Question

A car is fitted with an energy-conserving flywheel, which in operation is geared to the driveshaft so that it rotates at 237 rev/s when the car is traveling at \(86.5 \mathrm{~km} / \mathrm{h}\). The total mass of the car is \(822 \mathrm{~kg}\), the flywheel weighs \(194 \mathrm{~N}\), and it is a uniform disk \(1.08 \mathrm{~m}\) in diameter. The car descends a \(1500-\mathrm{m}\) long, \(5.00^{\circ}\) slope, from rest, with the flywheel engaged and no power supplied from the motor. Neglecting friction and the rotational inertia of the wheels, find \((a)\) the speed of the car at the bottom of the slope, \((b)\) the angular acceleration of the flywheel at the bottom of the slope, and \((c)\) the power being absorbed by the rotation of the flywheel at the bottom of the slope.

Short answer

The solution involves setting the potential energy change equal to the final kinetic energy (linear and rotational), and then solving for the linear speed. Once the linear speed is found, differentiate the relationship \(v=aω\) with respect to time, and set \(dv/dt = g \sin θ\) to find the angular acceleration. Finally, use the formula \(P = Iαω\) to find the power absorbed.

Step-by-step solution

Finding the car's linear velocity from rotational speed

First let’s find the linear speed the flywheel translates to. Using the relationship \(V = rω\), where V is linear speed, r is the radius of disk and ω is angular speed, we find linear speed = \((0.54 \mathrm{~m})(237 \, (2\pi) \mathrm{~rads/s})\). Convert this speed to kilometers per hour for comparison with the car’s speed.

Calculating the Potential and Kinetic Energy change

The car's initial potential energy as it starts to descend is \(mgh\). Here, m is the total mass of the car, g is the acceleration due to gravity and h is the height of hill which is \(1500 \mathrm{m} \cdot \sin(5^{\circ})\). As the car moves down, potential energy is converted to kinetic energy. Assuming no energy loss due to friction or air resistance, conservation of energy allows us to set the potential energy equal to the final kinetic energy, half of which goes to the translational motion of car, and the other half goes to the rotational motion of the flywheel. This gives us the equation \((1/2)mv^2 + (1/2)I\omega^2 = mgh\). Here, \(I\) is the moment of inertia of the flywheel which equals to \(1/2 \cdot (weight of flywheel/g) \cdot (diameter/2)^2 = 1/2 \cdot (194 \mathrm{N}/9.8 \mathrm{m/s^2})(0.54 \mathrm{m})^2).\)

Finding the speed of the car

Solve the equation from previous step to find v, the speed of the car at the bottom of the hill.

Finding the angular acceleration of the flywheel

Angular acceleration (α) or how quickly the flywheel speeds up, can be derived from the relationship between linear and angular speeds, i.e. \(v=aω\). Differentiating both sides with respect to time gives \(dv/dt= a(dω/dt) + αω\). Setting \(dv/dt = g \sin θ\), we get \(g \sin θ = a(dω/dt) + αω\). Solving for \(α\) gives us the angular acceleration at the bottom of the slope.

Finding the power absorbed by the rotation of the flywheel

The power absorbed by the rotation of the flywheel can be found using the formula \(P = Iαω\). Substitute the values from previous steps to calculate the power absorbed.

Key concepts

Energy Conservation

Energy Conservation is a fundamental principle in physics, which states that energy cannot be created or destroyed, only transformed from one form to another.

In our exercise involving a car and a flywheel, the car's initial potential energy when it is on top of the slope converts into kinetic energy as the car moves downhill. Let's break down this energy transformation:

• Potential Energy (PE): At the start, the car has potential energy due to its elevation on the slope. This energy is given by the formula \(mgh\), where \(m\) is the mass of the car, \(g\) is the acceleration due to gravity, and \(h\) is the height.
• Kinetic Energy (KE): As the car descends, its potential energy is converted into kinetic energy, which is the energy of motion. This energy is divided into translational kinetic energy (the car moving forward) and rotational kinetic energy (the flywheel spinning). The total kinetic energy can be described by the equation \((1/2)mv^2 + (1/2)I\omega^2 = mgh\).

This clear division of energy showcases how energy is conserved and merely changes form as the car moves down the slope.

Flywheel Dynamics

Flywheel Dynamics is at the heart of this exercise as the car uses a flywheel to conserve energy while descending.

A flywheel is a mechanical device specifically designed to efficiently store rotational energy. Here’s how it works in context:

• Mechanical Advantage: The flywheel allows the car to store energy while descending. As the flywheel spins, it takes part of the energy from the car’s motion, allowing the car to maintain its speed without additional power input.
• Moment of Inertia: For a uniform disk like our flywheel, the moment of inertia \(I\) is calculated using \(I = \frac{1}{2}mv^2 = \frac{1}{2} \cdot (\text{weight of flywheel}/g) \cdot (\text{radius})^2 \).
• Angular Momentum: As the flywheel spins faster, its angular momentum increases, which means the flywheel holds more energy. This energy can later be used to maintain the car’s speed when it's needed.

This application of a flywheel demonstrates how mechanical systems can help optimize energy usage and efficiency in vehicles.

Rotational Kinematics

Rotational Kinematics explores the relationship between the rotational motion of the flywheel and the linear motion of the car. Understanding this concept is crucial to solving the exercise effectively.

Key points of rotational kinematics include:

• Angular Velocity: The flywheel spins at 237 revolutions per second, where angular velocity \( \omega \) is a measure of how fast an object is rotating. This is converted from revolutions per second to radians per second through the relationship \( \omega = 237 \, \times \, 2\pi \).
• Linear Velocity: The relationship between the flywheel’s rotation and the linear speed of the car is given by \( V = r\omega \). This allows one to find the car's linear speed that corresponds to the flywheel's rotational speed.
• Angular Acceleration: When the car descends, any change in the flywheel's rotational speed is encompassed by angular acceleration \( \alpha \). This is derived from the linear acceleration of the car as it moves down the slope.

By understanding these interconnections, students can better solve kinematic problems involving rotational components.

Conservation of Energy

The Conservation of Energy concept is vital here to understand how energy is managed in this system.

Throughout the exercise:

• Transformation: Energy conservation is evidenced by the transform of potential energy to kinetic energy during the car's descent, maintaining a balance in the energy equation \( (1/2)mv^2 + (1/2)I\omega^2 = mgh \).
• Absorption: The power being absorbed by the flywheel’s rotation is calculated as \( P = I\alpha\omega \), reflecting how energy is converted from the car’s linear motion into the flywheel’s rotational motion.
• Balance: The careful balance of energy transformation ensures that no energy is lost in the process, aside from gravitational energy being directly used for the descent.

Embracing the conservation of energy helps students grasp how energy is continuously managed within a mechanical system.