Question

A certain wheel turns through 90 rev in \(15 \mathrm{~s}\), its angular speed at the end of the period being \(10 \mathrm{rev} / \mathrm{s} .(a)\) What was the angular speed of the wheel at the beginning of the 15 -s interval, assuming constant angular acceleration? \((b)\) How much time had elapsed between the time the wheel was at rest and the beginning of the 15 -s interval?

Short answer

The initial angular speed of the wheel was 4 revolutions per second and the time that had passed when the wheel was at rest was approximately 1.6 seconds.

Step-by-step solution

Calculation of angular acceleration

Use the formula for final speed in uniformly accelerated motion \( v_f = v_i + at \) which becomes \( \omega_f = \omega_i + \alpha t \), where \(\omega_f\) is the final angular speed, \(\omega_i\) is the initial angular speed, \(\alpha\) is the angular acceleration, and t is the time. We are given that \(\omega_f = 10 \)\( \mathrm{rev/s} \), t= 15 s and \(\omega_i\) has to be found. First we need to calculate the total angular distance covered (D) by the wheel from its formula \( D = \omega_i t + 0.5 \alpha t^2 \). We know D = 90 rev, t = 15 s and need to find \(\alpha\). And from calculus we know \( \omega_f = \omega_i + \alpha t \) so \( \alpha = (\omega_f - \omega_i) / t \) and we can isolate this to find \(\alpha = (D - \omega_i t) / 0.5 t^2 \).

Calculate initial angular speed

Next, generate an equation system based on the two equations for \(\alpha\) and solve for \(\omega_i\). The equation system would look like: \( \omega_i = 10 - 15 \alpha \) and \( \alpha = (90 - \omega_i * 15) / (0.5 * 15^2) \). After substitution and simplification, one finds that the initial angular speed, \(\omega_i\), was 4 revolution per second.

Find the time elapsed for the wheel to come to rest

For this step use the formula for final speed, \( \omega_f = \omega_i + \alpha t \). But when the wheel was at rest, \(\omega_f = 0 \mathrm{rev/s}\). So the formula becomes \(0 = 4 + \alpha t \). Solving this equation yields the time elapsed before the wheel was set into motion. Here, \(\alpha\) is the angular acceleration calculated in the first step. After substitution, you get that the time was approximately 1.6 seconds.

Key concepts

Angular Acceleration

Angular acceleration is a measure of how quickly the angular speed of an object changes over time. It's like the rotational equivalent of linear acceleration. When a wheel is speeding up or slowing down, it undergoes angular acceleration. This concept is crucial for understanding any scenario involving rotational motion, like our wheel problem.

To calculate angular acceleration (\( \alpha \)), we use the formula \( \alpha = \frac{\omega_f - \omega_i}{t} \), where \( \omega_f \) is the final angular speed, \( \omega_i \) is the initial angular speed, and \( t \) is the time over which the change occurs.

In our exercise, by rearranging the formula and substituting known values, the angular acceleration can be precisely determined. This foundational step helps to solve other aspects of the problem, like finding initial angular speed or the time taken to come to rest.

Angular Speed

Angular speed is how fast an object rotates or revolves relative to another point. It tells you how quickly an angle changes over time, often measured in revolutions per second or radians per second.

For the wheel, its angular speed at the end of the 15-second interval was given as 10 revolutions per second. This speed helps us calculate earlier speeds and the total angular movement.

Initial angular speed (\( \omega_i \)) can be thought of as the starting speed before any acceleration has affected the wheel. By using an understanding of uniformly accelerated motion, we find that the wheel's starting angular speed was 4 revolutions per second, showing a consistent change over the given time.

Uniformly Accelerated Motion

Uniformly accelerated motion occurs when an object experiences constant acceleration. In angular kinematics, this means the angular acceleration remains the same throughout the motion.

This condition simplifies calculations, allowing us to use straightforward equations of motion. These equations help predict the object's behavior across the timeline of motion, even in rotational scenarios.

Our wheel exercise assumes uniformly accelerated motion, meaning once we have the angular acceleration, predictions about angular speeds and distances become manageable relative to time.

Angular Distance

Angular distance refers to the amount of rotation an object undergoes. It's similar to linear distance but for circular paths. In our problem, the wheel turns through a complete 90 revolutions in 15 seconds.

To find the angular distance (\( D \)) in calculations, we use the formula \( D = \omega_i t + 0.5 \alpha t^2 \), akin to distance in linear motion equations. Knowing the wheel's angular distance informs us about its rotation, helping to guess initial speeds and calculate accelerations.

By understanding this distance, coupled with time and speed, all rotational parameters align to solve the equation of motion.

Equations of Motion

Equations of motion provide a structured way to relate various elements of kinematics, like speed, distance, time, and acceleration. They are essential for both linear and angular motion calculations.

In rotational scenarios, these equations adjust to handle angles instead of straightforward distances. We use forms like \( \omega_f = \omega_i + \alpha t \) and \( D = \omega_i t + 0.5 \alpha t^2 \), showing how components interlink during motion.

For the wheel, these equations enabled the calculation of unknown parameters, such as initial speed and the time it took to reach or stop from rest. Mastery of these formulas paves the way to effortlessly navigate and solve real-life angular kinematics problems.