Problem 9 - University Physics with Modern Physics

Question

The flywheel of an engine has moment of inertia 2.50 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 \(\mathrm{rev} / \mathrm{min}\) in 8.00 \(\mathrm{s}\) starting from rest?

Step-by-Step Solution

Answer

A torque of 13.1 N·m is required.

Step 1: Convert Angular Speed Units

First, we need to convert the final angular speed from revolutions per minute to radians per second. Since there are \(2\pi\) radians in one revolution and 60 seconds in a minute, we use the conversion \(\omega = 400 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{400 \times 2\pi}{60} \, \text{rad/s}\). This simplifies to \(\omega = \frac{400 \times 2\pi}{60} \, \text{rad/s} \approx 41.89\,\text{rad/s}\).

Step 2: Calculate Angular Acceleration

Since the object starts from rest, the initial angular speed \(\omega_0 = 0\). The angular acceleration \(\alpha\) can be found using \(\alpha = \frac{\Delta \omega}{\Delta t}\), where \(\Delta \omega\) is the change in angular speed and \(\Delta t\) is the time interval. Thus, \(\alpha = \frac{41.89 \, \text{rad/s} - 0 \, \text{rad/s}}{8 \, \text{s}} = \frac{41.89}{8} \, \text{rad/s}^2 \approx 5.24\,\text{rad/s}^2\).

Step 3: Use Torque Formula

Torque \(\tau\) is related to angular acceleration by the equation \(\tau = I \cdot \alpha\), where \(I\) is the moment of inertia. Here, \(I = 2.50 \, \text{kg} \cdot \text{m}^2\). Using the previously calculated \(\alpha = 5.24 \, \text{rad/s}^2\), we find \(\tau = 2.50 \, \text{kg} \cdot \text{m}^2 \times 5.24 \, \text{rad/s}^2 = 13.1 \, \text{N} \cdot \text{m}\).

Key Concepts

Angular Acceleration

Angular acceleration is the rate of change of angular velocity over time. Simply put, it tells us how quickly an object is speeding up or slowing down as it rotates. The formula for angular acceleration \(\alpha\) is given by:\[\alpha = \frac{\Delta \omega}{\Delta t}\]where \(\Delta \omega\) is the change in angular velocity and \(\Delta t\) is the time it takes for this change.
In the case of our exercise, since the flywheel starts from rest, the initial angular speed is zero. After 8 seconds, it reaches a certain speed. By inserting these values into the formula, you can find the angular acceleration.
Remember that if an object speeds up, the angular acceleration is positive. Conversely, if the object slows down, the angular acceleration is negative. Understanding this concept helps us predict how fast a rotating object will spin at any given moment.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. Think of it as the rotational "mass" of the object. Just as more massive objects are harder to move, objects with a high moment of inertia are harder to spin or stop spinning.
Mathematically, the moment of inertia \(I\) depends on the mass distribution of the object and is expressed in \(\text{kg} \cdot \text{m}^2\). In simple systems, you often see formulas like \(I = mr^2\) for point masses at a distance \(r\) from the rotation axis.
In this exercise, the flywheel has a moment of inertia of \(2.50 \, \text{kg} \cdot \text{m}^2\). This tells us how much torque is needed to achieve a certain angular acceleration. Combining moment of inertia with angular acceleration, you can calculate the torque required to make the flywheel spin.

Angular Speed Conversion

Angular speed or angular velocity is typically expressed in revolutions per minute (RPM) or radians per second (rad/s). Converting between these units is essential for solving many physics problems involving rotation.
Each revolution represents a full circle, equivalent to \(2\pi\) radians. Therefore, to convert angular speed from RPM to rad/s, you multiply by \(2\pi\) and divide by 60 (seconds per minute).
For the flywheel in the exercise, its speed was originally given as \(400\, \text{rev/min}\). By applying the conversion method, you end up with \(\approx 41.89\, \text{rad/s}\). This conversion step is crucial to ensure all units are consistent when calculating further properties like angular acceleration and torque.
Understanding the conversions helps simplify complex rotational problems by standardizing measurements into common units.