/*! 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 54 A holiday ornament in the shape ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 54

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.)

Short Answer

Expert verified
To calculate the period of the hollow sphere physical pendulum, use the formula for the period of a physical pendulum, the characteristics of the sphere, and the Parallel Axis Theorem to find the Moment of Inertia.

Step by step solution

01

Formula of a Physical Pendulum Period

The formula for the period of a physical pendulum is given by \(T = 2\pi \sqrt{\frac{I}{mgh}}\), where \(T\) is the period, \(m\) is the mass of the pendulum, \(g\) is the acceleration due to gravity, \(h\) is the height of the center of mass above the pivot, and \(I\) is the Moment of Inertia.
02

Moment of Inertia via Parallel Axes Theorem

The Parallel Axis Theorem can be used to calculate the moment of inertia of the ornament. It states that \(I = I_{\text_{cm}} + MD^2\), where \(I_{\text_{cm}}\) is the moment of inertia of the body about an axis passing through the center of mass and parallel to the given axis, \(M\) is the total mass of the body, and \(D\) is the distance between the two parallel axes, which is equal to the radius \(R\) in this case. For a hollow sphere, \(I_{\text_{cm}}\) is \(2/3 MR^2\). Therefore, \(I = 2/3 Mr^2 + Mr^2 = 5/3 Mr^2\).
03

Find Height of Center of Mass

The height of the center of mass above the pivot is simply the radius \(R\) of the sphere.
04

Substitute Values and Solve

Finally, substitute the values into the formula in Step 1 with \(m=0.015 kg\), \(R=0.05 m\), \(g=9.81 m/s^2\), and \(I=5/3 Mr^2\). Simplifying gives the period \(T\).

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.

Parallel-Axis Theorem
Understanding the parallel-axis theorem is crucial when calculating the moment of inertia for an object rotating about an axis that is not through its center of mass. This mathematical tool allows us to relate the moment of inertia about an axis through the center of mass (\(I_{\text{cm}}\text)\) to the moment of inertia about any parallel axis.The theorem states that the moment of inertia about the parallel axis (\(I\text)\) can be found using the formula \[\begin{equation} I = I_{\text{cm}} + MD^2 \end{equation}\]where
  • \(I_{\text{cm}}\text)\) is the moment of inertia about the center of mass,
  • \(M\text)\) is the mass of the object,
  • \(D\text)\) is the perpendicular distance between the two axes.
When applied to real-world problems, like calculating the swing of a holiday ornament or any other physical pendulum, the parallel-axis theorem is indispensable for determining how the distribution of mass around the pivot point affects the pendulum's motion.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. In simple terms, it's akin to mass in linear motion, but for rotation. The value of the moment of inertia depends on how the object's mass is distributed in relation to the axis of rotation.In the case of our holiday ornament, which is a hollow sphere, the moment of inertia about an axis through the center of mass is given by the formula\[\begin{equation} I_{\text{cm}} = \frac{2}{3}MR^2 \end{equation}\]Here, \(M\text)\) is the mass of the sphere, and \(R\text)\) is its radius. When the axis of rotation is moved away from the center of mass, such as when the sphere swings as a pendulum, the parallel-axis theorem must be used to calculate the new moment of inertia, involving both the sphere's geometry and the distance from the new axis to the center of mass.
Simple Harmonic Motion
Simple harmonic motion (SHM) describes a type of predictable, oscillating movement found in systems where the restoring force is directly proportional to the displacement and acts in the opposite direction. This is typical for pendulums and springs.

Characteristics of SHM

  • It is periodic, meaning it repeats in cycles.
  • It has a fixed frequency that depends on the system's properties, such as mass and stiffness for a spring or length and gravity for a pendulum.
  • The motion is sinusoidal, following sine or cosine functions over time.
The period of a physical pendulum exhibiting SHM, like the holiday ornament, is calculated using a formula related to the system's moment of inertia and the gravitational torque. For our ornament, it represents the time taken for one complete back and forth swing.
Center of Mass
The center of mass of an object is the point where its mass is considered to be concentrated for the purpose of analysis. In the context of rotational motion, it plays a pivotal role as it's the point about which the object's mass is evenly distributed.When dealing with the motion of rigid bodies, like our physical pendulum (the holiday ornament), its center of mass hangs directly below the pivot point when at rest. In calculating the period of oscillation, the height above the pivot point to the center of mass (\(h\text)\) is a crucial variable. Since the ornament hangs from a wire, the center of mass does not shift from the center of the sphere, thereby simplifying the calculation process. The entire mass of the ornament can be thought to act through this single point during its swinging motion.

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

Consider the system of two blocks and a spring shown in Fig. \(\mathrm{P} 14.66 .\) The horizontal surface is friction less, but there is static friction between the two blocks. The spring has force constant \(k=150 \mathrm{~N} / \mathrm{m} .\) The masses of the two blocks are \(m=0.500 \mathrm{~kg}\) and \(M=4.00 \mathrm{~kg} .\) You set the blocks into motion by releasing block \(M\) with the spring stretched a distance \(d\) from equilibrium. You start with small values of \(d,\) and then repeat with successively larger values. For small values of \(d,\) the blocks move together in SHM. But for larger values of \(d\) the top block slips relative to the bottom block when the bottom block is released. (a) What is the period of the motion of the two blocks when \(d\) is small enough to have no slipping? (b) The largest value \(d\) can have and there be no slipping is \(d=8.8 \mathrm{~cm} .\) What is the coefficient of static friction \(\mu_{\mathrm{s}}\) between the surfaces of the two blocks?

BIO Weighing a Virus. In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just \(30 \mathrm{nm}\) long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{V}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by \(f_{\mathrm{S}+\mathrm{V}} / f_{\mathrm{S}}=1 / \sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)},\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that it is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{~g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{~Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{~Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

DATA You hang various masses \(m\) from the end of a vertical, \(0.250 \mathrm{~kg}\) spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T=2 \pi \sqrt{m / k}\) with \(m+m_{\text {eff }},\) where \(m_{\text {eff }}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass \(m\) and measure the time for 10 complete oscillations, obtaining these data: $$ \begin{array}{l|lcccc} \boldsymbol{m}(\mathbf{k g}) & 0.100 & 0.200 & 0.300 & 0.400 & 0.500 \\ \hline \text { Time (s) } & 8.7 & 10.5 & 12.2 & 13.9 & 15.1 \end{array} $$ (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_{\text {eff }}\) of the spring's mass? (e) If a \(0.450 \mathrm{~kg}\) mass oscillates on the end of the spring, find its period, frequency, and angular frequency.

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\). (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium.) (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

A \(1.50 \mathrm{~kg},\) horizontal, uniform tray is attached to a vertical ideal spring of force constant \(185 \mathrm{~N} / \mathrm{m}\) and a \(275 \mathrm{~g}\) metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A,\) which is \(15.0 \mathrm{~cm}\) below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Radiation

Read Explanation

Nuclear Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Electromagnetism

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.