91Ó°ÊÓ

A flywheel (i.e., a massive disk capable of rotating about its central axis) has a moment of inertia of \(3.8 \mathrm{~kg} \cdot \mathrm{m}^{2}\). What constant torque is required to increase the wheel's frequency from \(2.0 \mathrm{rev} / \mathrm{s}\) to \(5.0 \mathrm{rev} / \mathrm{s}\) in \(6.0\) revolutions? Neglect friction. Given $$ \begin{array}{lll} \theta=12 \pi \mathrm{rad} & \omega_{i}=4.0 \pi \mathrm{rad} / \mathrm{s} & \text { and } & \omega_{f}=10 \pi \mathrm{rad} / \mathrm{s} \end{array} $$ we can write Work done on wheel = Change in \(\mathrm{KE}_{r}\) of wheel $$ \begin{aligned} \tau \theta &=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} I \omega_{i}^{2} \\ (\tau)(12 \pi \mathrm{rad}) &=\frac{1}{2}\left(3.8 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left[\left(100 \pi^{2}-16 \pi^{2}\right)(\mathrm{rad} / \mathrm{s})^{2}\right] \end{aligned} $$ which leads to \(\tau=42 \mathrm{~N} \cdot \mathrm{m}\). Notice in all of these problems that radians and seconds must be used.

Step by step solution

01

Convert initial given values

We are given the final and initial angular velocities in revolutions per second: \(\omega_i = 2.0\,\mathrm{rev/s}\) and \(\omega_f = 5.0\,\mathrm{rev/s}\). We need to convert these to radians per second by using the conversion \(1\,\mathrm{rev} = 2\pi\,\mathrm{rad}\). So,\[\omega_i = 4\pi\,\mathrm{rad/s} \]\[\omega_f = 10\pi\,\mathrm{rad/s} \]

02

Identify the total angular displacement (θ)

We know the wheel turns through a total of 6 revolutions, which we need to convert to radians:\[ \theta = 6\,\mathrm{rev} \times 2\pi\,\mathrm{rad/rev} = 12\pi\,\mathrm{rad} \]

03

Apply the work-energy principle

The work done on the flywheel results in a change in its rotational kinetic energy. The rotational kinetic energy is given by \( \frac{1}{2}I\omega^2 \), where \(I\) is the moment of inertia. Thus:\[ \tau \theta = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2 \]

04

Substitute into the work-energy equation

Substitute the known values into the equation:\[ \tau (12\pi) = \frac{1}{2}(3.8)\left[(10\pi)^2 - (4\pi)^2\right] \]This simplifies to\[ 12\pi \tau = \frac{1}{2}(3.8)(100\pi^2 - 16\pi^2) \]

05

Calculate the torque (Ï„)

Simplify:\[ 12\pi \tau = \frac{1}{2}(3.8)(84\pi^2) \]\[ 12\pi \tau = (1.9)(84\pi^2) \]\[ \tau = \frac{(1.9)(84\pi)}{12} \]\[ \tau \approx 42.0\,\mathrm{N}\cdot\mathrm{m} \]

06

Conclusion

The constant torque required to increase the wheel's frequency from 2.0 rev/s to 5.0 rev/s in 6 revolutions is \(42\,\mathrm{N}\cdot\mathrm{m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque

Torque is a measure of the rotational force applied to an object, causing it to rotate around an axis. It is akin to the concept of force in linear motion, but specifically for rotational movement. The formula for torque is given by \( \tau = F \cdot r \), where \( F \) is the force applied and \( r \) is the lever arm or the distance from the axis of rotation to the point where the force is applied.

• Torque is measured in newton-meters (N·m) in the SI unit system.
• It plays a crucial role in determining how quickly or slowly an object can be made to rotate.

In our problem, the torque is responsible for changing the angular velocity of the flywheel by overcoming its inertia and shuttling energy into the wheel. The equation \( \tau \theta = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2 \) shows how torque over an angular displacement (\( \theta \)) results in a change of rotational kinetic energy of the flywheel.

Angular Velocity

Angular velocity is the rate of change of angular displacement and is a vector quantity. It represents how quickly an object rotates around a specific axis.

• The standard unit for angular velocity in the SI system is radians per second (rad/s).
• It can also be expressed in revolutions per second (rev/s), but for calculations, converting to rad/s is essential because radians are the standard unit for angular measures in physics.

In the original exercise, the angular velocity of the flywheel increased from \( 4\pi \) rad/s to \( 10\pi \) rad/s. It's crucial to convert between these units accurately to solve rotational motion problems. One revolution is equivalent to \( 2\pi \) radians, a handy conversion for moving between revolutions and radians.

Kinetic Energy

Kinetic energy in rotational motion is known as rotational kinetic energy and is given by the formula \( \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

• Rotational kinetic energy depends on how mass is distributed relative to the axis of rotation.
• The greater the mass or the farther from the axis it is distributed, the more energy it takes to change its state of motion.

For the flywheel, increasing its rotational kinetic energy from the initial to the final state required work, facilitated by torque, and is expressed as a change in the expression \( \frac{1}{2} I \omega^2 \). This change in kinetic energy is how we calculate the work done on the flywheel in the exercise.

Conversion of Units

In physics, converting units accurately is a bedrock principle necessary for solving any problem correctly. In rotational dynamics, units such as revolutions, radians, and seconds frequently come into play.

• Since calculations need to be consistent, converting units like revolutions to radians is crucial.
• The conversion between these units allows dynamic quantities like angular displacement and angular velocity to be used in formulae consistently.

In our exercise, the challenge begins with converting the initial and final angular velocities from revolutions per second to radians per second using the equation \( 1\,\text{rev} = 2\pi\,\text{rad} \). This conversion ensures compatibility of units when calculating torque and kinetic energy changes, maintaining the integrity of the physics equations involved.