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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

A small 0.500 -kg object moves on a frictionless horizontal table in a circular path of radius \(1.00 \mathrm{~m} .\) The angular speed is \(6.28 \mathrm{rad} / \mathrm{s} .\) The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than \(105 \mathrm{~N}\), what is the radius of the smallest possible circle on which the object can move?

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?

A thin uniform rod is rotating at an angular velocity of \(7.0 \mathrm{rad} / \mathrm{s}\) about an axis that is perpendicular to the rod at its center. As the drawing indicates, the rod is hinged at two places, one-quarter of the length from each end. Without the aid of external torques, the rod suddenly assumes a "u" shape, with the arms of the "u" parallel to the rotation axis. What is the angular velocity of the rotating "u"?

Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

A pair of forces with equal magnitudes, opposite directions, and different lines of action is called a "couple." When a couple acts on a rigid object, the couple produces a torque that does not depend on the location of the axis. The drawing shows a couple acting on a tire wrench, each force being perpendicular to the wrench. Determine an expression for the torque produced by the couple when the axis is perpendicular to the tire and passes through (a) point \(\mathrm{A},\) (b) point \(\mathrm{B}\), and (c) point C. Express your answers in terms of the magnitude \(F\) of the force and the length \(L\) of the wrench.

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