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Chapter 14: Problem 42

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 165 oscillations in \(265 \mathrm{~s}\). What is its moment of inertia?

Short Answer

Expert verified
The moment of inertia of the machine part is \( 0.0292 kg \cdot m^2 \).

Step by step solution

01

Determine the period of oscillation

The period T is calculated as the total time divided by the number of oscillations. Using the given data, this can be calculated as \( T = \frac{265 s}{165} = 1.6061 s \).
02

Find the angular frequency

The angular frequency \( \omega \) is the inverse of the period, i.e., \( \omega = \frac{2\pi}{T} \). Substituting the value of T calculated in the previous step, we get \( \omega = \frac{2\pi}{1.6061 s} = 3.913 rad/s \).
03

Calculate the moment of inertia

The moment of inertia \( I \) of a body about an axis can be determined using the formula \( I = \frac{k}{\omega^2} \), where \( k \) is the torsion constant of the wire and \( \omega \) is the angular frequency. Substituting the given value of \( k = 0.450 N \cdot m / rad \) and the calculated value of \( \omega \) we get \( I = \frac{0.450 N \cdot m / rad}{(3.913 rad/s)^2} = 0.0292 kg \cdot m^2 \).

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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 is a value that describes a material's resistance to twisting when force is applied. It is directly related to the material's stability and strength under torsional motion. In the exercise, the torsion constant is given as 0.450 N鈰卪/rad. This tells us how much torque the wire can resist per unit radian of twist.
  • A higher torsion constant means that the wire offers more resistance to twisting.
  • This is important because greater resistance will affect the oscillation period and angular frequency of the object suspended.
  • Materials with high torsion constants are typically stronger or thicker鈥攊deal for applications requiring minimal twisting.
The torsion constant is a key factor in calculations involving rotational dynamics. It determines how the system behaves when rotational forces are applied.
Angular Frequency
Angular frequency is a measure of how quickly an object moves through its oscillatory motion. It tells us how many radians the object covers per second. The angular frequency is crucial in understanding the vibrations of the object.
In the exercise, you found the angular frequency using the formula:
\( \omega = \frac{2\pi}{T} \)
where \( T \) is the period of oscillation. For our exercise, the angular frequency was calculated to be 3.913 rad/s.
  • This value gives insight into the speed of oscillation.
  • A higher angular frequency means the object oscillates quicker within a given time.
  • It's related to the period, where a shorter period results in a higher angular frequency.
The angular frequency integrates the time element in oscillatory motion, serving as a bridge between the linear concept of frequency and rotational motion.
Oscillation Period
The oscillation period is the time it takes to complete one full cycle of oscillatory motion. It's a vital part of understanding how systems work when subjected to periodic forces. In the exercise, the oscillation period was determined by dividing the total time taken by the number of oscillations.
The calculation was given by:
\( T = \frac{265 s}{165} = 1.6061 s \)
This value points out how long it takes for the suspended part to complete one full oscillation.
  • The period is inversely related to the angular frequency, meaning if one is large, the other is small.
  • Knowing the period helps predict the timing of future oscillations and can support the design of mechanical systems.
  • A longer period indicates slower oscillations, which can affect the system's stability and response time.
Understanding the oscillation period helps in designing and analyzing systems that rely on precise timing and regular movement cycles.

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Most popular questions from this chapter

BIO "Seeing" Surfaces at the Nanoscale. One technique for making images of surfaces at the nanometer scale, including membranes and biomolecules, is dynamic atomic force microscopy. In this technique, a small tip is attached to a cantilever, which is a flexible, rectangular slab supported at one end, like a diving board. The cantilever vibrates, so the tip moves up and down in simple harmonic motion. In one operating mode, the resonant frequency for a cantilever with force constant \(k=1000 \mathrm{~N} / \mathrm{m}\) is \(100 \mathrm{kHz}\). As the oscillating tip is brought within a few nanometers of the surface of a sample (as shown in the figure), it experiences an attractive force from the surface. For an oscillation with a small amplitude (typically, \(0.050 \mathrm{nm}),\) the force \(F\) that the sample surface exerts on the tip varies linearly with the displacement \(x\) of the tip, \(|F|=k_{\text {surf }} x,\) where \(k_{\text {surf }}\) is the effective force constant for this force. The net force on the tip is therefore \(\left(k+k_{\text {surf }}\right) x\), and the frequency of the oscillation changes slightly due to the interaction with the surface. Measurements of the frequency as the tip moves over different parts of the sample's surface can provide information about the sample.

\(\mathrm{A}\) mass is oscillating with amplitude \(A\) at the end of a spring. How far (in terms of \(A\) ) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?

A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{~kg}\) and radius \(R=0.050 \mathrm{~m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

A \(2.00 \mathrm{~kg}\) bucket containing \(10.0 \mathrm{~kg}\) of water is hanging from a vertical ideal spring of force constant \(450 \mathrm{~N} / \mathrm{m}\) 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?

BIO Weighing Astronauts. This procedure has been used to "weigh" astronauts in space: A \(42.5 \mathrm{~kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes \(1.30 \mathrm{~s}\) to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes \(2.54 \mathrm{~s}\) for one cycle. What is the mass of the astronaut?
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