Problem 44 - University Physics with Modern Physics

Question

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.)

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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: \( \frac{d\theta}{dt} = -\Theta \omega \sin(\omega t) \).

Step 3: 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: \( \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cos(\omega t) \).

Step 4: 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.

Step 5: 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 \( -\Theta \omega^2 \).

Step 6: 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 \( \frac{d^2\theta}{dt^2} = -\Theta \omega^2 \cdot \left(-\frac{1}{2}\right) = \frac{\Theta \omega^2}{2} \).

Key Concepts

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.