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Chapter 13: Problem 38

The balance wheel of a watch vibrates with an angular amplitude \(\Theta,\) angular frequency \(\omega\) , and phase angle \(\phi=0\) . (a) Find expressions for the angular velocity \(d \theta / d t\) and angular acceleration \(d^{2} \theta / d t^{2}\) as functions of time. (b) Find the balance wheel's angular velocity and angular acceleration when its angular displacement is \(\Theta,\) and when its angular displacement is \(\Theta / 2\) and \(\theta\) is decreasing. (Hint: Sketch a graph of \(\theta\) versus \(t\) )

Short Answer

Expert verified
The expressions are: \(\frac{d\theta}{dt} = -\omega \Theta \sin(\omega t)\) and \(\frac{d^2\theta}{dt^2} = -\omega^2 \Theta \cos(\omega t)\). At \(\theta = \Theta\), velocity is 0, acceleration is \(-\omega^2 \Theta\). At \(\theta = \Theta/2\) (decreasing), velocity is \(\frac{\sqrt{3}}{2}\omega\Theta\), acceleration is \(-\frac{1}{2}\omega^2\Theta\).

Step by step solution

01

Angular Displacement Equation

For a harmonic oscillator such as a balance wheel, the angular displacement as a function of time is given by:\[\theta(t) = \Theta \cos(\omega t + \phi)\]Since the phase angle \(\phi\) is 0, this simplifies to:\[\theta(t) = \Theta \cos(\omega t)\]
02

Find Angular Velocity

Angular velocity is the derivative of angular displacement with respect to time. Differentiating \(\theta(t)\) with respect to time gives:\[\frac{d\theta}{dt} = \frac{d}{dt}(\Theta \cos(\omega t)) = -\omega \Theta \sin(\omega t)\]
03

Find Angular Acceleration

Angular acceleration is the derivative of angular velocity with respect to time. Differentiating \(\frac{d\theta}{dt}\) with respect to time gives:\[\frac{d^2\theta}{dt^2} = \frac{d}{dt}(-\omega \Theta \sin(\omega t)) = -\omega^2 \Theta \cos(\omega t)\]
04

Angular Velocity at Maximum Displacement

When \(\theta = \Theta\), the balance wheel is at maximum displacement, and from the cosine function, \(\cos(\omega t) = 1\). At this point, the angular velocity \(\frac{d\theta}{dt} = 0\) since \(\sin(\omega t) = 0\).
05

Angular Acceleration at Maximum Displacement

When \(\theta = \Theta\), the angular acceleration is:\[\frac{d^2\theta}{dt^2} = -\omega^2 \Theta \cos(\omega t) = -\omega^2 \Theta\]
06

Angular Displacement at \(\theta = \Theta/2\)

When \(\theta = \Theta / 2\), we need to find the angular velocity and acceleration as the displacement decreases. From \(\theta(t) = \Theta \cos(\omega t)\), solve for \(\omega t\):\[\frac{\Theta}{2} = \Theta \cos(\omega t) \implies \cos(\omega t) = \frac{1}{2}\]This gives two solutions, \(\omega t = \pi/3\) or \(\omega t = 5\pi/3\). In the decreasing phase, \(\omega t = 5\pi/3\).
07

Angular Velocity at \(\theta = \Theta/2\) (Decreasing)

Using \(\omega t = 5\pi/3\), substitute into \(\frac{d\theta}{dt}\):\[\frac{d\theta}{dt} = -\omega \Theta \sin(5\pi/3) = -\omega \Theta (-\sqrt{3}/2) = \frac{\sqrt{3}}{2}\omega\Theta\]
08

Angular Acceleration at \(\theta = \Theta/2\) (Decreasing)

At \(\omega t = 5\pi/3\):\[\frac{d^2\theta}{dt^2} = -\omega^2 \Theta \cos(5\pi/3) = -\omega^2 \Theta \cdot \frac{1}{2} = -\frac{1}{2}\omega^2\Theta\]

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Key Concepts

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

Angular Velocity
In the world of rotational motion, angular velocity is an essential concept. It refers to how fast an object rotates or spins around an axis. This speed is determined by the change in angular displacement over time.
In terms of a harmonic oscillator, which includes systems like a balance wheel in a watch, angular velocity is represented mathematically by the derivative of angular displacement \(\theta(t)\). For the given oscillator, this is expressed as:
  • \(\frac{d\theta}{dt} = -\omega \Theta \sin(\omega t)\)
Here:
  • \(\omega\) is the angular frequency
  • \(\Theta\) is the angular amplitude
  • \(\sin(\omega t)\) describes the sine wave nature of the motion
The negative sign indicates the direction of motion, showing that as the displacement curve cycles through the sine function, the angular velocity moves through positive and negative values.
Angular Acceleration
Angular acceleration defines the rate of change of angular velocity. It describes how quickly an object speeds up or slows down in its rotational path. For dynamic systems like the balance wheel, angular acceleration is a crucial property that allows it to function properly.
By differentiating the angular velocity \(\frac{d\theta}{dt}\), we obtain the angular acceleration:
  • \(\frac{d^2\theta}{dt^2} = -\omega^2 \Theta \cos(\omega t)\)
In this equation:
  • \(\omega^2\) represents the square of the angular frequency
  • \(\cos(\omega t)\) demonstrates the cosine wave, indicating the oscillatory nature of acceleration
The function reveals how the acceleration fluctuates as the rotating system continues its motion, aiding in the understanding of complex rotational dynamics.
Harmonic Oscillator
A harmonic oscillator is a system that experiences periodic motion. It is a core concept in physics and engineering, describing everything from simple pendulums to molecular vibrations.
This system can often be described by two main trigonometric functions: sine and cosine. These functions display how the energy moves back and forth over time. In the balance wheel example, the motion is dominated by:
  • \(\Theta \cos(\omega t)\) for displacement
  • \(-\omega \Theta \sin(\omega t)\) for velocity
These equations illustrate how a harmonic oscillator can maintain a regular cycle of motion. This predictability is what allows devices like watches to keep precise time.
Angular Displacement
In rotational motion, angular displacement represents the angle through which an object moves over a period of time. It is a vital measurement in understanding how far, and often how the object has traveled in a circular path.
For the balance wheel, angular displacement is expressed by:
  • \(\theta(t) = \Theta \cos(\omega t)\)
Here:
  • \(\Theta\) is the maximum angle the wheel achieves, or the angular amplitude
  • \(\omega t\) helps define the time-dependent oscillatory pattern
Understanding angular displacement allows us to predict other variables like velocity and acceleration, as changes in \(\theta\) lead to differences in these derivative-related quantities.

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Most popular questions from this chapter

When displaced from equilibrium, the two liydrogen atoms in an \(\mathrm{H}_{2}\) molecule are acted on by a restoring force \(F_{x}=-k x\) with \(k=580 \mathrm{N} / \mathrm{m}\) . Calculate the oscillation frequency of the \(\mathrm{H}_{2}\) molecule. (Hint: The mass of a hydrogen atom is 1.008 atomic mass units, or 1 u; see Appendix E. As in Example 13.7 (Section \(13.4 ),\) use \(m / 2\) instead of \(m\) in the expression for \(f\) )

You measure the period of a physical pendulum about one pivot point to be \(T\) . Then you find another pivot point on the opposite side of the center of mass that gives the same period. The two points are separated by a distance \(L\) . Use the parallel-axis theorem to show that \(g=L(2 \pi / T)^{2}\) . (This result shows a way that you can measure \(g\) without knowing the mass or any moments of inertia of the physical pendulum.)

You want to find the moment of inertia of a complicated machine part about an axis through its center of mass. You suspend it from a wire along this axis. The wire has a torsion constant of 0.450 \(\mathrm{N} \cdot \mathrm{m} / \mathrm{rad}\) You twist the part a small amount about this axis and let it go, timing 125 oscillations in 265 \(\mathrm{s}\) . What is the moment of inertia you want to find?

A harmonic oscillator consists of a \(0.500-\mathrm{kg}\) mass attached to an ideal spring with force constant 140 \(\mathrm{N} / \mathrm{m}\) . Find (a) the period; (b) the frequency; (c) the angular frequency of the oscillations.

A \(0.0200-\mathrm{kg}\) bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{c})\) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;(\mathrm{d})\) the speed of the bolt when \(x=-0.180 \mathrm{m} .\)
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