Problem 10 - University Physics with Modern Physics

Question

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s. (a) Find the angular acceleration in rev/s\(^2\) and the number of revolutions made by the motor in the 4.00-s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

Step-by-Step Solution

Answer

Angular acceleration is -1.25 rev/s², and 23.32 revolutions occur. It takes an additional 2.664 seconds to stop.

Step 1: Convert angular velocities to consistent units

First, convert the given angular velocities from revolutions per minute (rev/min) to revolutions per second (rev/s). For this, we'll use the conversion factor: \(1 \text{ min} = 60 \text{ s}\).Initial angular velocity, \( \omega_i = 500 \text{ rev/min} = \frac{500}{60} \text{ rev/s} \approx 8.33 \text{ rev/s} \).Final angular velocity, \( \omega_f = 200 \text{ rev/min} = \frac{200}{60} \text{ rev/s} \approx 3.33 \text{ rev/s} \).

Step 2: Calculate angular acceleration

The angular acceleration \( \alpha \) is given by the formula:\[ \alpha = \frac{\omega_f - \omega_i}{\Delta t} \]where \( \Delta t = 4.00 \text{ s} \) is the time interval.Substituting the values, we find:\[ \alpha = \frac{3.33 \text{ rev/s} - 8.33 \text{ rev/s}}{4.00 \text{ s}} = \frac{-5.00 \text{ rev/s}}{4.00 \text{ s}} = -1.25 \text{ rev/s}^2 \].

Step 3: Calculate the number of revolutions during deceleration

We can find the number of revolutions, \( \theta \), using the equation:\[ \theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2 \]Substituting the known values, we get:\[ \theta = 8.33 \text{ rev/s} \times 4.00 \text{ s} + \frac{1}{2} \times (-1.25 \text{ rev/s}^2) \times (4.00 \text{ s})^2 \]\[ \theta = 33.32 \text{ rev} - 10.00 \text{ rev} = 23.32 \text{ rev} \].

Step 4: Calculate time to come to rest

To find the additional time \( t' \) for the fan to come to a complete stop, use:\[ \omega_f = \omega_i + \alpha t \]We know \( \omega_f = 0 \underline{\phantom{xxx}} \text{rev/s (at rest)}, \omega_i = 3.33 \text{ rev/s}\), and \(\alpha = -1.25 \text{ rev/s}^2\).Rearranging to solve for \( t' \):\[ 0 = 3.33 \text{ rev/s} + (-1.25 \text{ rev/s}^2) \times t' \]\[ 1.25 \text{ rev/s}^2 \times t' = 3.33 \text{ rev/s} \]\[ t' = \frac{3.33 \text{ rev/s}}{1.25 \text{ rev/s}^2} \approx 2.664 \text{ s} \].

Key Concepts

Angular Velocity

Angular velocity is a measure of how fast an object rotates or revolves around a point in a circular path. It's similar to linear velocity, but instead of measuring the speed along a straight line, it describes the rate of rotation in a circle. Angular velocity is usually denoted by the symbol \( \omega \) and is measured in units like revolutions per second (rev/s) or radians per second (rad/s).

When you have an object like a fan, and you measure how quickly it spins in complete circles over a given time, you're working with angular velocity. In the given example, the fan's initial angular velocity was 500 revolutions per minute, which we converted to approximately 8.33 revolutions per second. This tells you how quickly the fan's blades are spinning when the fan is first turned off.

To understand this better, imagine a second hand on a clock. If it completes one full turn (revolution) every 60 seconds, its angular velocity would be 1/60 rev/s. However, for analyses, angular velocity is often expressed in more manageable unit proportions, like revolutions per second.

Angular Acceleration

Angular acceleration describes how angular velocity changes over time. It's the rotational counterpart to linear acceleration and is denoted by the symbol \( \alpha \). The units for angular acceleration are typically revolutions per second squared (rev/s\(^2\)) or radians per second squared (rad/s\(^2\)).

In this problem, the fan slows down uniformly, meaning it has a constant angular acceleration. This angular acceleration is negative because the fan is decelerating, or slowing down. We calculated it as \( \alpha = -1.25 \) rev/s\(^2\). This negative sign indicates a reduction in angular velocity, unlike positive acceleration, which implies an increase.

A practical way to visualize angular acceleration is by thinking about a carousel slowing to a stop. As it slows, each second, it spins a little slower than before. This change in spin rate every second is what we call angular acceleration. If it slows at the same rate throughout, that's uniform angular deceleration.

Revolutions per Second

Revolutions per second (rev/s) is a unit of angular velocity that tells us how many complete circular turns are made in one second. It's a straightforward metric for understanding the speed of rotation in situations involving rotation or circular motion.

In our exercise, we initially convert 500 revolutions per minute to revolutions per second to standardize units. After conversion, we found that the fan began its slowing process at 8.33 rev/s and ended at 3.33 rev/s over the 4-second interval. This conversion and use of consistent units are crucial for calculating other variables such as angular acceleration and distance in terms of revolutions.

If you consider something familiar, like a bicycle wheel spinning once every second, that wheel has an angular velocity of 1 rev/s. Such a clear-cut unit allows for easy comparison between different rotational systems.

Uniform Deceleration

Uniform deceleration is when an object's speed decreases at a constant rate over time. In terms of rotational motion, this is often referred to as uniform angular deceleration. When an object, like a fan, is slowing its spin smoothly and steadily, it's experiencing uniform deceleration.

The exercise solution showed that the fan experienced an angular deceleration of -1.25 rev/s\(^2\), maintaining this rate till the point of complete rest. This constancy is what defines the 'uniform' nature of the deceleration. It means there wasn't any unexpected speeding up or slowing down other than the normal deceleration predicted.

Envision a smoothly stopping car. As the brake pedal is applied steadily, the car slows at a constant rate rather than jerking or abruptly halting. Similarly, in rotational mechanics, uniform deceleration translates to a predictable reduction in speed, leading to easy calculations of dynamics like total revolutions or time to stop, as illustrated in the exercise.