Problem 17 - University Physics with Modern Physics

Question

A safety device brings the blade of a power mower from an initial angular speed of \(\omega_1\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_3\) that was three times as great, \(\omega_3 = 3\omega_1\)?

Step-by-Step Solution

Answer

18 revolutions.

Step 1: Understand the Problem

We need to find out how many revolutions it takes for the blade to stop from an angular speed that is three times the initial speed, given it takes 1 revolution to stop from the initial speed \( \omega_1 \).

Step 2: Use Kinematic Equation

The relevant kinematic equation for angular motion relates the final angular velocity \( \omega_f \), initial angular velocity \( \omega_i \), angular acceleration \( \alpha \), and angular displacement \( \theta \): \[ \omega_f^2 = \omega_i^2 + 2\alpha \theta \]. Here \( \omega_f = 0 \) (the blade stops), \( \omega_i = \omega_1 \), and \( \theta = 1 \text{ revolution} \).

Step 3: Find Angular Acceleration

Since we know \( \theta = 1 \text{ revolution} = 2\pi \text{ radians} \) when the blade starts stopping from \( \omega_1 \), we can solve the previous step's equation: \( 0 = \omega_1^2 + 2\alpha (2\pi) \), which yields \( \alpha = -\frac{\omega_1^2}{4\pi} \).

Step 4: Calculate New Angular Displacement

Now find \( \theta' \) when \( \omega_i = 3\omega_1 \) using the same kinematic equation: \( 0 = (3\omega_1)^2 + 2\alpha\theta' \). Replacing \( \alpha \), we have \( 9\omega_1^2 = \frac{\omega_1^2}{2\pi} \theta' \). Simplify to get \( \theta' = \frac{9\cdot4\pi}{2\pi} = 18 \text{ revolutions} \).

Key Concepts

Angular Speed

Angular speed is a measure of the rate at which an object rotates or spins about a central point. It is a vector quantity and is commonly represented by the Greek letter \( \omega \). Angular speed is critical when analyzing rotational motion, such as the spinning blade of a power mower. In this context, the initial angular speed is denoted as \( \omega_1 \).

• Angular speed is measured in radians per second (rad/s).
• It describes how quickly the angle is changing as the object spins.

Understanding angular speed is crucial for predicting how an object will behave as it slows down or speeds up. Higher angular speed indicates faster rotation, which in turn affects how quickly the object may come to rest or change direction.

Angular Acceleration

Angular acceleration represents how quickly the angular speed of an object is changing over time. It is denoted by the Greek letter \( \alpha \) and is an essential factor when evaluating rotational dynamics.

• Angular acceleration is measured in radians per second squared (rad/s²).
• It can be positive (speeding up) or negative (slowing down).

In the problem, the blade experiences a constant angular acceleration as it slows to a halt. Calculating this acceleration involves using the kinematic equation that links initial and final angular speeds, angular acceleration, and angular displacement.

Kinematic Equations in Angular Motion

Kinematic equations are vital tools for analyzing motion, and they can be adapted for use in angular motion contexts. These equations help to establish relationships between angular speed, angular acceleration, and angular displacement. In our problem, the equation used is:\[ \omega_f^2 = \omega_i^2 + 2\alpha \theta \]

• \( \omega_f \) is the final angular speed (zero when the object stops).
• \( \omega_i \) is the initial angular speed.
• \( \alpha \) is the angular acceleration.
• \( \theta \) is the angular displacement.

In the exercise, this equation helped determine the number of revolutions the blade makes before stopping. By substituting known values, we can solve for the unknowns such as total revolutions required to bring a faster spinning blade to a complete halt.