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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a thin rim with radius \(0.55 \mathrm{cm},\) connected to the balance szaff by thin spokes of negligible mass. The total mass of the balance wheel is 0.90 \(\mathrm{g}\) . (a) What is the moment of inertia of the balance wheel about its shaft? (b) What is the torsion constant of the hairspring?

Short Answer

Expert verified
Moment of inertia: \(2.72 \times 10^{-8} \text{ kg m}^2\), Torsion constant: \(4.30 \times 10^{-6} \text{ N m/rad}\).

Step by step solution

01

Understand the problem

We have a balance wheel with a radius of 0.55 cm and a mass of 0.90 g. It operates with four ticks per second, each tick is half a period. We need to find the moment of inertia and the torsion constant of the hairspring.
02

Convert units

First, convert the radius from cm to meters (0.55 cm = 0.0055 m) and the mass from grams to kilograms (0.90 g = 0.0009 kg).
03

Moment of inertia for a thin rim

The moment of inertia (I) for a thin rim is given by the formula \( I = mr^2 \), where \( m \) is the mass and \( r \) is the radius. Substituting the given values, we get \( I = 0.0009 \text{ kg} \times (0.0055 \text{ m})^2 \).
04

Calculate moment of inertia

Calculate the moment of inertia. \[ I = 0.0009 \times 0.0055^2 = 0.0009 \times 0.00003025 = 2.7225 \times 10^{-8} \text{ kg m}^2 \].
05

Calculate angular frequency

Given that there are four ticks per second, and each tick represents half a cycle, the full cycle frequency is \( 2 \text{ Hz} \). Angular frequency \( \omega \) is \( \omega = 2\pi \times 2 = 4\pi \text{ rad/s} \).
06

Relate angular frequency to torsion constant

The formula relating angular frequency \( \omega \), moment of inertia \( I \), and torsion constant \( \kappa \) is \( \omega = \sqrt{\frac{\kappa}{I}} \). Rearranging gives \( \kappa = I \omega^2 \).
07

Calculate torsion constant

Substitute the values into the equation \( \kappa = 2.7225 \times 10^{-8} \text{ kg m}^2 \times (4\pi)^2 \text{ rad}^2/\text{s}^2 \).
08

Final calculation of torsion constant

Perform the calculation: \( \kappa = 2.7225 \times 10^{-8} \times 16\pi^2 = 2.7225 \times 10^{-8} \times 157.91 = 4.297 \times 10^{-6} \text { N m/rad} \).

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Key Concepts

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

Torsion Constant
The torsion constant, denoted as \( \kappa \), is a measure of the stiffness of the hairspring in a balance wheel system. It tells us how much torque, or twisting force, is needed to produce a certain angle of twist in the spring.
The relationship between the torsion constant and angular motion is described by the formula:
  • \( \tau = \kappa \theta \)
where \( \tau \) is the torque applied and \( \theta \) is the angular displacement. A larger torsion constant means the spring is stiffer, and more force is required to twist it by a certain angle.
In the context of a balance wheel connected to a hairspring, the torsion constant is crucial for determining how the wheel behaves under oscillation. The stiffness of the spring affects the wheel's frequency of oscillation, which is important for the accuracy of timekeeping devices such as watches and clocks. A precise torsion constant ensures that the balance wheel ticks consistently, allowing for the reliable measurement of time.
Balance Wheel Dynamics
Balance wheel dynamics is all about understanding the movement and forces within a balance wheel system. A balance wheel acts like a pendulum in a mechanical timekeeping device. It controls the movement and accuracy of clocks and watches. The balance wheel itself is a circular object that oscillates back and forth.
The dynamics of this wheel involve:
  • Mass distribution: The mass is focused on the rim in this case, resembling a thin hoop.
  • Moment of inertia: Determines resistance to angular acceleration. It's calculated as \( I = mr^2 \), where \( m \) is mass and \( r \) is radius.
  • Oscillation: The movement back and forth in a regular pattern. This oscillation influences the timekeeping accuracy.
Understanding these dynamics helps engineers and watchmakers design mechanisms that keep precise and reliable time.
Watchmakers pay attention to balance wheel dynamics to ensure every tick is consistent, providing dependable time measurement. By tuning factors like moment of inertia and ensuring proper torsion constant, they can achieve optimal functioning of the clock or watch mechanism.
Angular Frequency
Angular frequency is an essential concept related to the oscillation of rotating systems, such as balance wheels. It gives us insight into how fast an object rotates, which ties directly into its rotational speed and periodic motion. In the realm of physics and clocks:
  • It is denoted by \( \omega \) and is measured in rad/s (radians per second).
  • Calculated as \( \omega = 2\pi f \), where \( f \) is the frequency in Hz (hertz).
  • In this case, with a frequency of 2 Hz, we have \( \omega = 4\pi \) rad/s.
Angular frequency not only tells us about speed but also links with other fundamental concepts, such as the torsion constant.
The angular frequency can also define the energy cycled in the system and relates to the precision needed in timekeeping mechanisms like clocks.
By understanding and controlling angular frequency, designers can optimize for precise timing, which is crucial for both the artistry and functionality of timepieces.

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Most popular questions from this chapter

A rocket is accelerating upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) from the launchpad on the earth. Inside a small \(1.50-\mathrm{kg}\) ball hangs from the ceiling by a light \(1.10-\mathrm{m}\) wire. If the ball is displaced \(8.50^{\circ}\) from the vertical and released, find the amplitude and period of the resulting swings of this pendulum.

A \(2.00-\mathrm{kg}\) 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 unhappy 0.300 -kg rodent, moving on the end of a spring with force constant \(k=2.50 \mathrm{N} / \mathrm{m}\) , is acted on by a damping force \(F_{x}=-b v_{x}\) (a) If the constant \(b\) has the value \(0.900 \mathrm{kg} / \mathrm{s},\) what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

A machine part is undergoing SHM with a frequency of 5.00 Hz and amplitude 1.80 \(\mathrm{cm} .\) How long does it take the part to go from \(x=0\) to \(x=-1.80 \mathrm{cm} ?\)

A mass \(m\) is attached to one end of a massess spring with a force constant \(k\) and an unstretched length \(l_{0}\) . The other end of the spring is free to turn about a nail driven into a frictionless, horizontal surface (Fig. 13.44\() .\) The mass is made to revolve in a circle with an angular frequency of revolution \(\omega^{\prime}\) , (a) Calculate the length \(l\) of the spring as a function of \(\omega^{\prime} .\) (b) What happens to the result in part (a) when \(\omega^{\prime}\) approaches the natural frequency \(\omega=\sqrt{k} / m\) of the mass-spring system? (ff your result bothers you, remember that massless springs and frictionless surfaces don't exist as such, but are only approximate descriptions of real springs and surfaces. Also, Hooke's law is only an approximation of the way real springs behave; the greater the elongation of the spring, the greater the deviation from Hooke's law.)

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