/*! 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 72 A rock stuck in the tread of a 6... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 72

A rock stuck in the tread of a 60.0 -cm-diameter bicycle wheel has a tangential speed of \(3.00 \mathrm{m} / \mathrm{s} .\) When the brakes are applied, the rock's tangential deceleration is \(1.00 \mathrm{m} / \mathrm{s}^{2}\) a. What are the magnitudes of the rock's angular velocity and angular acceleration at \(t=1.50 \mathrm{s} ?\) b. At what time is the magnitude of the rock's acceleration equal to \(g\)?

Short Answer

Expert verified
a. The angular velocity at \( t=1.50 \, s \) is \( 8.50 \, rad/s \) and the angular acceleration is \( -3.33 \, rad/s^2 \). b. The magnitude of the rock's acceleration is equal to \( g \) at \( t=0.306 \, s \).

Step by step solution

01

Determine the Angular Velocity

Angular velocity (\( \omega \)) at any point can be defined as tangential speed divided by radius. In this case, the diameter is given. So first, the radius should be determined, which is half the diameter. Thus, \( r = 60.0 \, cm / 2 = 30.0 \, cm = 0.30 \, m \). So, let's calculate the angular velocity (\( \omega \)): \( \omega = v / r = 3.00 \, m/s / 0.30 \, m = 10.0 \, rad/s \). Now, after braking, the speed decreases. Angular deceleration is considered to be a negative acceleration. Thus, for \( t=1.50 \, s \), we find the new angular velocity (\( \omega' \)) using the equation \( \omega' = \omega + a * t \), where \( a = -1.00 \, m/s^2 \). Thus, \( \omega' = 10.0 \, rad/s - 1.00 \, m/s^2 * 1.50 \, s = 8.50 \, rad/s \).
02

Determine Angular Acceleration

Angular acceleration (\( \alpha \)) is the change in angular speed divided by time. Since we have defined our deceleration as a negative acceleration, we can write the expression as \( \alpha = - a / r = -1.00 \, m/s^2 / 0.30 \, m = -3.33 \, rad/s^2 \).
03

Calculate When Acceleration Equals g

We can find when the magnitude of the rock's acceleration equals \( g \) by setting the acceleration equal to \( g \) and solving for \( t \). Acceleration ( \( a \) ) at any time can be found using formula \( a = v - gt \). Substitute \( a \) to \( g \) in the equation and solve for \( t \) : \( 3.00 \, m/s - 9.81 \, m/s^2 \cdot t = 0 \) which gives \( t = 3.00 \, m/s / 9.81 \, m/s^2 = 0.306 \, s \). Therefore, the acceleration equals \( g \) at \( t = 0.306 \, s \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Tangential Speed
In the realm of rotational motion, tangential speed is a measure of how fast a point on a rotating object travels in its circular path. It's the linear speed at a point located a certain distance from the axis of rotation. Typically, it is calculated as the product of the angular velocity and the radius of rotation. In our exercise, a rock lodged in a bicycle wheel tread exemplifies a point on a rotating object. At a tangential speed of 3.00 m/s, the rock moves swiftly when the bicycle is in motion, making a complete circle along the wheel's circumference depending on the speed of rotation.

Understanding tangential speed is crucial because it links circular motion to linear motion, enabling students to predict and calculate the speed of any point on a rotating object. Notably, tangential speed is always perpendicular to the radius and may change if either the angular velocity or radius changes - an important consideration in real-life applications like adjusting the speed of a record player or the wheels of a vehicle.
Angular Deceleration
Imagine you're riding your bike and begin to apply the brakes. The slowing down of your bike's wheels is due to angular deceleration. Angular deceleration, also known as angular retardation, is the rate at which an object's angular velocity decreases over time. It's the rotational equivalent of linear deceleration. For an object to decelerate tangentially, there must be an opposite force applied to its motion, such as friction from brake pads. In our problem, when the rock experiences a tangential deceleration of 1.00 m/s虏, this refers to the linear deceleration at the rock's position along the wheel's edge. To find angular deceleration, we divide the tangential deceleration by the radius of the wheel.

By grasping the concept of angular deceleration, students can solve problems involving changing rotational speeds. This can apply to scenarios ranging from the careful stopping of a spinning satellite to the controlled deceleration of a car coming to a full stop.
Acceleration Due to Gravity
Acceleration due to gravity, denoted as 'g', is a critical concept in physics, representing the rate at which objects accelerate when dropped in a vacuum near Earth's surface. It is approximately 9.81 m/s虏. This acceleration is caused by the gravitational pull of the Earth and acts on all objects regardless of their mass. In the context of our exercise, the rock on the bicycle wheel will eventually accelerate downward at this rate when it becomes separated from the wheel. Moreover, when the wheel's tangential acceleration opposite to the direction of motion equals the magnitude of 'g', it experiences an equivalent rate of change in velocity as a freely falling object.

Understanding 'g' allows students to analyze the motion of projectiles, the free fall of objects, and even the forces experienced on rollercoasters. It is a universal force experienced by all matter under Earth's gravitational influence and a key concept for many practical applications in engineering and science.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A magnetic computer disk \(8.0 \mathrm{cm}\) in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at \(600 \mathrm{rad} / \mathrm{s}^{2}\) for \(\frac{1}{2} \mathrm{s},\) then coasts at a steady angular velocity for another \(\frac{1}{2}\) s. What is the speed of the dot at \(t=1.0 \mathrm{s} ?\) Through how many revolutions has the disk turned?

Consider a pendulum swinging back and forth on a string. Use a motion diagram analysis and a written explanation to answer the following questions. a. At the lowest point in the motion, is the velocity zero or nonzero? Is the acceleration zero or nonzero? If these vectors aren't zero, which way do they point? b. At the end of its arc, when the pendulum is at the highest point on the right or left side, is the velocity zero or nonzero? Is the acceleration zero or nonzero? If these vectors aren't zero, which way do they point?

A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. a. What is the acceleration at the end of a test tube that is \(10 \mathrm{cm}\) from the axis of rotation? b. For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of \(1.0 \mathrm{m}\) and stopped in a 1.0 -ms-long encounter with a hard floor?

A child in danger of drowning in a river is being carried downstream by a current that flows uniformly with a speed of \(2.0 \mathrm{m} / \mathrm{s}\) The child is 200 m from the shore and 1500 m upstream of the boat dock from which the rescue team sets out. If their boat speed is \(8.0 \mathrm{m} / \mathrm{s}\) with respect to the water, at what angle from the shore should the pilot leave the shore to go directly to the child?

King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of \(50 \mathrm{m} / \mathrm{s}\) and an angle of \(30^{\circ} .\) A cannonball that was accidentally dropped hits the moat below in \(1.5 \mathrm{s}\) a. How far from the castle wall does the cannonball hit the ground? b. What is the ball's maximum height above the ground?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Famous Physicists

Read Explanation

Modern Physics

Read Explanation

Classical Mechanics

Read Explanation

Engineering Physics

Read Explanation

Astrophysics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.