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Chapter 11: Problem 76

Two spheres of equal mass and radius. are rolling across the floor with the same speed. Sphere 1 is a a uniform solid; sphere 2 is hollow. Is the work required to stop sphere 1 greater than, less than, or equal to the work required to stop sphere \(2 ?\) (b) Choose the best explanation from among the following: I. Sphere 2 has the greater moment of inertia and hence the greater rotational kinetic energy. II. The spheres have equal mass and speed; therefore, they have the same kinetic energy. III. The hollow sphere has less kinetic energy.

Short Answer

Expert verified
The work required to stop sphere 2 is greater; statement I is correct.

Step by step solution

01

Understand the Problem

We need to compare the work required to stop two spheres with equal mass and radius: one is solid and the other is hollow. Both spheres are moving at the same speed. We will determine which sphere requires more work to stop by examining their kinetic energy.
02

Determine the Kinetic Energy Components

Each sphere has both translational kinetic energy and rotational kinetic energy. The total kinetic energy for a rolling sphere is given by \( KE = KE_{translational} + KE_{rotational} \). Translational kinetic energy is \( KE_{translational} = \frac{1}{2}mv^2 \). Rotational kinetic energy is given by \( KE_{rotational} = \frac{1}{2}I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
03

Calculate Moment of Inertia

For a solid sphere, the moment of inertia \( I_1 \) is \( \frac{2}{5}mr^2 \). For a hollow sphere, \( I_2 \) is \( \frac{2}{3}mr^2 \). The angular velocity \( \omega \) for both spheres can be related to the translational velocity \( v \) by \( \omega = \frac{v}{r} \).
04

Calculate Rotational Kinetic Energy for Each Sphere

Using \( \omega = \frac{v}{r} \), calculate the rotational kinetic energy: - For the solid sphere: \( KE_{rotational, 1} = \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2 \).- For the hollow sphere: \( KE_{rotational, 2} = \frac{1}{2} \cdot \frac{2}{3}mr^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{3}mv^2 \).
05

Total Kinetic Energy Comparison

Add the translational and rotational kinetic energies:- Solid sphere: \( KE_1 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \).- Hollow sphere: \( KE_2 = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2 \).Compare these values to see that \( KE_2 > KE_1 \).
06

Evaluate the Statements

Given that the hollow sphere has greater total kinetic energy, the correct explanation (from provided options) is I: 'Sphere 2 has the greater moment of inertia and hence the greater rotational kinetic energy.' Therefore, more work is required to stop sphere 2.

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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 a key concept in rotational motion, similar to mass in linear motion. It measures how difficult it is to change the rotational speed of an object. For example, with our two spheres, the solid and the hollow one, the moment of inertia determines the resistance each offers to changes in its rotational state.
  • Solid sphere: Its moment of inertia is given by \( \frac{2}{5}mr^2 \). This is because the mass is more uniformly distributed.
  • Hollow sphere: Here, the moment of inertia is \( \frac{2}{3}mr^2 \). This higher value reflects the mass being spread further from the axis.
Thus, the hollow sphere requires more effort to alter its rotational speed.
Work-Energy Principle
The work-energy principle is a cornerstone of mechanics, tying together work and kinetic energy. It states that work done on an object results in a change in its kinetic energy. This principle helps us understand how energy is transferred and transformed.
For our spheres, to stop them, we perform work equal to their total kinetic energy. The sphere with more total kinetic energy requires more work for stopping. So, if the hollow sphere has more kinetic energy, it will take more work to halt it.
Translational Kinetic Energy
When an object moves linearly, it possesses translational kinetic energy. This forms part of the total kinetic energy when the object is also rotating. It is calculated using the formula \( KE_{translational} = \frac{1}{2}mv^2 \).
For both spheres, since they have the same mass and speed, their translational kinetic energy remains identical. Thus, differences in their total kinetic energy arise from the rotational aspects, despite equal translational energy.
Rolling Motion
Rolling motion combines both translational and rotational motion. When an object rolls, every point on it follows a circular path.
For the two spheres:
  • Both have similar translational kinetic energy.
  • Their rotational kinetic energy depends on how mass is distributed, evident from their different moments of inertia.
With rolling motion, understanding rotational contributions helps us grasp how each sphere's kinematic parameters contribute to energy distributions. In exercises like this, it becomes clear why one sphere would require more force to stop based on its motion dynamics.

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Most popular questions from this chapter

Calculate the angular momentum of the Earth about its own axis, due to its daily rotation. Assume that the Earth is a uniform sphere.

A turntable with a moment of inertia of \(5.4 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) rotates freely with an angular speed of \(33 \frac{1}{3}\) rpm. Riding on the rim of the turntable, \(15 \mathrm{cm}\) from the center, is a cute, \(32-g\) mouse. (a) If the mouse walks to the center of the turntable, will the turntable rotate faster, slower, or at the same rate? Explain. (b) Calculate the angular speed of the turntable when the mouse reaches the center.

A \(0.015-\mathrm{kg}\) record with a radius of \(15 \mathrm{cm}\) rotates with an angular speed of \(33 \frac{1}{3} \mathrm{rpm} .\) Find the angular momentum of the record.

A person exerts a tangential force of \(36.1 \mathrm{N}\) on the rim of a disk- shaped merry-go-round of radius \(2.74 \mathrm{m}\) and mass \(167 \mathrm{kg}\). If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of \(32.5^{\circ} ?\)

A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is \(4.1 \mathrm{kg} \cdot \mathrm{m}^{2} .\) A second student tosses a \(1.5-\mathrm{kg}\) mass with a speed of \(2.7 \mathrm{m} / \mathrm{s}\) to the student on the stool, who catches it at a distance of \(0.40 \mathrm{m}\) from the axis of rotation. What is the resulting angular speed of the student and the stool?
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