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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.

Short Answer

Expert verified
The cylinder's acceleration is \( \ddot{x} = -\frac{mg\sin(\alpha)}{m + \frac{I}{R^2}} \).

Step by step solution

01

Define the Kinetic Energy

The kinetic energy of the rolling cylinder is the sum of translational and rotational components. Using the formulas given, the translational kinetic energy is \( T_{trans} = \frac{1}{2} mv^2 \), and the rotational kinetic energy is \( T_{rot} = \frac{1}{2} I \omega^2 \). For rolling without slipping, the velocity \( v = \dot{x} \) and the angular velocity \( \omega = \frac{\dot{x}}{R} \). Substituting for \( \omega \) into the rotational energy, we have \( T_{rot} = \frac{1}{2} I \left(\frac{\dot{x}}{R}\right)^2 \). Therefore, the total kinetic energy \( T \) is \( T = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\left(\frac{\dot{x}}{R}\right)^2 \).
02

Define the Potential Energy

The potential energy \( V \) of the cylinder when it has fallen a distance \( x \) down the incline is \( V = mgx\sin(\alpha) \), as the component of gravitational force acting along the incline is \( mg \sin(\alpha) \).
03

Calculate the Lagrangian

The Lagrangian \( L \) is obtained by taking the difference between the kinetic energy and potential energy: \( L = T - V = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\left(\frac{\dot{x}}{R}\right)^2 - mgx\sin(\alpha) \).
04

Write the Lagrange's Equation

Lagrange's equation is given by \( \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0 \). First, calculate \( \frac{\partial L}{\partial \dot{x}} = m \dot{x} + \frac{I}{R^2}\dot{x} \). Differentiating this with respect to time, we get \( \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) = (m + \frac{I}{R^2})\ddot{x} \). Next, compute \( \frac{\partial L}{\partial x} = -mg\sin(\alpha) \). Use these results in Lagrange's equation: \((m + \frac{I}{R^2})\ddot{x} + mg\sin(\alpha) = 0\).
05

Solve for the Acceleration

Rearrange the equation from Step 4 to solve for \( \ddot{x} \): \( \ddot{x} = -\frac{mg\sin(\alpha)}{m + \frac{I}{R^2}} \). This is the linear acceleration of the cylinder down the incline.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept in physics that deals with the energy an object possesses due to its motion. For a rolling cylinder, kinetic energy is divided into two types: translational and rotational.
  • Translational kinetic energy is the energy due to the motion of the cylinder's center of mass. It is given by the formula: \( T_{trans} = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity.
  • Rotational kinetic energy is the energy due to the cylinder spinning around its own axis. It is calculated using: \( T_{rot} = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
In the case of a rolling cylinder without slipping, the linear velocity \( v \) relates to the angular velocity \( \omega \) through the radius \( R \), as \( \omega = \frac{v}{R} \). Hence, the total kinetic energy \( T \) of the cylinder becomes: \[T = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \left(\frac{\dot{x}}{R}\right)^2\] This combined energy tells us how much energy the cylinder has because of its overall motion.
Potential Energy
Potential energy is associated with the position of an object in a force field, typically a gravitational field. In the scenario of a cylinder rolling down an incline, the potential energy is influenced by its position along the slope.
  • The potential energy \( V \) of the cylinder is given by \( V = mgx\sin(\alpha) \).
  • Here, \( x \) is the distance the cylinder has traveled down the incline.
  • \( \alpha \) is the angle of the incline with respect to the horizontal, while \( m \) and \( g \) represent the mass of the cylinder and gravitational acceleration, respectively.
Potential energy essentially quantifies the energy stored within the cylinder because of its elevated position on the slope. As it rolls downward, this energy is converted into kinetic energy, specifically the translational and rotational energies discussed before.
Lagrange's Equation
Lagrange's equation is a powerful tool in mechanics that allows us to derive equations of motion for systems. It is particularly useful when dealing with systems possessing complex constraints like the rolling cylinder. This equation links the kinetic and potential energies through the concept of the Lagrangian, \( L \), defined as the difference between kinetic energy \( T \) and potential energy \( V \):
\[ L = T - V \] For the rolling cylinder, the Lagrangian is computed as: \[ L = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\left(\frac{\dot{x}}{R}\right)^2 - mgx\sin(\alpha) \]Using Lagrange's equation \( \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0 \), we can derive the motion equation for the cylinder. This involves calculating the derivative of the Lagrangian with respect to velocity \( \dot{x} \) and position \( x \), and setting up the equation:
  • The term \( \frac{\partial L}{\partial \dot{x}} = m \dot{x} + \frac{I}{R^2}\dot{x} \) is differentiated over time \( t \).
  • Simultaneously, \( \frac{\partial L}{\partial x} = -mg\sin(\alpha) \) is calculated.
  • Inserting these into Lagrange's equation gives the acceleration \( \ddot{x} \): \[ \ddot{x} = -\frac{mg\sin(\alpha)}{m + \frac{I}{R^2}} \]
This result provides the acceleration with which the cylinder rolls down the incline. Understanding Lagrange's formulation helps simplify complex systems and is a cornerstone in classical mechanics.

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Most popular questions from this chapter

The center of a long frictionless rod is pivoted at the origin, and the rod is forced to rotate in a horizontal plane with constant angular velocity \(\omega\). Write down the Lagrangian for a bead threaded on the rod, using \(r\) as your generalized coordinate, where \(r, \phi\) are the polar coordinates of the bead. (Notice that \(\phi\) is not an independent variable since it is fixed by the rotation of the rod to be \(\phi=\omega t\).) Solve Lagrange's equation for \(r(t) .\) What happens if the bead is initially at rest at the origin? If it is released from any point \(r_{\mathrm{o}}>0,\) show that \(r(t)\) eventually grows exponentially. Explain your results in terms of the centrifugal force \(m \omega^{2} r\).

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.

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) 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.

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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