/*! 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 57 A pulley, with a rotational iner... [FREE SOLUTION] | 91Ó°ÊÓ

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A pulley, with a rotational inertia of \(1.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its axle and a radius of \(10 \mathrm{~cm},\) is acted on by a force applied tangentially at its rim. The force magnitude varies in time as \(F=0.50 t+0.30 t^{2},\) with \(F\) in newtons and \(t\) in seconds. The pulley is initially at rest. At \(t=3.0 \mathrm{~s}\) what are its (a) angular acceleration and (b) angular speed?

Short Answer

Expert verified
(a) Angular acceleration: 420 rad/s²; (b) Angular speed: 1260 rad/s at t = 3 s.

Step by step solution

01

Understand Force Variation Over Time

The force applied to the pulley is given as a function of time: \( F(t) = 0.50t + 0.30t^2 \). This function describes how the force increases as time progresses.
02

Calculate Torque at Specific Time

Torque \( \tau \) is calculated by multiplying the force \( F \) by the radius \( r \) of the pulley. At \( t = 3.0 \text{ s} \), the force is \( F = 0.50(3) + 0.30(3)^2 = 0.50 \times 3 + 0.30 \times 9 = 1.5 + 2.7 = 4.2\, \text{N} \). Thus, torque is \( \tau = r \cdot F = 0.10 \cdot 4.2 = 0.42 \text{ Nm} \).
03

Determine Angular Acceleration

The angular acceleration \( \alpha \) is found using the relation \( \tau = I \cdot \alpha \), where \( I \) is the moment of inertia. With \( I=1.0 \times 10^{-3} \text{ kg} \cdot \text{m}^2 \), solve for \( \alpha \): \( \alpha = \frac{\tau}{I} = \frac{0.42}{1.0 \times 10^{-3}} = 420 \text{ rad/s}^2 \).
04

Integrate Angular Acceleration to Find Angular Speed

Since the pulley is initially at rest, its initial angular velocity \( \omega_0 = 0 \). The angular speed \( \omega \) at time \( t \) is obtained by integrating the angular acceleration: \( \omega = \int \alpha \, dt = \alpha \cdot t + \omega_0 \). At \( t = 3.0 \text{ s} \), \( \omega = 420 \times 3 = 1260 \text{ rad/s} \).

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Key Concepts

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

Torque
Torque is a fundamental concept in rotational dynamics. It measures how much a force acting on an object causes it to rotate. Think of torque as a twist or a spin generated by force applied at a certain distance from the axis of rotation. This distance is known as the radius of the pulley.

Torque (\( \tau \)) is calculated by multiplying the force (\( F \)) applied to the rim of a pulley by its radius (\( r \)). Mathematically:
  • \( \tau = r \cdot F \)
In the original exercise, the force changes with time, modeled as \( F(t) = 0.50t + 0.30t^2 \). By plugging in the value at \( t = 3 \text{s} \), we calculate the force and subsequently the torque. Understanding that torque is central to controlling the rotational motion helps break down complex mechanical problems into more manageable parts.
Angular Acceleration
Angular acceleration describes how quickly an object's rotational speed changes. Imagine a spinning wheel that speeds up; the rate of this speed increase is its angular acceleration.

In rotational dynamics, angular acceleration (\( \alpha \)) is linked to torque and the object's moment of inertia with the formula:
  • \( \tau = I \cdot \alpha \)
Where \( I \) is the moment of inertia. Solving for \( \alpha \) gives:
  • \( \alpha = \frac{\tau}{I} \)
For the pulley in the exercise, calculating torque allows us to find angular acceleration. This value helps understand how fast the pulley begins to spin faster at the given point in time.
Moment of Inertia
Moment of inertia is the rotational equivalent of mass in linear dynamics. It's a measure of an object's resistance to changes in its rotation. A heavier or larger object will have a higher moment of inertia, making it harder to start or stop spinning.

Represented by \( I \), moment of inertia depends on both the mass distribution in the object and the axis of rotation:
  • \( I = \sum m r^2 \)
where \( m \) is mass and \( r \) is the radius.

In the pulley system exercise, the moment of inertia was given as \( 1.0 \times 10^{-3} \text{ kg} \cdot \text{m}^2 \). This relatively small value indicates that the pulley can be easily accelerated by applying torque. It directly ties into determining angular acceleration, linking back to how the system responds dynamically to applied forces.
Pulley System
A pulley system is a simple machine used to change the direction or magnitude of a force. It's particularly effective in rotating systems where forces are applied tangentially, like the exercise example.

In the context of rotational dynamics, pulleys allow for efficient force application to produce torque. The setup involves:
  • A wheel or drum equipped to rotate around an axle
  • A force applied to the rim
These elements combine to create motion through torque and change the angular acceleration of the wheel.

Understanding pulley systems in terms of these dynamics helps uncover the principles driving everyday machinery. In our exercise, applying the force strategy to the rim and observing how it evolves over time is crucial for analyzing resultant rotational motion.

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Most popular questions from this chapter

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