/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 CALC The balance wheel of a watc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 44

CALC 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
(a) \( \frac{d\theta}{dt} = -\Theta \omega \sin(\omega t) \), \( \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(\omega t) \); (b) At \( \theta = \Theta \), \( v = 0 \) and \( a = -\Theta \omega^2 \); At \( \theta = \Theta/2 \), \( v = -\Theta \omega \cdot \frac{\sqrt{3}}{2} \), \( a = \frac{\Theta \omega^2}{2} \).

Step by step solution

01

Angular Displacement Equation

Since the phase angle \( \phi = 0 \), the angular displacement \( \theta \) of the balance wheel can be expressed as a simple harmonic motion: \( \theta(t) = \Theta \cos(\omega t) \) where \( \Theta \) is the angular amplitude, and \( \omega \) is the angular frequency.
02

Find Angular Velocity

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

Find Angular Acceleration

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

Angular Velocity at \( \theta = \Theta \)

When \( \theta = \Theta \), the cosine term is 1, meaning \( \omega t = 0 \) (modulo \( 2\pi \)). Then, \( \frac{d\theta}{dt} = -\Theta \omega \sin(0) = 0 \). The angular velocity is zero.
05

Angular Acceleration at \( \theta = \Theta \)

For \( \theta = \Theta \), \( \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(0) = -\Theta \omega^2 \). The angular acceleration is \( -\Theta \omega^2 \).
06

Angular Velocity and Acceleration at \( \theta = \frac{\Theta}{2} \) while \( \theta \) Decreases

For \( \theta = \frac{\Theta}{2} \), substitute into \( \Theta \cos(\omega t) = \frac{\Theta}{2} \) to solve for \( \cos(\omega t) = \frac{1}{2} \) (i.e., \( \omega t = \frac{\pi}{3} \) or \( \omega t = \frac{5\pi}{3} \), but since \( \theta \) is decreasing, we use \( \omega t = \frac{2\pi}{3} \)). Substitute this into locomotion equations to find: \( \frac{d\theta}{dt} = -\Theta \omega \cdot \frac{\sqrt{3}}{2} \), and \( \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cdot \left(-\frac{1}{2}\right) = \frac{\Theta \omega^2}{2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Angular Displacement
Angular displacement refers to the angle through which a point or line has been rotated in a specified direction around a specified axis. In the case of harmonic motion, like the balance wheel of a watch, the angular displacement is given by the function \( \theta(t) = \Theta \cos(\omega t) \). Here, \( \Theta \) stands for the angular amplitude, \( \omega \) is the angular frequency, and \( t \) is the time.
The equation represents simple harmonic motion, indicating how far the balance wheel has turned from its resting position at any given moment in time. The balance wheel's position varies sinusoidally over time, returning to its original position and repeating this cycle.
  • Amplitude \( \Theta \): The maximum angle reached from the resting position.
  • Frequency \( \omega \): The rate at which the balance wheel oscillates back and forth.
  • Note: Since phase angle \( \phi \) is zero, it simplifies our expression without any phase shift.
Angular Velocity
Angular velocity is the rate of change of angular displacement with respect to time. It indicates how fast the angle of rotation is changing, effectively describing how quickly an object is spinning or rotating. To find the angular velocity \( \frac{d\theta}{dt} \), we differentiate the angular displacement equation with respect to time:
\[ \frac{d\theta}{dt} = -\Theta \omega \sin(\omega t) \].
This suggests that the angular velocity of our balance wheel varies over time and is influenced by the angular amplitude, \( \omega \), and the time \( t \).
  • Maximum Angular Velocity: Occurs when \( \sin(\omega t) = \pm 1 \).
  • Value: The maximum absolute value is \( \Theta \omega \), but with a negative sign indicating the direction of rotation.
Angular Acceleration
This is the rate at which angular velocity changes with time. Angular acceleration provides insights into how the rotational speed is either increasing or decreasing over time. It is found by differentiating the angular velocity \( \frac{d\theta}{dt} = -\Theta \omega \sin(\omega t) \).
Thus, our expression for angular acceleration \( \frac{d^2\theta}{dt^2} \) becomes:
\[ \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(\omega t) \].
The equation tells us that angular acceleration is influenced by the frequency of oscillation and the current angle's position. This acceleration changes in a sinusoidal pattern as well.
  • Maximum Angular Acceleration: Occurs when \( \cos(\omega t) = \pm 1 \).
  • At these points, it reaches \( \pm \Theta \omega^2 \).
  • The acceleration's sign indicates whether the object is speeding up or slowing down its rotation in that direction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

CALC A 2.00 -kg bucket containing 10.0 \(\mathrm{kg}\) of water is hanging from a vertical ideal spring of force constant 125 \(\mathrm{N} / \mathrm{m}\) and oscillating up and down with an amplitude of 3.00 \(\mathrm{cm} .\) Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 \(\mathrm{g} / \mathrm{s}\) . When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

An object is undergoing SHM with period 0.900 s and amplitude 0.320 \(\mathrm{m} .\) At \(t=0\) the object is at \(x=0.320 \mathrm{m}\) and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x=0.320 \mathrm{m}\) to \(x=0.160 \mathrm{m}\) and (b) from \(x=0.160 \mathrm{m}\) to \(x=0 .\)

CP CALC An approximation for the potential energy of a KCl molecule is \(U=A\left[\left(R_{0}^{7} / 8 r^{8}\right)-1 / r\right],\) where \(R_{0}=2.67 \times\) \(10^{-10} \mathrm{m}, A=2.31 \times 10^{-28} \mathrm{J} \cdot \mathrm{m},\) and \(r\) is the distance between the two atoms. Using this approximation: (a) Show that the radial component of the force on each atom is \(F_{r}=A\left[\left(R_{0}^{7} / r^{9}\right)-1 / r^{2}\right].\) (b) Show that \(R_{0}\) is the equilibrium separation. (c) Find the minimum potential energy. (d) Use \(r=R_{0}+x\) and the first two terms of the binomial theorem \(\quad\) (Eq. 14.28\()\) to show that \(F_{r} \approx-\left(7 A / R_{0}^{3}\right) x,\) so that the molecule's force constant is \(k=7 A / R_{0}^{3} .\) (e) With both the \(K\) and \(C 1\) atoms vibrating in opposite directions on opposite sides of the molecule's center of mass, \(m_{1} m_{2} /\left(m_{1}+m_{2}\right)=3.06 \times 10^{-26} \mathrm{kg}\) is the mass to use in cal- culating the frequency. Calculate the frequency of small-amplitude vibrations.

A 1.50 -kg mass on a spring has displacement as a function of time given by the equation $$x(t)=(7.40 \mathrm{cm}) \cos \left[\left(4.16 \mathrm{s}^{-1}\right) t-2.42\right]$$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{s} ;\) (f) the force on the mass at that time.

CALC Jerk. A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Linear Momentum

Read Explanation

Medical Physics

Read Explanation

Astrophysics

Read Explanation

Force

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.