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Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius. One has the shape of a hoop and the other the shape of a solid disk. Each wheel starts from rest and has a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each makes the same number of revolutions in the same time. (a) Which wheel, if either, has the greater angular acceleration? (b) Which, if either, has the greater moment of inertia? (c) To which wheel, if either, is a greater net external torque applied? Explain your answers.

Short Answer

Expert verified
(a) Both have the same angular acceleration. (b) The hoop has a greater moment of inertia. (c) The hoop has a greater net external torque applied.

Step by step solution

01

Understanding Angular Acceleration

The equation relating angular displacement \( \theta \), initial angular velocity \( \omega_0 \), angular acceleration \( \alpha \), and time \( t \) is \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \). Since both wheels start from rest \( \omega_0 = 0 \) and make the same number of revolutions in the same time, \( \alpha t^2 \) must be the same for both wheels. Thus, both wheels have the same angular acceleration \( \alpha \).
02

Comparing Moments of Inertia

The moment of inertia \( I \) for a hoop is given by \( I_{\text{hoop}} = mR^2 \), while for a solid disk it is \( I_{\text{disk}} = \frac{1}{2}mR^2 \). Given that both wheels have the same mass \( m \) and radius \( R \), the hoop has a larger moment of inertia. Therefore, \( I_{\text{hoop}} > I_{\text{disk}} \).
03

Analyzing the Net External Torque

According to Newton's second law for rotation, torque \( \tau \) is related to the moment of inertia \( I \) and angular acceleration \( \alpha \) by \( \tau = I\alpha \). Since \( \alpha \) is the same for both wheels, the wheel with the larger moment of inertia will require a greater net external torque. Thus, the hoop, having a larger moment of inertia, has a greater net external torque applied, so \( \tau_{\text{hoop}} > \tau_{\text{disk}} \).

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Key Concepts

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

Angular Acceleration
Angular acceleration refers to the rate of change of angular velocity over time. It is a crucial concept in understanding how objects rotate. The formula for angular displacement \( \theta \) in terms of angular acceleration \( \alpha \) and time \( t \) is:
  • \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \)
Here, \( \omega_0 \) is the initial angular velocity. When an object starts from rest, like our wheels, \( \omega_0 = 0 \). Therefore, the equation simplifies to \( \theta = \frac{1}{2} \alpha t^2 \).
In the problem scenario, both wheels make the same number of revolutions in the same time, meaning their angular displacements are equal.
Therefore, their angular accelerations must also be equal.
Angular acceleration helps us understand the motion dynamics of rotating bodies and is key in determining how quickly these objects spin up or down.
Net External Torque
Net external torque relates to the effectiveness of a force acting at a distance from an axis of rotation, causing an object to rotate. Torque \( \tau \) is calculated as:
  • \( \tau = I \alpha \)
where \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration.
The moment of inertia represents an object's resistance to change in its rotation, similar to mass in linear motion.
In the given problem, the hoop has a larger moment of inertia \( I_{\text{hoop}} = mR^2 \), compared to the disk \( I_{\text{disk}} = \frac{1}{2}mR^2 \). Thus, with the same angular acceleration, it requires a greater net external torque to achieve the same rotational motion as the disk.
This means that in practice, more force is necessary to rotate or stop a hoop compared to a solid disk when subjected to the same angular conditions.
Newton's Second Law for Rotation
Newton's Second Law for rotation is a fundamental principle that relates to rotational motion, and it is expressed as:
  • \( \tau = I \alpha \)
This equation states that the torque applied to an object is the product of its moment of inertia \( I \) and the angular acceleration \( \alpha \) it undergoes. Moments of inertia differ based on an object's shape and mass distribution.
The law indicates that greater torque is needed for objects with larger moments of inertia to achieve the same angular acceleration. In our exercise, the hoop, compared to the disk, requires more torque due to its larger moment of inertia.
Understanding this law is crucial for solving rotational dynamics problems because it highlights how both the object's physical properties and the applied forces influence its rotational behavior. It echoes the linear form of Newton’s Second Law \( F = ma \) in the domain of rotation.

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Most popular questions from this chapter

Interactive Solution \(9.47\) at offers a model for solving problems of this type. A solid sphere is rolling on a surface. What fraction of its total kinetic energy is in the form of rotational kinetic energy about the center of mass?

A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{~kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{~s}\), and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{~m} .\) The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{k}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?

Concept Questions The drawing shows a rectangular piece of wood. The forces applied to corners \(\mathrm{B}\) and \(\mathrm{D}\) have the same magnitude and are directed parallel to the long and short sides of the rectangle. An axis of rotation is shown perpendicular to the plane of the rectangle at its center. (a) Relative to this axis, which force produces the torque with the greater magnitude? (b) A force is to be applied to corner A, directed along the short side of the rectangle. The net torque produced by the three forces is zero. How is the force at corner A directed- from A toward B or away from B? Give your reasoning. Problem The magnitudes of the forces at corners \(\mathrm{B}\) and \(\mathrm{D}\) are each \(12 \mathrm{~N}\). The long side of the rectangle is twice as long as the short side. Find the magnitude and direction of the force applied to corner \(\mathrm{A}\).

Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of \(125 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) When one person is standing at a distance of \(1.50 \mathrm{~m}\) from the center, the carousel has an angular velocity of \(0.600 \mathrm{rad} / \mathrm{s}\). However, as this person moves inward to a point located \(0.750 \mathrm{~m}\) from the center, the angular velocity increases to \(0.800 \mathrm{rad} / \mathrm{s}\). What is the person's mass?

The drawing shows a bicycle wheel resting against a small step whose height is \(h=0.120 \mathrm{~m}\). The weight and radius of the wheel are \(W=25.0 \mathrm{~N}\) and \(r=0.340 \mathrm{~m}\). A horizontal force \(\overrightarrow{\mathbf{F}}\) is applied to the axle of the wheel. As the magnitude of \(\overrightarrow{\mathbf{F}}\) increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. What is the magnitude of the force when this happens?

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