Problem 39 - Fundamentals of Physics

Question

The angular momentum of a flywheel having a rotational inertia of \(0.140 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis decreases from 3.00 to \(0.800 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) in \(1.50 \mathrm{~s}\). (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the average power of the flywheel?

Step-by-Step Solution

Answer

(a) 1.47 N·m (b) 16.07 rad (c) 33.47 J (d) 22.31 W

Step 1: Understanding the Problem

The problem involves several aspects of rotational motion: angular momentum, torque, angle, work, and power. We will solve each part methodically using the given data: rotational inertia \( I = 0.140 \, \mathrm{kg} \cdot \mathrm{m}^2 \), initial angular momentum \( L_i = 3.00 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s} \), final angular momentum \( L_f = 0.800 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s} \), and time \( t = 1.50 \, \mathrm{s} \).

Step 1: Calculate the average torque

Torque \( \tau \) is related to the change in angular momentum \( \Delta L = L_f - L_i \) and the time interval \( t \) by the equation \( \tau = \frac{\Delta L}{t} \). Calculate \( \Delta L = 0.800 - 3.00 = -2.20 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s} \), and then find the torque: \[ \tau = \frac{-2.20}{1.50} = -1.47 \, \mathrm{N} \cdot \mathrm{m} \]. The magnitude of the average torque is \( 1.47 \, \mathrm{N} \cdot \mathrm{m} \).

Step 2: Find the angle turned by the flywheel

Assume constant angular acceleration \( \alpha \). The final and initial angular velocities \( \omega_f \) and \( \omega_i \) are related to angular momentum by \( L = I \omega \), so \( \omega_i = \frac{3.00}{0.140} = 21.43 \, \mathrm{rad/s} \) and \( \omega_f = \frac{0.800}{0.140} = 5.71 \, \mathrm{rad/s} \). The angular acceleration \( \alpha \) is \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{5.71 - 21.43}{1.50} = -10.81 \, \mathrm{rad/s}^2 \). The angle \( \theta \) is given by \( \theta = \omega_i t + \frac{1}{2} \alpha t^2 \):\[ \theta = 21.43 \times 1.50 + \frac{1}{2} \times (-10.81) \times (1.50)^2 = 16.07 \, \mathrm{rad} \].

Step 3: Calculate the work done on the wheel

The work done is related to the change in kinetic energy \( \Delta KE \) of the flywheel. The initial and final kinetic energies are \( \frac{1}{2}I\omega_i^2 \) and \( \frac{1}{2}I\omega_f^2 \). Calculate \( \Delta KE = \frac{1}{2} \times 0.140 \times (5.71^2 - 21.43^2) = -33.47 \, \mathrm{J} \), indicating 33.47 J of work was done on the wheel.

Step 4: Find the average power

Power is the work done per unit time. Calculate it using \( P = \frac{\Delta KE}{t} = \frac{-33.47}{1.50} = -22.31 \, \mathrm{W} \). The average power is 22.31 W.

Key Concepts

Rotational Inertia

Rotational inertia, also known as the moment of inertia, is a measure of an object's resistance to changes in its rotation. Just like mass is a measure of an object’s resistance to changes in linear motion, rotational inertia applies to rotational motion.
It depends on the mass of the object and how that mass is distributed relative to the axis of rotation. More mass or mass further from the axis means higher rotational inertia, making it harder to start or stop spinning.
For a specific flywheel, rotational inertia is given as \(0.140 \, \mathrm{kg} \cdot \mathrm{m}^2\). This value tells us how much torque is needed to change its rotational speed.

• High rotational inertia: harder to change rotational speed.
• Low rotational inertia: easier to change rotational speed.

This concept is crucial when calculating angular momentum which affects how flywheels and similar rotating objects behave.

Torque

Torque is the rotational equivalent of force. While force is needed to change linear velocity, torque is needed to change angular velocity. It measures how much a force acting on an object causes that object to rotate.
Similar to force making an object accelerate linearly, torque causes angular acceleration. It is calculated by the change in angular momentum over time, like in the following example:

• Given a change in angular momentum \(-2.20 \, \mathrm{kg} \cdot \mathrm{m}^2 / \mathrm{s}\), and a time of 1.50 seconds.

Torque \( \tau \) is given by the formula:
\[\tau = \frac{\Delta L}{t} = \frac{-2.20}{1.50} = -1.47 \, \mathrm{N} \cdot \mathrm{m} \]
The negative sign indicates the direction of torque is opposite to the assumed direction of rotation. The absolute value tells the magnitude. This concept is key to understanding how rotational systems change speed.

Angular Acceleration

Angular acceleration is how quickly an object changes its angular velocity. It is the rotational counterpart to linear acceleration.
This concept assumes constant angular acceleration for simplicity. We calculate angular acceleration \( \alpha \) during a specific interval:
Initial and final angular velocities, \( \omega_i \) and \( \omega_f \), are derived from the initial and final angular momentums and rotational inertia.

• Initial angular velocity \( \omega_i = \frac{3.00}{0.140} = 21.43 \, \mathrm{rad/s} \)
• Final angular velocity \( \omega_f = \frac{0.800}{0.140} = 5.71 \, \mathrm{rad/s} \)

The formula for angular acceleration is:
\[\alpha = \frac{\omega_f - \omega_i}{t} = \frac{5.71 - 21.43}{1.50} = -10.81 \, \mathrm{rad/s}^2 \]
A negative angular acceleration indicates the flywheel is slowing down. This key concept aids in calculating the change in angular motion over time.

Kinetic Energy

Kinetic energy in rotational motion is analogous to the kinetic energy of translational motion. It depends on both rotational inertia and angular velocity. For rotating objects, kinetic energy (\( KE \)) is given by:
\[ KE = \frac{1}{2} I \omega^2 \]
In our flywheel example, find:

• Initial kinetic energy using \( \omega_i\)
• Final kinetic energy using \( \omega_f\)
• These values help calculate work done on/by the flywheel

Work done on the flywheel is the change in kinetic energy (\( \Delta KE \)):
\[ \Delta KE = \frac{1}{2} \times 0.140 \times (5.71^2 - 21.43^2) = -33.47 \, \mathrm{J} \]
The negative result shows work was done against the flywheel's motion, reducing its kinetic energy. This concept links to how energy transforms in rotational systems.