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Chapter 14: Problem 56

CP 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
The period of the pendulum is approximately 0.72 seconds.

Step by step solution

01

Understanding the Problem Statement

We need to find the period of a pendulum made from a hollow sphere. The mass is given as \( M = 0.015 \, \text{kg} \) and the radius is \( R = 0.050 \, \text{m} \). The sphere is displaced and released, acting like a physical pendulum. The period of such a pendulum depends on the moment of inertia about the pivot point.
02

Recall the Parallel-Axis Theorem

The parallel-axis theorem states that the moment of inertia about any axis parallel to an axis through the center of mass is \( I = I_{CM} + Md^2 \), where \( I_{CM} \) is the moment of inertia at the center of mass and \( d \) is the distance from the center of mass to the new axis.
03

Find the Moment of Inertia at the Center of Mass

The moment of inertia for a hollow sphere about its center of mass is \( I_{CM} = \frac{2}{3}MR^2 \). For our sphere, substitute \( M = 0.015 \, \text{kg} \) and \( R = 0.050 \, \text{m} \): \[ I_{CM} = \frac{2}{3} \times 0.015 \, \text{kg} \times (0.050 \, \text{m})^2 \] Calculate \( I_{CM} \).
04

Use the Parallel-Axis Theorem to Find the Moment of Inertia About the Pivot

The pivot is at the surface of the sphere. Therefore, \( d = R = 0.050 \, \text{m} \). Using the parallel-axis theorem: \[ I = I_{CM} + M R^2 = \frac{2}{3}MR^2 + MR^2 \] Simplify to find:\[ I = \frac{2}{3}MR^2 + \frac{3}{3}MR^2 = \frac{5}{3}MR^2 \] Substitute the values for \( M \) and \( R \) and calculate \( I \).
05

Calculate the Period of the Physical Pendulum

The period \( T \) of a physical pendulum is given by \[ T = 2\pi \sqrt{\frac{I}{MgR}} \] Substitute \( I \) from the previous step, \( M = 0.015 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( R = 0.050 \, \text{m} \) into this formula and calculate \( T \).
06

Calculation and Solution

Using the values obtained:1. Calculate \( I_{CM} = \frac{2}{3} \times 0.015 \, \text{kg} \times (0.050 \, \text{m})^2 = 2.5 \times 10^{-5} \, \text{kg} \cdot \text{m}^2 \).2. Use the parallel-axis theorem: \( I = \frac{5}{3} \times 0.015 \, \text{kg} \times (0.050 \, \text{m})^2 = 6.25 \times 10^{-5} \, \text{kg} \cdot \text{m}^2 \).3. Compute the period: \[ T = 2\pi \sqrt{\frac{6.25 \times 10^{-5} \, \text{kg} \cdot \text{m}^2}{0.015 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.050 \, \text{m}}} \] Simplifying gives \( T \approx 0.72 \, \text{s} \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is an important concept in physics that quantifies how difficult it is to change the rotational motion of an object. Think of it like the rotational counterpart to mass in linear motion. It depends not just on the mass of the object, but also on how the mass is distributed in relation to the axis of rotation.
For a physical pendulum like the hollow sphere in our exercise, the moment of inertia is crucial in determining how it swings when disturbed. The more mass that is further from the rotational axis, the higher the moment of inertia, which means the pendulum will resist changes in its rotational motion more.
The moment of inertia for a hollow sphere about its center of mass is given by the formula: \( I_{CM} = \frac{2}{3}MR^2 \). Here, \( M \) is the mass and \( R \) is the radius of the sphere. This formula tells us that even though a hollow sphere may be light, the distribution of its mass can make it more difficult to spin than a solid sphere of the same mass.
Parallel-Axis Theorem
The parallel-axis theorem helps us find the moment of inertia of an object about any axis that is parallel to and a distance away from an axis through its center of mass. This theorem is particularly handy when dealing with systems like a pendulum, where the axis of rotation is not through the center of mass.
The theorem states: \( I = I_{CM} + Md^2 \), where:
  • \( I \) is the moment of inertia about the new axis.
  • \( I_{CM} \) is the moment of inertia through the center of mass.
  • \( M \) is the mass of the object.
  • \( d \) is the distance between the center of mass and the new axis.
In the context of the hollow sphere pendulum, the pivot point is at the surface of the sphere and not its center, making the parallel-axis theorem extremely useful. This allows us to adjust our calculations from the center of mass to the surface, ensuring an accurate determination of the pendulum's motion.
Hollow Sphere
A hollow sphere is a spherical object where the mass is distributed along its surface, with an empty interior. This unique structure affects its moment of inertia, just as it does in our exercise.
Since all of a hollow sphere's mass is further out from the center than in a solid sphere, it will have a higher moment of inertia when considering the same mass and radius. The formula for the moment of inertia about its center of mass is \( I_{CM} = \frac{2}{3}MR^2 \), reflecting this distinct mass distribution.
In the pendulum problem, the hollow nature of the sphere influences how it swings when used as a pendulum. Understanding this property helps us correctly calculate its moment of inertia around the pivot, ensuring precise predictions of its swinging period. This analysis underscores the importance of accounting for structural differences in physical problems involving rotation.

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Most popular questions from this chapter

When a body of unknown mass is attached to an idealspring with force constant \(120 \mathrm{N} / \mathrm{m},\) it is found to vibrate with a frequency of 6.00 \(\mathrm{Hz}\) . Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the body.

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.120 \(\mathrm{m} .\) The maximum speed of the block is 3.90 \(\mathrm{m} / \mathrm{s}\) . What is the maximum magnitude of the acceleration of the block?

A 2.50 -kg rock is attached at the end of a thin, very light rope 1.45 \(\mathrm{m}\) long. You start it swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after 10\(\frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

CP Two uniform solid spheres, each with mass \(M=0.800 \mathrm{kg}\) and radius \(R=0.0800 \mathrm{m},\) are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k=160 \mathrm{N} / \mathrm{m}\) has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

CP A block with mass \(M\) rests on a frictionless surface and is connected to a horizontal spring of force constant \(k\) . The other end of the spring is attached to a wall (Fig. Pl4.72). A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{\mathrm{s}}\) . Find the maximum amplitude of oscillation such that the top block will not slip on the bot tom block.
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