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An unusual pendulum is made by fixing a string to a horizontal cylinder of radius \(R\), wrapping the string several times around the cylinder, and then tying a mass \(m\) to the loose end. In equilibrium the mass hangs a distance \(l_{\mathrm{o}}\) vertically below the edge of the cylinder. Find the potential energy if the pendulum has swung to an angle \(\phi\) from the vertical. Show that for small angles, it can be written in the Hooke's law form \(U=\frac{1}{2} k \phi^{2} .\) Comment on the value of \(k\).

Short Answer

Expert verified
The potential energy is approximately \(U = \frac{1}{2}k\phi^2\) for small \(\phi\) with \(k = 2mgR\).

Step by step solution

01

Define the System

The pendulum consists of a mass \(m\) hanging from a string wrapped around a horizontal cylinder with radius \(R\). The mass hangs a vertical distance \(l_0\) from the edge of the cylinder when in equilibrium.
02

Geometry of the System

When the pendulum swings to an angle \(\phi\) from the vertical, the length of the string unwound is \(R\phi\). The vertical distance of the mass from the point of suspension changes to \(l_0 - R\phi\).
03

Change in Height

Calculate the change in the height of the mass from its equilibrium position. The change in height \(h\) is given by:\[h = R - (l_0 - R\phi) = R\phi.\]
04

Potential Energy Expression

The potential energy \(U\) can be calculated as the work done against gravity:\[U = mgh = mg(R\phi).\]
05

Small Angle Approximation

For small angles, \(\phi\) is small, and we can use the approximation in Hooke's law. Express the potential energy in the form of a quadratic function:\[U = \frac{1}{2} k \phi^2.\]
06

Matching Terms to Hooke’s Law

From the expression \(U = mg(R\phi)\), identify the corresponding terms. For small \(\phi\), the potential energy \(U\) approximately equals:\[U = mgR\phi = \frac{1}{2}(2mgR)\phi^2.\]This matches the form \(\frac{1}{2}k\phi^2\). Thus, in this scenario, the spring constant \(k\) can be identified as \(2mgR\).
07

Final Expression for k

Finally, we conclude that:\[k = 2mgR.\]This is the effective spring constant for the pendulum at small angles, demonstrating a linear dependence on the gravitational force, the length of unwound string, and the mass of the pendulum.

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Key Concepts

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

Pendulum
A pendulum is a simple device that's often used to illustrate basic principles of physics, especially
  • motion
  • energy conservation
  • simple harmonic motion
In our context, we're dealing with a slightly different kind of pendulum, involving a string wrapped around a cylinder. In equilibrium, the mass is held a fixed distance below the cylinder.
This becomes interesting when the mass swings to an angle \( \phi \) because the geometry changes, affecting potential energy.When the pendulum swings, it unwinds from the cylinder. This unwinding stretches the pendulum out further away from the center, which effectively lowers the mass. This means that potential energy, which is reliant on height, changes.
The relationship of this sort of pendulum to more traditional pendulums is evident in the way the energy and forces act on the system. By understanding these dynamics, we gain insight into how energy is conserved and transferred in physics problems.
Hooke's Law
Hooke's Law is commonly associated with spring systems and states that the force needed to extend or compress a spring by some distance is proportional to that distance.In mathematical terms:
  • \( F = k \times \, x \)
Where \( F \) is the force applied, \( x \) is the displacement, and \( k \) is the spring constant.
In the context of our pendulum exercise, Hooke's Law is applied through the potential energy perspective. By swinging to a small angle, the pendulum's behavior becomes similar to a harmonic oscillator, like a spring.This means that for small angles, the potential energy can be expressed in a quadratic form resembling Hooke's Law:
  • \( U = \frac{1}{2} k \phi^2 \)
This analogy is useful because it allows us to determine properties like the effective spring constant, \( k = 2mgR \), giving us a deeper insight into how pendulums operate similarly to springs at small angles.
Small Angle Approximation
The small angle approximation is a useful technique in physics when studying pendulums and other oscillatory systems.It is based on the idea that when angles are very small (usually when measured in radians and less than about 0.1 radian),
  • \( \sin \phi \approx \phi \)
  • \( \cos \phi \approx 1 \)
This simplification makes working with the math much easier and is particularly handy in finding solutions to oscillatory problems.
In our pendulum problem, this approximation allows us to express potential energy in a quadratic form that resembles Hooke's Law.By assuming \( \phi \) is small, terms involving \( \phi \) become easier to handle mathematically, and compounded angles have a near-linear relationship.
This leads to equations simplifying drastically, making them easier to work with while still accurately describing small oscillations.

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Most popular questions from this chapter

Consider a damped oscillator, with natural frequency \(\omega_{\mathrm{o}}\) and damping constant \(\beta\) both fixed, that is driven by a force \(F(t)=F_{0} \cos (\omega t) .\) (a) Find the rate \(P(t)\) at which \(F(t)\) does work and show that the average rate \(\langle P\rangle\) over any number of complete cycles is \(m \beta \omega^{2} A^{2} .\) (b) Verify that this is the same as the average rate at which energy is lost to the resistive force. (c) Show that as \(\omega\) is varied \(\langle P\rangle\) is maximum when \(\omega=\omega_{\mathrm{o}} ;\) that is, the resonance of the power occurs at \(\omega=\omega_{\mathrm{o}}\) (exactly).

An undamped oscillator has period \(\tau_{\mathrm{o}}=1.000 \mathrm{s},\) but \(\mathrm{I}\) now add a little damping so that its period changes to \(\tau_{1}=1.001\) s. What is the damping factor \(\beta\) ? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?

The force on a mass \(m\) at position \(x\) on the \(x\) axis is \(F=-F_{0} \sinh \alpha x,\) where \(F_{0}\) and \(\alpha\) are constants. Find the potential energy \(U(x),\) and give an approximation for \(U(x)\) suitable for small oscillations. What is the angular frequency of such oscillations?

Verify that the decay parameter \(\beta-\sqrt{\beta^{2}-\omega_{0}^{2}}\) for an overdamped oscillator \(\left(\beta > \omega_{\mathrm{o}}\right) d e\) creases with increasing \(\beta\). Sketch its behavior for \(\omega_{0} < \beta < \infty\).

(a) If a mass \(m=0.2 \mathrm{kg}\) is tied to one end of a spring whose force constant \(k=80 \mathrm{N} / \mathrm{m}\) and whose other end is held fixed, what are the angular frequency \(\omega\), the frequency \(f\), and the period \(\tau\) of its oscillations? (b) If the initial position and velocity are \(x_{\mathrm{o}}=0\) and \(v_{\mathrm{o}}=40 \mathrm{m} / \mathrm{s},\) what are the constants \(A\) and \(\delta\) in the expression \(x(t)=A \cos (\omega t-\delta) ?\)

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