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Chapter 15: Problem 59

The balance wheel of an old-fashioned watch oscillates with angular amplitude \(\pi\) rad and period \(0.600 \mathrm{~s}\). Find (a) the maximum angular speed of the wheel, (b) the angular speed at displacement \(\pi / 2 \mathrm{rad}\), and (c) the magnitude of the angular acceleration at displacement \(\pi / 4 \mathrm{rad}\).

Short Answer

Expert verified
(a) \(32.93 \, \text{rad/s}\), (b) \(28.52 \, \text{rad/s}\), (c) \(86.25 \, \text{rad/s}^2\)

Step by step solution

01

Identify Given Values

Extract the given values from the problem statement. We are given the angular amplitude \(A = \pi\) radians and the period \(T = 0.600\) seconds.
02

Find the Angular Frequency

The angular frequency \(\omega\) is related to the period by \(\omega = \frac{2\pi}{T}\). Substitute the given period to find \(\omega\). \[\omega = \frac{2\pi}{0.600}\approx 10.47 \, \text{rad/s}\]
03

Calculate Maximum Angular Speed

The maximum angular speed \(\omega_{\text{max}}\) is given by \(\omega_{\text{max}} = \omega A\). Substitute \(\omega\) and \(A\) to find the maximum angular speed.\[\omega_{\text{max}} = 10.47 \times \pi \approx 32.93 \, \text{rad/s}\]
04

Determine Angular Speed at \(\pi/2\) rad

Use the formula for angular speed in simple harmonic motion: \(v = \omega \sqrt{A^2 - x^2}\), where \(x\) is the displacement. Here, \(x = \pi/2\).\[v = 10.47 \sqrt{\pi^2 - (\pi/2)^2} \] \[v = 10.47 \sqrt{\pi^2 - \pi^2/4} = 10.47 \sqrt{3\pi^2/4}\] \[v \approx 28.52 \, \text{rad/s}\]
05

Calculate Angular Acceleration at \(\pi/4\) rad

The angular acceleration \(a\) in simple harmonic motion is \(a = -\omega^2 x\). For \(x = \pi/4\):\[a = -(10.47)^2 \times \pi/4\]\[a = -(109.74)\times \pi/4 \approx -86.25 \, \text{rad/s}^2\]Since we need the magnitude, we take the absolute value: \[|a| = 86.25 \, \text{rad/s}^2\]

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Key Concepts

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

Angular Amplitude
Angular amplitude refers to the maximum angle reached by an oscillating object measured in radians. In the context of simple harmonic motion, it represents the furthest point the object can reach from its equilibrium position.
For the balance wheel of an old-fashioned watch, as given in our example, the angular amplitude is \( \pi \) radians. This indicates that the wheel can swing a total angle of \( \pi \) radians from one extreme end to the other.
  • Importance: The angular amplitude is critical because it informs us about the range of motion of the system.
  • Relation: It is a factor in determining the maximum angular speed, as it multiplies with angular frequency.
Understanding angular amplitude helps in analyzing the extent of oscillation, which can be essential when assessing the performance or design of oscillating systems.
Angular Frequency
Angular frequency, denoted by \( \omega \), is a key concept in describing oscillations. It measures how quickly an object oscillates and is expressed in radians per second. Angular frequency is derived from the period \( T \) using the formula \( \omega = \frac{2\pi}{T} \). For our problem, the period given is 0.600 seconds, leading to an angular frequency of approximately 10.47 radians per second.
  • Significance: Angular frequency is pivotal in calculating other parameters such as angular speed and acceleration.
  • Role in Motion: It directly relates to how fast the cyclic motion repeats itself.
A thorough grasp of angular frequency enables students to predict the dynamic behavior of oscillating systems, whether they're clock mechanisms or oscillating springs.
Maximum Angular Speed
The maximum angular speed, denoted as \( \omega_{\text{max}} \), is the highest speed reached by a rotating object during its oscillation.
It can be calculated using the formula \( \omega_{\text{max}} = \omega A \), where \( \omega \) is the angular frequency and \( A \) is the angular amplitude.From our example, we have \( \omega \approx 10.47 \) rad/s and \( A = \pi \) rad, resulting in a maximum angular speed of about 32.93 rad/s.
  • Use: Knowing the maximum angular speed helps in designing components that can sustain the maximum kinetic load.
  • Behavior Interpretation: It indicates the peak energy point during the oscillation cycle.
Comprehending maximum angular speed is essential for applications that involve precise timekeeping or any rotational dynamics.
Angular Acceleration
Angular acceleration measures how quickly the angular speed changes with time in an oscillating system.
It is represented by \( a \) and is calculated using \( a = -\omega^2 x \), where \( x \) is the displacement from equilibrium. In our problem, at a displacement of \( \pi/4 \) rad, the angular acceleration becomes approximately -86.25 rad/s虏.
  • Magnitude: We often take the absolute value to understand its effect without considering direction, resulting in 86.25 rad/s虏.
  • Application: Angular acceleration information helps in understanding how forces within the system change over time.
Understanding angular acceleration is vital in assessing the stresses and forces acting in oscillating systems, beneficial for building robust mechanical designs.

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Most popular questions from this chapter

33 The amplitude of a lightly. damped oscillator decreases by \(4.0 \%\) during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

The position function \(x=(6.0 \mathrm{~m}) \cos [(3 \pi \mathrm{rad} / \mathrm{s}) t+\pi / 3 \mathrm{rad}]\) gives the simple harmonic motion of a body. At \(t=2.1 \mathrm{~s}\), what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

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