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Consider a double Atwood machine constructed as follows: A mass 4 \(m\) is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of \(3 m\) at one end and \(m\) at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass \(4 m\) when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side.

Short Answer

Expert verified
The acceleration of mass 4m is \(\frac{g}{2}\). The top pulley rotates due to the coupled motion of the system.

Step by step solution

01

Identify Generalized Coordinates

Let's choose two generalized coordinates to describe the system. 1. Let \(x_1\) be the vertical position of the mass \(4m\) from a reference point (e.g., ground level).2. Let \(x_2\) be the vertical position of the mass \(3m\) from a different, secondary reference point.
02

Express Other Positions in Terms of Coordinates

Since the pulley is massless and frictionless, we can express the position of mass \(m\) in terms of \(x_2\). If the string is of length \(L\), then:\[x_{2m} = L - x_2\]This implies that the motion of mass \(m\) is directly tied to the mass \(3m\).
03

Set Up Kinetic and Potential Energy

The kinetic energy (T) of the system can be expressed as:\[ T = \frac{1}{2} (4m) \dot{x_1}^2 + \frac{1}{2} (3m) \dot{x_2}^2 + \frac{1}{2} m (L - x_2)^2 \]The potential energy (V) can be expressed by the gravitational potential energy:\[ V = 4mgx_1 + 3mgx_2 + mg(L-x_2) \]
04

Construct the Lagrangian

The Lagrangian \( \mathcal{L} \) is given by the kinetic energy minus the potential energy:\[ \mathcal{L} = T - V = \frac{1}{2} (4m) \dot{x_1}^2 + \frac{1}{2} (3m) \dot{x_2}^2 + \frac{1}{2} m (L - x_2)^2 - \left( 4mgx_1 + 3mgx_2 + mg(L-x_2) \right) \]
05

Apply Lagrange's Equations

Lagrange's equation is given by:\[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0 \]Apply these to each coordinate, \(x_1\) and \(x_2\) to get the equations of motion.For \(x_1\):\[ 8m\ddot{x_1} = 4mg \quad (1) \]For \(x_2\):\[ 6m\ddot{x_2} + m\ddot{x_2} = 3mg - mg \quad (2)\]
06

Derive the Accelerations

From (1), solving for \(\ddot{x_1}\):\[ \ddot{x_1} = \frac{g}{2} \]From (2), solving for \(\ddot{x_2}\):\[ \ddot{x_2} = \frac{2g}{7} \]
07

Explain Pulley Rotation

Although the top pulley carries equal weights (3m and m) on each side, acceleration occurs because the Lagrangian approach considers the multiple-pulley system as a whole. The motion of the pulleys ties the mass 4m and the 3m-m difference together, causing rotation as they strive to minimize energy.

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Key Concepts

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

Lagrangian mechanics
Lagrangian mechanics is a powerful method for analyzing dynamic systems using energy rather than force considerations. This approach involves constructing a Lagrangian function, denoted as \( \mathcal{L} \), which is formulated as the difference between the system's kinetic energy \( T \) and potential energy \( V \). Lagrangian mechanics streamlines the process of finding equations of motion, especially for complex systems like a double Atwood machine.

The double Atwood machine involves a system of masses and pulleys. Here, the Lagrangian is crucial to understanding how the system behaves when set into motion. The kinetic and potential energies are expressed in terms of generalized coordinates, which reflect the system's unique nature. It allows easier solving of problems involving multiple interconnected components, thanks to reducing the problem to differential equations via Lagrange's equations of motion.

In practice, finding the Lagrangian begins with acknowledging all forms of energy within the system: kinetic due to motion, and gravitational potential due to position. By expressing \( \mathcal{L} = T - V \), one encapsulates the system's dynamics. The principle of stationary action tells us that the path taken by the system will make this equation stationary, leading to easily derived equations of motion.
generalized coordinates
Generalized coordinates are a key concept in analyzing mechanics, providing a more flexible and efficient way to describe the dynamics of systems with constraints or multiple moving parts. These coordinates can simplify complex systems into a manageable form by highlighting essential variables.

In a double Atwood machine, the system has multiple masses and pulleys. Instead of tracking each mass's position directly, one can use generalized coordinates like \( x_1 \) and \( x_2 \) as chosen in the problem's solution. These coordinates describe the vertical movements of the masses without directly considering pulley complexities. This allows for the description of related movements through a smaller number of parameters.

Choosing appropriate generalized coordinates can greatly simplify setting up the Lagrangian. Typically, one may opt for position coordinates tied to specific system components. From these coordinates, all other necessary positions or velocities are derived. It's about finding coordinates that reduce the complexity of equations, focusing on pivotal movements or constraints, often resulting in more manageable equations of motion.
equations of motion
Equations of motion describe the physical laws determining the system's behavior over time. In Lagrangian mechanics, these equations are derived using the Euler-Lagrange equation, a fundamental equation that comes from the action principle: \[ \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q}_i} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0\]

For a double Atwood machine, this approach simplifies finding how different masses accelerate and how the system evolves. With the Lagrangian established, one calculates the partial derivatives with respect to each generalized coordinate \( q_i \) and their respective velocities \( \dot{q}_i \). These derivatives lead to differential equations that describe the path and behavior of the system.

In this particular problem, solving the equations of motion provided outputs for accelerations: \( \ddot{x_1} = \frac{g}{2} \) for mass \( 4m \) and \( \ddot{x_2} = \frac{2g}{7} \) for mass \( 3m \). Despite equal weights on a pulley, differing accelerations arise because the Lagrangian method takes into account the system's energy as a whole. This perspective allows for understanding intricate behaviors like pulley rotations, which can be counterintuitive when only forces are considered.

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Most popular questions from this chapter

Consider a particle of mass \(m\) and charge \(q\) moving in a uniform constant magnetic field \(\mathbf{B}\) in the \(z\) direction. (a) Prove that \(\mathbf{B}\) can be written as \(\mathbf{B}=\nabla \times \mathbf{A}\) with \(\mathbf{A}=\frac{1}{2} \mathbf{B} \times \mathbf{r} .\) Prove equivalently that in cylindrical polar coordinates, \(\mathbf{A}=\frac{1}{2} B \rho \hat{\phi}\). (b) Write the Lagrangian (7.103) in cylindrical polar coordinates and find the three corresponding Lagrange equations. (c) Describe in detail those solutions of the Lagrange equations in which \(\rho\) is a constant.

A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

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.

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 \(\sqrt{(g+a) / l}\).

(a) Write down the Lagrangian for a particle moving in three dimensions under the influence of a conservative central force with potential energy \(U(r),\) using spherical polar coordinates \((r, \theta, \phi)\). (b) Write down the three Lagrange equations and explain their significance in terms of radial acceleration, angular momentum, and so forth. (The \(\theta\) equation is the tricky one, since you will find it implies that the \(\phi\) component of \(\ell\) varies with time, which seems to contradict conservation of angular momentum. Remember, however, that \(\ell_{\phi}\) is the component of \(\ell\) in a variable direction.) (c) Suppose that initially the motion is in the equatorial plane (that is, \(\theta_{0}=\pi / 2\) and \(\dot{\theta}_{0}=0\) ). Describe the subsequent motion. (d) Suppose instead that the initial motion is along a line of longitude (that is, \(\dot{\phi}_{0}=0\) ). Describe the subsequent motion.

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