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Chapter 8: Problem 53

A motorcycle accelerates uniformly from rest and reaches a linear speed of \(22.0 \mathrm{m} / \mathrm{s}\) in a time of \(9.00 \mathrm{s}\). The radius of each tire is \(0.280 \mathrm{m}\) What is the magnitude of the angular acceleration of each tire?

Short Answer

Expert verified
The angular acceleration is \(8.71\, \mathrm{rad/s^2}\).

Step by step solution

01

Understand the Problem

We are given the linear acceleration details of a motorcycle and need to find the angular acceleration of its tires. The given values are initial speed \(u = 0\, \mathrm{m/s}\), final speed \(v = 22.0\, \mathrm{m/s}\), and time \(t = 9.00\, \mathrm{s}\). We also know the radius of the tire, \(r = 0.280\, \mathrm{m}\). The relationship between linear acceleration \(a\) and angular acceleration \(\alpha\) is \(a = r\alpha\).
02

Calculate Linear Acceleration

First, we calculate the linear acceleration \(a\) of the motorcycle using the formula for acceleration: \ \[ a = \frac{v - u}{t} \] Plugging in the given values: \ \[ a = \frac{22.0 \, \mathrm{m/s} - 0 \, \mathrm{m/s}}{9.00 \, \mathrm{s}} \] \ \[ a = \frac{22.0}{9.00} \, \mathrm{m/s^2} = 2.44 \, \mathrm{m/s^2} \]
03

Relate Linear and Angular Acceleration

The linear acceleration \(a\) is related to the angular acceleration \(\alpha\) by the equation: \ \[ a = r \alpha \] To find \(\alpha\), rearrange the equation: \ \[ \alpha = \frac{a}{r} \]
04

Calculate Angular Acceleration

Substitute the known values into the rearranged equation to find \(\alpha\): \ \[ \alpha = \frac{2.44 \, \mathrm{m/s^2}}{0.280 \, \mathrm{m}} \] \ \[ \alpha = 8.71 \, \mathrm{rad/s^2} \]

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Key Concepts

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

Linear Acceleration
Linear acceleration is the rate at which an object's velocity changes over time in a straight line. It is an important concept in physics, particularly when discussing movement along a path or trajectory. Here is what you need to know about linear acceleration:
  • It is denoted by the symbol \( a \).
  • Measured in meters per second squared (\( \mathrm{m/s^2} \)).
  • Calculated using the formula \[ a = \frac{v - u}{t} \] where \( v \) is the final velocity, \( u \) is the initial velocity, and \( t \) is the time over which the velocity changes.
In the context of the exercise, the motorcycle's linear acceleration is found to be \( 2.44 \, \mathrm{m/s^2} \). This indicates how quickly the bike has sped up along the roadway. Understanding linear acceleration helps us predict how long it will take for objects to speed up or slow down.
Uniform Acceleration
Uniform acceleration occurs when an object experiences a constant acceleration. This means the rate of change of velocity remains the same throughout its motion.
  • If an object is under uniform acceleration, its velocity changes at a consistent rate over time.
  • This consistency makes it easier to use kinematic equations to predict future motion.
  • For example, the formula for calculating velocity under uniform acceleration is \( v = u + at \).
In our motorcycle example, the vehicle accelerates uniformly from rest. This uniformity simplifies calculations and ensures the acceleration rate doesn鈥檛 vary unpredictably. Such scenarios are a common simplification in physics problems, making them easier to analyze.
Relation between Linear and Angular Motion
The relationship between linear and angular motion is pivotal in understanding how wheels and rotating objects function.
  • Linear motion involves movement in a straight path, while angular motion involves rotation around an axis.
  • These two types of motion are interrelated in that linear acceleration of the edges of a rotating object, like a wheel, can be determined by its angular acceleration.
  • Mathematically, the relation is given by \( a = r \alpha \), where \( a \) is the linear acceleration, \( r \) is the radius of the rotating object, and \( \alpha \) is the angular acceleration.
In practical terms, for the motorcycle in our exercise, knowing the tire radius and linear acceleration allows us to calculate the angular acceleration of the wheels as \( 8.71 \, \mathrm{rad/s^2} \). This illustrates how linear speeds translate to rotational dynamics, providing a comprehensive understanding of movement for both straight and circular paths.

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Most popular questions from this chapter

A Ferris wheel rotates at an angular velocity of \(0.24 \mathrm{rad} / \mathrm{s}\). Starting from rest, it reaches its operating speed with an average angular acceleration of \(0.030 \mathrm{rad} / \mathrm{s}^{2} .\) How long does it take the wheel to come up to operating speed?

The drive propeller of a ship starts from rest and accelerates at \(2.90 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) for \(2.10 \times 10^{3} \mathrm{s} .\) For the next \(1.40 \times 10^{3} \mathrm{s}\) the propeller rotates at a constant angular speed. Then it decelerates at \(2.30 \times 10^{-3} \mathrm{rad} / \mathrm{s}^{2}\) until it slows (without reversing direction) to an angular speed of \(4.00 \mathrm{rad} / \mathrm{s}\). Find the total angular displacement of the propeller.

A bicycle is rolling down a circular portion of a path; this portion of the path has a radius of 9.00 m. As the drawing illustrates, the angular displacement of the bicycle is 0.960 rad. What is the angle (in radians) through which each bicycle wheel (radius \(=0.400 \mathrm{m}\) ) rotates?

After 10.0 s, a spinning roulette wheel at a casino has slowed down to an angular velocity of \(+1.88 \mathrm{rad} / \mathrm{s} .\) During this time, the wheel has an angular acceleration of \(-5.04 \mathrm{rad} / \mathrm{s}^{2} .\) Determine the angular displacement of the wheel.

An automatic dryer spins wet clothes at an angular speed of \(5.2 \mathrm{rad} / \mathrm{s} .\) Starting from rest, the dryer reaches its operating speed with an average angular acceleration of \(4.0 \mathrm{rad} / \mathrm{s}^{2} .\) How long does it take the dryer to come up to speed?
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