle \(\alpha\) with the horizontal. Write down the Lagrangian in terms of coordinates \(x,\) measured horizontally across the slope, and \(y\), measured down the slope. (Treat the system as two-dimensional, but include the gravitational potential energy.) Find the two Lagrange equations and show that they are what you should have expected.

Write down the Lagrangian for a cylinder (mass \(m\), radius \(R\), and moment of inertia \(I\) ) that rolls without slipping straight down an inclined plane which is at an angle \(\alpha\) from the horizontal. Use as your generalized coordinate the cylinder's distance \(x\) measured down the plane from its starting point. Write down the Lagrange equation and solve it for the cylinder's acceleration \(\ddot{x}\). Remember that \(T=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2},\) where \(v\) is the velocity of the center of mass and \(\omega\) is the angular velocity.

Lagrange's equations in the form discussed in this chapter hold only if the forces (at least the nonconstraint forces) are derivable from a potential energy. To get an idea how they can be modified to include forces like friction, consider the following: A single particle in one dimension is subject to various conservative forces (net conservative force \(=F=-\partial U / \partial x)\) and a nonconservative force (let's call it \(F_{\text {fric }}\) ). Define the Lagrangian as \(\mathcal{L}=T-U\) and show that the appropriate modification is $$\frac{\partial \mathcal{L}}{\partial x}+F_{\mathrm{fric}}=\frac{d}{d t} \frac{\partial \mathcal{L}}{\partial \dot{x}}.$$

Two equal masses, \(m_{1}=m_{2}=m,\) are joined by a massless string of length \(L\) that passes through a hole in a frictionless horizontal table. The first mass slides on the table while the second hangs below the table and moves up and down in a vertical line. (a) Assuming the string remains taut, write down the Lagrangian for the system in terms of the polar coordinates \((r, \phi)\) of the mass on the table. (b) Find the two Lagrange equations and interpret the \(\phi\) equation in terms of the angular momentum \(\ell\) of the first mass. (c) Express \(\dot{\phi}\) in terms of \(\ell\) and eliminate \(\dot{\phi}\) from the \(r\) equation. Now use the \(r\) equation to find the value \(r=r_{0}\) at which the first mass can move in a circular path. Interpret your answer in Newtonian terms. (d) Suppose the first mass is moving in this circular path and is given a small radial nudge. Write \(r(t)=r_{0}+\epsilon(t)\) and rewrite the \(r\) equation in terms of \(\epsilon(t)\) dropping all powers of \(\epsilon(t)\) higher than linear. Show that the circular path is stable and that \(r(t)\) oscillates sinusoidally about \(r_{\mathrm{o}}\) What is the frequency of its oscillations?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free