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Chapter 10: Problem 18

Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass \(5.00 \mathrm{~kg}\) and radius \(0.120 \mathrm{~m} .\) For each the translational speed of the center of mass is \(4.00 \mathrm{~m} / \mathrm{s}\). Sphere \(A\) is a uniform solid sphere and sphere \(B\) is a thin-walled, hollow sphere. How much work, in joules, must be done on each sphere to bring it to rest? For which sphere is a greater magnitude of work required? Explain. (The spheres continue to roll without slipping as they slow down.

Short Answer

Expert verified
The work required to bring sphere A to rest is 70 Joules, while sphere B requires 100 Joules. Thus, more work is required to bring sphere B to rest.

Step by step solution

01

Compute the kinetic energy for sphere A

First, calculate the total kinetic energy for a solid sphere that includes both translational and rotational kinetic energy. The equation for total kinetic energy can be expressed as \( KE = KE_{translational} + KE_{rotational} \). The translational kinetic energy is given by \( \frac{1}{2}mv^2 \) and the rotational kinetic energy by \( \frac{1}{2}I\omega^2 \). For a sphere rolling without slipping, \( \omega = \frac{v}{r} \). Substituting that into the kinetic energy equation and using the moment of inertia for a solid sphere \( I = \frac{2}{5}mr^2 \), we get: \[ KE_{A} = \frac{1}{2}mv^2 + \frac{1}{2} * \frac{2}{5}mr^2 * (\frac{v}{r})^2 = \frac{1}{2}vg(5 + 2) = 7 * \frac{1}{2} * 5 * (4)^2 = 70 J \].
02

Compute the kinetic energy for sphere B

Similarly, calculate the total kinetic energy for a thin-walled, hollow sphere. The process is similar, just using the moment of inertia for a hollow spherical shell \( I = \frac{2}{3}mr^2 \), we get: \[ KE_{B} = \frac{1}{2}mv^2 + \frac{1}{2} * \frac{2}{3}mr^2 * (\frac{v}{r})^2 = \frac{1}{2}mv(3+2) = 5 * \frac{1}{2} * 5 * (4)^2 = 100 J \].
03

Compare the work required to stop each sphere

The amount of work required to bring each sphere to rest is equal to its initial kinetic energy. So, the work required to stop sphere A is 70 J, and for sphere B, it's 100 J. Therefore, more work is required to bring sphere B to rest.

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Key Concepts

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

Translational Kinetic Energy
Translational kinetic energy (TKE) is the energy associated with the linear motion of an object. When a student is asked to determine the kinetic energy of a rolling object, it's crucial to understand that this form of energy is calculated using the equation \( KE_{translational} = \frac{1}{2}mv^2 \) where \( m \) is the object's mass, and \( v \) is the velocity of its center of mass.

For objects moving along a straight path, such as our two spheres rolling on the floor, TKE can be thought of as the energy needed to keep the object moving in its translational motion. Calculating the TKE tells us how much energy is invested in the sphere's forward motion, excluding any rotation.
Rotational Kinetic Energy
In the context of rolling objects, rotational kinetic energy (RKE) is just as important as translational kinetic energy. This energy is linked to the object's rotation around an axis. The RKE is given by the equation \( KE_{rotational} = \frac{1}{2}I\omega^2 \) where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

An interesting point here is that the angular velocity is related to the translational speed for rolling objects that are not slipping. The relation \( \omega = \frac{v}{r} \) allows us to combine translational and rotational motion into a comprehensive understanding of the object's kinetic energy. For educational purposes, it's crucial for students to see how both translational and rotational energies together form the total kinetic energy of a rolling sphere.
Moment of Inertia
The moment of inertia (I) is effectively the rotational equivalent of mass for translational motion. It measures how difficult it is to change an object's rotational speed around a certain axis. Students should recognize that the moment of inertia depends not just on the mass of an object, but also on how this mass is distributed with respect to the axis of rotation.

For example, in the exercise, sphere A had a moment of inertia \( I_A = \frac{2}{5}mr^2 \) because it's a uniform solid sphere, while sphere B's moment of inertia was different, \( I_B = \frac{2}{3}mr^2 \) for a thin-walled, hollow sphere. This difference in distribution of mass significantly affects the amount of work needed to bring each sphere to rest.
Work-Energy Principle
The work-energy principle is a foundational concept in physics, stating that the work done on an object is equal to the change in its kinetic energy. The principle helps us understand that to stop a rolling object, work must be applied to remove its kinetic energy.

In our exercise, to determine the work needed to bring the spheres to rest, you equate the work to the initial kinetic energy of each sphere. Since kinetic energy is the sum of translational and rotational energy, the work also reflects both components. The larger the initial kinetic energy, the greater the work required to stop the object. This principle allows students to predict and compare the work needed to stop different objects based on their kinetic energies.

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Most popular questions from this chapter

A \(12.0 \mathrm{~kg}\) box resting on a horizontal, frictionless surface is attached to a \(5.00 \mathrm{~kg}\) weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass \(2.00 \mathrm{~kg}\) and diameter \(0.500 \mathrm{~m} .\) After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

A \(5.00 \mathrm{~kg}\) ball is dropped from a height of \(12.0 \mathrm{~m}\) above one end of a uniform bar that pivots at its center. The bar has mass 8.00 \(\mathrm{kg}\) and is \(4.00 \mathrm{~m}\) in length. At the other end of the bar sits another 5.00 \(\mathrm{kg}\) ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?

A hollow, spherical shell with mass \(2.00 \mathrm{~kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of static friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to \(4.00 \mathrm{~kg}\) ?

Example 10.7 discusses a uniform solid sphere rolling with- out slipping down a ramp that is at an angle \(\beta\) above the horizontal. Now consider the same sphere rolling without slipping up the ramp. (a) In terms of \(g\) and \(\beta\), calculate the acceleration of the center of mass of the sphere. Is your result larger or smaller than the acceleration when the sphere rolls down the ramp, or is it the same? (b) Calculate the friction force (in terms of \(M, g,\) and \(\beta\) ) for the sphere to roll without slipping as it moves up the incline. Is the result larger, smaller, or the same as the friction force required to prevent slipping as the sphere rolls down the incline?

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel \(0.650 \mathrm{~m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects \(0.200 \mathrm{~m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is \(8.00 \mathrm{~kg} ;\) its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at \(5.00 \mathrm{rev} / \mathrm{s}\). Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at \(0.300 \mathrm{rev} / \mathrm{s}\). (d) At what rate must the shaft rotate in order that it may be supported at one end only?
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