/*! 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 18 A mass \(m\) is suspended from a... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\frac{1}{2} I \omega^{2} .\)]

Short Answer

Expert verified
The acceleration of mass \(m\) is \(\ddot{y} = \frac{mg}{m + \frac{I}{R^2}}\).

Step by step solution

01

Define the Coordinates

Let's choose the vertical position of the mass, denoted by \( y \), as the generalized coordinate. The mass moves downward as it unwinds the string, so \( y \) increases as the mass descends.
02

Express Velocities in Terms of y

The velocity of the mass is \( v_m = \frac{dy}{dt} \). The angular velocity of the cylinder \( \omega \) is related to the linear velocity of the string by \( v_m = R\omega \), which gives us \( \omega = \frac{\dot{y}}{R} \).
03

Write the Lagrangian

The Lagrangian \( L \) is given by the kinetic energy minus the potential energy, \( L = T - V \). Kinetic energy \( T \) has two components: the translational kinetic energy of the mass \( \frac{1}{2} m \dot{y}^2 \) and the rotational kinetic energy of the cylinder \( \frac{1}{2} I \omega^2 \). Potential energy \( V = mgy \), where \( g \) is the acceleration due to gravity. Thus,\[ L = \frac{1}{2} m \dot{y}^2 + \frac{1}{2} I \left(\frac{\dot{y}}{R}\right)^2 - mgy. \]
04

Derive the Lagrange Equation of Motion

The Lagrange equation is given by:\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right) - \frac{\partial L}{\partial y} = 0. \]Calculate \( \frac{\partial L}{\partial \dot{y}} = m \dot{y} + \frac{I \dot{y}}{R^2} \) and \( \frac{\partial L}{\partial y} = -mg \). Differentiate \( \frac{\partial L}{\partial \dot{y}} \) with respect to time to get \( (m + \frac{I}{R^2}) \ddot{y} \). Substituting these into the Lagrange equation gives:\[ (m + \frac{I}{R^2}) \ddot{y} = mg \].
05

Solve for the Acceleration \(\ddot{y}\)

Rearrange the equation from Step 4 to solve for \( \ddot{y} \):\[ \ddot{y} = \frac{mg}{m + \frac{I}{R^2}}. \]

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Key Concepts

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

Lagrange Equation of Motion
The Lagrange Equation of Motion is a foundational principle in Lagrangian Mechanics, providing a method for finding the equations of motion for a system. It is particularly useful in systems involving complex constraints or where forces are not easily defined. The Lagrange equation is expressed as:\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0, \]where \(q\) represents the generalized coordinates, variables that describe the configuration of the system. In our exercise, the vertical position \(y\) of the mass acts as this generalized coordinate.
This equation involves derivatives of the Lagrangian \(L\), which is the difference between the kinetic energy \(T\) and the potential energy \(V\):
  • \(T\) accounts for both translational and rotational kinetic energy.
  • \(V\) represents the potential energy due to gravity.
By substituting the expressions for \(\frac{\partial L}{\partial \dot{y}}\) and \(\frac{\partial L}{\partial y}\) from the original solution, we apply the equation to derive the equation of motion of the system. This allows us to ultimately determine how the system behaves over time, finding the acceleration of the mass.
Kinetic Energy
Kinetic Energy is a crucial concept that differentiates the types of movements in a dynamic system. In our problem, it consists of two forms:
  • Translational kinetic energy, which is the kinetic energy due to the motion of the mass itself, given by the formula \( \frac{1}{2} m \dot{y}^2 \).
  • Rotational kinetic energy, referring to the energy of the rotating cylinder, expressed as \( \frac{1}{2} I \omega^2 \).
These energies add up to form the total kinetic energy of the system and are used in the Lagrangian to calculate the system's behavior.
In this calculation, it's crucial to note how both translational and rotational components are related. The angular velocity \( \omega \) of the cylinder is connected to the linear speed \( \dot{y} \) of the string and mass. Hence, \( \omega = \frac{\dot{y}}{R} \) is substituted into the kinetic energy expression to create a unified relationship in the Lagrangian. This link between translational and rotational movements helps us describe the entire system's dynamics comprehensively.
Rotational Dynamics
Rotational Dynamics deals with the motion of objects that rotate about an axis. In our exercise, the cylinder exhibits rotational dynamics as it spins due to the string unwinding.
The moment of inertia \(I\) plays a vital role in this process, defined as the measure of an object's resistance to changes in its rotational motion. The larger the moment of inertia, the harder it is to alter the rotation. For our cylinder, its rotational kinetic energy is deduced using this moment of inertia.
Angular velocity \(\omega\) is another key term, representing how fast the object rotates. It's linked to linear velocity through the cylinder's radius \(R\), vital in expressing the system's dynamics as the string causes rotation. The broader the radius, the greater the rotational effect for the same linear speed. The exercise shows how understanding these parameters aids in deriving the equation of motion and predicting how the system will react when forced by the descending mass. This intertwined understanding of rotational and linear elements is what makes Lagrangian mechanics such a powerful method for solving complex motion problems.

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Most popular questions from this chapter

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.

Consider a mass \(m\) moving in two dimensions with potential energy \(U(x, y)=\frac{1}{2} k r^{2},\) where \(r^{2}=x^{2}+y^{2} .\) Write down the Lagrangian, using coordinates \(x\) and \(y,\) and find the two Lagrange equations of motion. Describe their solutions. [This is the potential energy of an ion in an "ion trap," which can be used to study the properties of individual atomic ions.]

[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

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 mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

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