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An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

Short Answer

Expert verified
(a) Prove the potential energy formula: \(U(\phi)=mgl(1-\cos\phi)\). (b) Total energy: \(E=\frac{1}{2}ml^2\dot{\phi}^2+mgl(1-\cos\phi)\), motion equation: \(\ddot{\phi}+\frac{g}{l}\sin\phi=0\). (c) Period \(\tau_\mathrm{o}=2\pi\sqrt{l/g}\) for small \(\phi\).

Step by step solution

01

Potential Energy of the Pendulum

The potential energy of a pendulum in relation to its equilibrium position is given by the height difference. If the pendulum is deflected by an angle \( \phi \), the height difference is \( l - l \cos \phi \) where \( l \) is the length of the pendulum rod. Therefore, the potential energy \( U \) is calculated as:\[ U(\phi) = mg(l - l \cos \phi) = mgl(1 - \cos \phi) \]This confirms the potential energy formula \( U(\phi) = m g l (1 - \cos \phi) \).
02

Total Energy Expression

The total energy \( E \) of the pendulum is the sum of its kinetic and potential energy. The kinetic energy \( T \) can be expressed in terms of the angular velocity \( \dot{\phi} \):\[ T = \frac{1}{2} m (l \dot{\phi})^2 = \frac{1}{2} m l^2 \dot{\phi}^2 \]Thus, the total energy \( E \) is:\[ E = T + U = \frac{1}{2} m l^2 \dot{\phi}^2 + mgl(1 - \cos \phi) \]
03

Differentiate Total Energy to Find Equation of Motion

The derivative of the total energy \( E \) with respect to time \( t \) should equal zero, as energy is conserved when there are no non-conservative forces. Thus, differentiating \( E \) with respect to \( t \):\[ \frac{dE}{dt} = m l^2 \dot{\phi} \ddot{\phi} + mgl \sin \phi \cdot \dot{\phi} = 0 \]Simplifying, we get:\[ \dot{\phi} ( m l^2 \ddot{\phi} + mgl \sin \phi ) = 0 \]For non-trivial solutions, \( m l^2 \ddot{\phi} + mgl \sin \phi = 0 \), which results in the equation of motion:\[ \ddot{\phi} + \frac{g}{l} \sin \phi = 0 \]
04

Linearize the Equation of Motion

For small angles, \( \sin \phi \approx \phi \). This makes the equation of motion simpler:\[ \ddot{\phi} + \frac{g}{l} \phi = 0 \]This is a simple harmonic motion equation, with solutions of the form \( \phi(t) = A \cos(\omega t + \delta) \), where \( \omega = \sqrt{\frac{g}{l}} \).
05

Find the Period of the Motion

The angular frequency \( \omega \) relates to the period \( \tau \) by \( \omega = \frac{2\pi}{\tau} \). Thus, \( \tau = \frac{2\pi}{\omega} \):\[ \tau_{\mathrm{o}} = 2\pi \sqrt{\frac{l}{g}} \]This is the period of the pendulum for small angles, confirming that the motion is periodic.

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Key Concepts

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

Potential Energy
In a simple pendulum system, potential energy plays a critical role in understanding the dynamics of motion. Potential energy is the energy stored due to an object's position in a force field, in this case, the gravitational field. For our pendulum, the potential energy is related to the height difference from its equilibrium position.
When the pendulum swings to an angle \(\phi\), it rises to a height given by \(l - l \cos \phi\) where \(l\) is the length of the rod. Thus, the potential energy \(U\) becomes:
  • \(U(\phi) = mg(l - l \cos \phi)\)
  • Simplified as \(U(\phi) = mgl(1 - \cos \phi)\)
This formula shows that potential energy increases as the pendulum is lifted away from its lowest point. It reaches a maximum when the pendulum is at its maximum swing height, illustrating how gravitational energy changes with the pendulum's position.
Equation of Motion
The equation of motion is instrumental in determining how the pendulum moves over time. For the simple pendulum, this equation arises from the laws of mechanics and describes how angular position \(\phi\) changes.
By starting from the total mechanical energy, which is conserved, we derive this motion equation. The total energy includes both the pendulum's kinetic energy and its potential energy. When expressing this as:
  • \(E = \frac{1}{2} ml^2 \dot{\phi}^2 + mgl(1 - \cos \phi)\)
Differentiating \(E\) with respect to time and realizing energy conservation gives the equation of motion:
  • \(\ddot{\phi} + \frac{g}{l} \sin \phi = 0\)
This equation describes the pendulum's behavior, relating angular acceleration \(\ddot{\phi}\), angular displacement \(\phi\), and gravitational force factors.
Small Angle Approximation
In examining the simple pendulum, the small angle approximation is highly useful. It simplifies complex equations allowing us to solve the pendulum's dynamics with ease. The approximation assumes \(\phi\) to be small, letting us use the approximation \(\sin \phi \approx \phi\).
By applying this to the equation of motion:
  • \(\ddot{\phi} + \frac{g}{l} \phi = 0\)
we transform the problem into a form resembling simple harmonic motion (SHM). This linearization holds for small oscillations where \(\phi\) remains minor. It helps simplify the understanding of pendulum motion, reducing complex trigonometric functions to manageable linear forms.
Simple Harmonic Motion
With the small angle approximation, the simple pendulum's behavior mimics simple harmonic motion (SHM). SHM is a type of periodic motion seen in various physical systems where a restoring force is proportional to the displacement. For the pendulum, this manifests when the equation of motion is simplified to:
  • \(\ddot{\phi} + \frac{g}{l} \phi = 0\)
Solving this differential equation yields solutions in trigonometric functions typical of SHM:
  • \(\phi(t) = A \cos(\omega t + \delta)\)
  • where \(\omega = \sqrt{\frac{g}{l}}\)
The frequency \(\omega\) relates to the period \(\tau_o\) by the expression \(\tau_o = 2\pi \sqrt{\frac{l}{g}}\). This period defines how the pendulum swings back and forth in a repeated cycle, a hallmark of SHM. This insight into the pendulum's movement, understood through the simple lens of SHM, illustrates its periodic nature.

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Most popular questions from this chapter

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a y^{2}+2 b y z+c z^{2},(\mathbf{b}) g(x, y, z)=\cos \left(a x y^{2} z^{3}\right),(\mathbf{c}) h(x, y, z)=a r,\) where \(a, b,\) and \(c\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}} .\) Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

For a system of \(N\) particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, \(U=\sum_{\alpha} U_{\alpha}=M g Y\) where \(M=\sum m_{\alpha}\) is the total mass and \(\mathbf{R}=(X, Y, Z)\) is the position of the \(\mathrm{CM},\) with the \(y\) coordinate measured vertically up. [Hint: We know from Problem 4.5 that \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

Verify that the gravitational force \(-G M m \hat{\mathbf{r}} / r^{2}\) on a point mass \(m\) at \(\mathbf{r},\) due to a fixed point mass \(M\) at the origin, is conservative and calculate the corresponding potential energy.

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