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Chapter 11: Problem 12

A rigid rod of length \(2 a\) is suspended by two light, inextensible strings of length \(l\) joining its ends to supports also a distance \(2 a\) apart and level with each other. Using the longitudinal displacement \(x\) of the centre of the rod, and the transverse displacements \(y_{1}, y_{2}\) of its ends, as generalized co-ordinates, find the Lagrangian function (for small \(\left.x, y_{1}, y_{2}\right)\). Determine the normal modes and frequencies. (Hint: First find the height by which each end is raised, the co-ordinates of the centre of mass and the angle through which the rod is turned.)

Short Answer

Expert verified
Question: Determine the Lagrangian function and find the normal modes and frequencies for a rigid rod of length \(2a\) and mass \(M\), suspended by two strings, each of length \(l\), attached to its endpoints. Assume small displacements.

Step by step solution

01

Find the height by which each end is raised

Given that the rod has length \(2a\), the strings attached to the rod have length \(l\). Since the displacements are small, we can apply Pythagorean theorem to find the height change for each end point \(A\) and \(B\) respectively: \(\Delta h_A := l - \sqrt{l^2 - (a - x+ y_1)^2}\), \(\Delta h_B := l - \sqrt{l^2 - (a - x- y_2)^2}\)
02

Determine the coordinates of the center of mass

The center of mass is located at the midpoint of the rod. Suppose the midpoint of the rod is \((x_m, y_m)\). Because the center of mass is at the midpoint of the rod and moves with longitudinal displacement \(x\), we have \((x_m, y_m) = (x, \frac{y_1 + y_2}{2})\)
03

Figure out the angle \(\theta\) through which the rod is turned

As the rod is displaced, it turns by a small angle \(\theta\). We can find this angle by using the arctangent function and noting that the length of the rod doesn't change: \(\tan{\theta} \approx \theta = \frac{y_1 - y_2}{2a}\)
04

Find the Lagrangian function

The kinetic energy \(T\) of the rod can be expressed as follows: \(T = \frac{1}{2}M\dot{x_m}^2 + \frac{1}{2}M\dot{y_m}^2 + \frac{1}{2}I\dot{\theta}^2\) Where \(M\) is the mass of the rod, and \(I=\frac{1}{3}ML^2\) is the moment of inertia. Similarly, the potential energy \(V\) of the rod can be represented as: \(V = Mg\left(\frac{(\Delta h_A + \Delta h_B)}{2}\right)\) Now, we can find the Lagrangian function \(L\), which is the difference between the kinetic energy \(T\) and the potential energy \(V\): \(L = T - V\)
05

Determine the normal modes and frequencies

Using the Lagrangian found in Step 4, write down the Euler-Lagrange equations for \(x, y_1,\) and \(y_2\). Then, linearize those equations by substituting the expressions for \(\theta, \Delta h_A,\) and \(\Delta h_B\). Finally, assume a harmonic time dependence, and find the normal modes of oscillation and the corresponding frequencies. By following these steps, you will be able to determine the Lagrangian function, as well as the normal modes and frequencies of the given system.

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Most popular questions from this chapter

Abead of mass \(m\) slides on a smooth circular hoop of mass \(M\) and radius \(a\), which is pivoted at a point on its rim so that it can swing freely in its plane. Write down the Lagrangian in terms of the angle of inclination \(\theta\) of the diameter through the pivot and the angular position \(\varphi\) of the bead relative to a fixed point on the hoop. Find the frequencies of the normal modes, and sketch the configuration of hoop and bead at the extreme point of each.

A simple pendulum of mass \(m\), whose period when suspended from a rigid support is \(1 \mathrm{~s}\), hangs from a supporting block of mass \(2 m\) which can move along a horizontal line (in the plane of the pendulum), and is restricted by a harmonic-oscillator restoring force. The period of the oscillator (with the pendulum removed) is \(0.1 \mathrm{~s}\). Find the periods of the two normal modes. When the pendulum bob is swinging in the slower mode with amplitude \(100 \mathrm{~mm}\), what is the amplitude of the motion of the supporting block?

A spring of negligible mass, and spring constant (force/extension) \(k\), supports a mass \(m\), and beneath it a second, identical spring, carrying a second, identical mass. Using the vertical displacements \(x\) and \(y\) of the masses from their positions with the springs unextended as generalized co- ordinates, write down the Lagrangian function. Find the position of equilibrium, and the normal modes and frequencies of vertical oscillations.

Show that a stretched string is equivalent mathematically to an infinite number of uncoupled oscillators, described by the co-ordinates $$ q_{n}(t)=\sqrt{\frac{2}{l}} \int_{0}^{l} y(x, t) \sin \frac{n \pi x}{l} \mathrm{~d} x $$ Determine the amplitudes of the various normal modes in the motion described in Chapter 10, Problem 14. Why, physically, are the modes for even values of \(n\) not excited?

A particle moves under a conservative force with potential energy \(V(\boldsymbol{r})\). The point \(\boldsymbol{r}=\mathbf{0}\) is a position of equilibrium, and the axes are so chosen that \(x, y, z\) are normal co- ordinates. Show that, if \(V\) satisfies Laplace's equation, \(\boldsymbol{\nabla}^{2} V=0\) (see \(\left.\S 6.7\right)\), then the equilibrium is necessarily unstable, and hence that stable equilibrium under purely gravitational and electrostatic forces is impossible. (Of course, dynamic equilibrium stable periodic motion - can occur. Note also that the two- dimensional stable equilibrium of Problem 11 does not contradict this result because there is another force imposed, confining the charge to the horizontal plane.)
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