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Chapter 8: Problem 57

A motorcycle, which has an initial linear speed of \(6.6 \mathrm{m} / \mathrm{s}\), decelerates to a speed of \(2.1 \mathrm{m} / \mathrm{s}\) in \(5.0 \mathrm{s}\). Each wheel has a radius of \(0.65 \mathrm{m}\) and is rotating in a counterclockwise (positive) direction. What are (a) the constant angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2}\) ) and (b) the angular displacement (in rad) of each wheel?

Short Answer

Expert verified
(a) The angular acceleration is -1.384 rad/s虏. (b) The angular displacement is 33.45 rad.

Step by step solution

01

Convert Linear Speeds to Angular Speeds

The angular speed \(\omega\) can be found from the linear speed \(v\) and the radius \(r\) using the formula:\[ \omega = \frac{v}{r} \]Using the initial speed \(v_i = 6.6 \ \text{m/s}\) and radius \(r = 0.65 \ \text{m}\):\[ \omega_i = \frac{6.6}{0.65} = 10.15 \ \text{rad/s} \]Using the final speed \(v_f = 2.1 \ \text{m/s}\) and the same radius:\[ \omega_f = \frac{2.1}{0.65} = 3.23 \ \text{rad/s} \]
02

Calculate the Angular Acceleration

Angular acceleration \(\alpha\) is the change in angular speed over time \(t\):\[ \alpha = \frac{\omega_f - \omega_i}{t} \]Substituting in the values:\[ \alpha = \frac{3.23 - 10.15}{5.0} = -1.384 \ \text{rad/s}^2 \]
03

Calculate the Angular Displacement

Angular displacement \(\theta\) can be calculated using the formula for uniformly accelerated motion:\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]Substitute \(\omega_i = 10.15 \ \text{rad/s}\), \(t = 5.0 \ \text{s}\), and \(\alpha = -1.384 \ \text{rad/s}^2\):\[ \theta = 10.15 \times 5.0 + \frac{1}{2} \times (-1.384) \times (5.0)^2 \]\[ \theta = 50.75 - 17.3 = 33.45 \ \text{rad} \]

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Key Concepts

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

Linear Speed
Linear speed is a measure of how fast an object moves along a path. It describes the rate of change of the object's position with respect to time. In simple terms, it's how fast something is moving in a straight line.
  • The formula for linear speed is given by: \[ v = \frac{\text{distance}}{\text{time}} \]
  • In the context of a rotating wheel, the linear speed at any point on the circumference can be directly related to the wheel's rotational motion.
  • In the exercise, the motorcycle's initial and final linear speeds are important for transforming linear motion into rotational motion.
Understanding linear speed helps us connect how translation (or moving in a straight line) relates to rotation.
Angular Speed
Angular speed, often denoted by \( \omega \), is the rate at which an object rotates or revolves. It's the rotational counterpart to linear speed and describes how quickly an angle is changing as an object rotates around an axis.
  • Angular speed can be calculated by dividing the linear speed by the radius of the rotation path:\[ \omega = \frac{v}{r} \]
  • This formula demonstrates how an increase in linear speed or a decrease in radius will result in a higher angular speed.
  • In the exercise, converting the motorcycle's linear speeds to angular speeds is crucial for understanding its wheel's rotational motion.
Grasping the concept of angular speed helps explain the connection between how fast something spins and its linear velocity.
Angular Displacement
Angular displacement is a measure of how much an object has rotated or turned and is usually expressed in radians. It tells us about the change in an object's angle as it rotates, providing insight into the extent of rotation over a certain period of time.
  • The formula considering constant angular acceleration is:\[ \theta = \omega_i t + \frac{1}{2} \alpha t^2 \]
  • This formula accounts for initial angular speed, time, and angular acceleration, thus effectively linking rotational motion to the decision made in the exercise about calculating the motorcycle wheel's displacement.
  • Using these parameters, we can determine the wheel's displacement during the given period, providing a complete picture of the rotational movement.
Understanding angular displacement is key to solving problems involving rotational motion, as it gives a direct measure of how much rotation has occurred.

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Most popular questions from this chapter

Conceptual Example 2 provides some relevant background for this problem. A jet is circling an airport control tower at a distance of \(18.0 \mathrm{km}\) An observer in the tower watches the jet cross in front of the moon. As seen from the tower, the moon subtends an angle of \(9.04 \times 10^{-3}\) radians. Find the distance traveled (in meters) by the jet as the observer watches the nose of the jet cross from one side of the moon to the other.

A baseball pitcher throws a baseball horizontally at a linear speed of \(42.5 \mathrm{m} / \mathrm{s}\) (about \(95 \mathrm{mi} / \mathrm{h}\) ). Before being caught, the baseball travels a horizontal distance of \(16.5 \mathrm{m}\) and rotates through an angle of \(49.0 \mathrm{rad}\). The baseball has a radius of \(3.67 \mathrm{cm}\) and is rotating about an axis as it travels, much like the earth does. What is the tangential speed of a point on the "equator" of the baseball?

The penny-farthing is a bicycle that was popular between 1870 and \(1890 .\) As the drawing shows, this type of bicycle has a large front wheel and a small rear wheel. During a ride, the front wheel (radius \(=1.20 \mathrm{m})\) makes 276 revolutions. How many revolutions does the rear wheel (radius \(=0.340 \mathrm{m}\) ) make?

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?

At the local swimming hole, a favorite trick is to run horizontally off a cliff that is \(8.3 \mathrm{m}\) above the water. One diver runs off the edge of the cliff, tucks into a "ball," and rotates on the way down with an average angular speed of 1.6 rev/s. Ignore air resistance and determine the number of revolutions she makes while on the way down.
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