/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Wheel and shaft \(^{*}\) A wheel... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 11

Wheel and shaft \(^{*}\) A wheel is attached to a fixed shaft, and the system is free to rotate without friction. To measure the moment of inertia of the wheelshaft system, a tape of negligible mass wrapped around the shaft is pulled with a known constant force \(F\). When a length \(L\) of tape has unwound, the system is rotating with angular speed \(\omega_{0}\). Find the moment of inertia \(I_{0}\) of the system.

Short Answer

Expert verified
The moment of inertia is \( I_0 = \frac{2FL}{\omega_0^2} \).

Step by step solution

01

Analyze the forces and motion

Considering the tape, when it is pulled with a constant force, it provides a constant torque on the shaft. Since there is no friction, all the force translates into rotating the wheel.
02

Relate force and torque

The torque \( \tau \) applied by the force \( F \) on the shaft is given by \( \tau = F \cdot r \), where \( r \) is the radius of the shaft. Torque is responsible for angular acceleration.
03

Use rotational kinematics

The relationship between angular acceleration \( \alpha \) and torque \( \tau \) is given by \( \tau = I_0 \cdot \alpha \), where \( I_0 \) is the moment of inertia which we want to find. Thus, \( \alpha = \frac{\tau}{I_0} = \frac{F \cdot r}{I_0} \).
04

Compute work done

The work done by the force can also be used to find the moment of inertia. The work done \( W \) as the tape unwinds a length \( L \) is \( W = F \times L \).
05

Relate work done to kinetic energy

The work done is equal to the change in rotational kinetic energy, which can be expressed as \( \frac{1}{2} I_0 \omega_0^2 \). Therefore, equating work and kinetic energy, we have \( F \cdot L = \frac{1}{2} I_0 \omega_0^2 \).
06

Solve for the moment of inertia

From \( F \cdot L = \frac{1}{2} I_0 \omega_0^2 \), we solve for \( I_0 \) to get \( I_0 = \frac{2FL}{\omega_0^2} \).

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Key Concepts

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

Understanding Angular Acceleration
Angular acceleration is a fundamental concept when you're dealing with rotating objects, like in our wheel-shaft system exercise. It refers to how quickly the angular velocity of an object changes over time. Let's break it down for clarity:
  • Definition: Angular acceleration, denoted by \( \alpha \), is the rate of change of angular velocity. It indicates how fast something is speeding up or slowing down in its rotational motion.
  • Units: Like linear acceleration is measured in meters per second squared (m/s") ), angular acceleration is measured in radians per second squared (rad/s虏).
  • Relation to Torque: Torque, which we'll discuss shortly, directly influences angular acceleration through the equation \( \tau = I \cdot \alpha \). This shows that the torque applied to an object will change its angular acceleration.
As we computed in the exercise, the relationship between the angular acceleration and the torque also depends on another important property of the system: moment of inertia, \( I_0 \). The moment of inertia serves as the rotational analog of mass in linear systems.Having a grasp of angular acceleration helps in predicting how quickly an object will reach a certain angular speed and is crucial when solving problems involving rotating systems.
Rotational Kinematics Made Simple
Rotational kinematics provides the tools and formulas to describe the motion of rotating objects. Just as linear kinematics describes motion through space, rotational kinematics describes motion around an axis. Here's what you need to know:
  • Key Variables: In rotational kinematics, the key quantities include angular displacement (\( \theta \)), angular velocity (\( \omega \)), and angular acceleration (\( \alpha \)).
  • Equations of Motion: The equations used in rotational motion are quite similar to those in linear motion but involve angular terms. A common equation is \( \omega = \omega_0 + \alpha t \), which shows how angular velocity changes with angular acceleration over time.
  • Energy Considerations: Rotational kinetic energy is also a part of rotational kinematics. The equation \( KE = \frac{1}{2} I \omega^2 \) describes how the rotational motion can contribute to the overall energy of the system.
When a piece of tape unwinds off the wheel-shaft system, the rotational kinematics allows us to determine the final angular velocity based on time and initial conditions.Understanding these principles is central to solving problems like the one given, where the unwinding tape contributes to the rotational speed of the wheel.
The Role of Torque in Rotational Dynamics
Torque is a key player in rotational dynamics, similar to how force is in linear dynamics. It is the "twist" applied to an object that causes it to rotate. Here's a simple explanation:
  • Definition: Torque (\( \tau \)) is the measure of the force causing an object to rotate about an axis. The equation is \( \tau = F \times r \), where \( F \) is the applied force and \( r \) is the distance from the axis of rotation.
  • Effect on Motion: Torque changes an object's angular momentum, causing angular acceleration (discussed earlier). Therefore, it's critical in changing the rotational state of an object.
  • Units: Torque is measured in Newton-meters (Nm). It's important to remember that it's a vector quantity, meaning it has both a direction and magnitude.
In our exercise, when the tape was pulled with a force, it generated torque to spin the wheel. Without this torque, the wheel-shaft system wouldn鈥檛 achieve any angular acceleration.Understanding torque allows us to see how rotational forces work and is essential for analyzing systems in mechanical physics. By mastering torque, you can predict how rotating bodies behave under various forces.

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Most popular questions from this chapter

Two drums Drum \(A\) of mass \(M\) and radius \(R\) is suspended from a drum \(B\) also of mass \(M\) and radius \(R\), which is free to rotate around its axis. The suspension is in the form of a massless metal tape wound around the outside of each drum, and free to unwind, as shown. Gravity is directed downward. Both drums are initially at rest. Find the initial acceleration of drum \(A\), assuming that it moves straight down.

Yo-yo motion A yo-yo of mass \(M\) has an axle of radius \(b\) and a spool of radius \(R\). Its moment of inertia can be taken to be \(M R^{2} / 2\) and the thickness of the string can be neglected. The yo-yo is released from rest. (a) What is the tension in the cord as the yo-yo descends and as it ascends? (b) The center of the yo-yo descends distance \(h\) before the string is fully unwound. Assuming that it reverses direction with uniform spin velocity, find the average force on the string while the yo-yo turns around.

Ring and bug" A ring of mass \(M\) and radius \(R\) lies on its side on a frictionless table. It is pivoted to the table at its rim. A bug of mass \(m\) walks around the ring with speed \(v\), starting at the pivot. What is the rotational velocity of the ring when the bug is (a) halfway around? (b) back at the pivot?

Disk pendulum A physical pendulum is made of a uniform disk of mass \(M\) and radius \(R\) suspended from a rod of negligible mass. The distance from the pivot to the center of the disk is \(l\). What value of \(l\) makes the period a minimum?

Bead and rod A bead of mass \(m\) slides without friction on a rod that is made to rotate at a constant angular speed \(\omega\). Neglect gravity. (a) Show that \(r=r_{0} e^{\text {ott }}\) is a possible motion of the bead, where \(r_{0}\) is the initial distance of the bead from the pivot. (b) For the motion described in part \((a)\), find the force exerted on the bead by the rod. (c) For the motion described above, find the power exerted by the agency that is turning the rod and show by direct calculation that this power equals the rate of change of kinetic energy of the bead.
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