/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 A passenger bus in Zurich, Switz... [FREE SOLUTION] | 91Ó°ÊÓ

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A passenger bus in Zurich, Switzerland, derived its motive power from the energy stored in a large flywheel. The wheel was brought up to speed periodically, when the bus stopped at a station, by an electric motor, which could then be attached to the electric power lines. The flywheel was a solid cylinder with mass 1000 \(\mathrm{kg}\) and diameter 1.80 \(\mathrm{m}\) ; its top angular speed was 3000 rev/min. (a) At this angular speed, what is the kinetic energy of the flywheel? (b) If the average power required to operate the bus is \(1.86 \times 10^{4} \mathrm{W},\) how long could it operate between stops?

Short Answer

Expert verified
(a) The kinetic energy is \(1.80 \times 10^7 \text{ J}\). (b) The bus can operate for 968 seconds between stops.

Step by step solution

01

Calculating the Moment of Inertia

The moment of inertia \(I\) for a solid cylinder rotating about its central axis is given by the formula: \[ I = \frac{1}{2} m r^2 \] where \( m = 1000 \text{ kg} \) is the mass, and \( r = \frac{1.80}{2} \text{ m} = 0.90 \text{ m} \) is the radius. Substituting these values in:\[ I = \frac{1}{2} \times 1000 \times (0.90)^2 = 405 \text{ kg} \cdot \text{m}^2 \].
02

Converting Angular Speed to Radians per Second

The given angular speed of the flywheel is \( 3000 \text{ rev/min} \). To convert this to radians per second, use:\[ 3000 \text{ rev/min} \times \frac{2 \pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = 300 \pi \text{ rad/s} \].
03

Calculating the Kinetic Energy

The rotational kinetic energy \( KE \) of the flywheel can be calculated using:\[ KE = \frac{1}{2} I \omega^2 \] where \( I = 405 \text{ kg} \cdot \text{m}^2 \) and \( \omega = 300\pi \text{ rad/s} \). Substituting these values in:\[ KE = \frac{1}{2} \times 405 \times (300\pi)^2 = 1.80 \times 10^7 \text{ J} \].
04

Calculating the Operating Time Between Stops

To find how long the bus can operate before needing to recharge, use the energy available and the power required:\[ \text{time} = \frac{\text{energy}}{\text{power}} = \frac{1.80 \times 10^7 \text{ J}}{1.86 \times 10^4 \text{ W}} \approx 968 \text{ seconds} \].

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Key Concepts

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

Understanding Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics. It represents the distribution of mass in a rotating object and affects how easily that object can change its rotational speed. Just like mass affects linear motion, moment of inertia affects rotational motion.
The formula for the moment of inertia of a solid cylinder rotating about its central axis is given by:
  • \[ I = \frac{1}{2} m r^2 \]
  • Where,
    • \(m\) is the mass of the cylinder, and
    • \(r\) is the radius.
For our flywheel in the exercise, the mass \( m \) is 1000 kg and the radius \( r \) is half of the diameter, 0.90 m. Substituting these values into the formula gives the moment of inertia as 405 kg·m².
A larger moment of inertia means the flywheel is more resistant to changes in its rotational speed, important for storing kinetic energy efficiently. The moment of inertia is crucial in determining how much energy the flywheel can store and release, a key aspect of its energy efficiency in powering the bus.
Exploring Rotational Kinetic Energy
Rotational kinetic energy is the energy possessed by an object due to its rotation. This type of energy is analogous to the linear kinetic energy of an object moving in a straight line. It's crucial in applications like our flywheel, which depends on rotational kinetic energy to perform its function as an energy storage device.
The formula for calculating rotational kinetic energy is:
  • \[ KE = \frac{1}{2} I \omega^2 \]
  • Where,
    • \(I\) is the moment of inertia, and
    • \(\omega\) is the angular speed in radians per second.
In our example, the flywheel's kinetic energy at its angular speed of 3000 rev/min (converted to 300π rad/s) is calculated using its moment of inertia of 405 kg·m². The result is an energy of 1.80 x 10^7 Joules.
This energy is what powers the bus, illustrating the efficiency and capacity of flywheel energy storage for sustained power supply.
Concept of Angular Speed Conversion
Angular speed indicates how fast something is rotating, typically given in revolutions per minute (rev/min) but often needs to be converted to radians per second (rad/s) for calculations in physics. Understanding this conversion is essential for correctly determining the energies involved in rotational motion.
The conversion factor is rooted in the relationship between revolutions and radians:
  • 1 revolution = \(2\pi\) radians
When converting from rev/min to rad/s, use the formula:
  • \[\text{Angular speed in rad/s} = \text{Angular speed in rev/min} \times \frac{2 \pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}\]
This calculation was used to convert the flywheel's angular speed of 3000 rev/min to 300Ï€ rad/s in the exercise, which is necessary to compute the rotational kinetic energy. Such conversions are vital in any task involving rotational dynamics.
Understanding Power and Energy Calculations
Power and energy are core components of physics that describe how much work an object can do over time, and how fast that energy is expended, respectively. For a system like our flywheel-powered bus, understanding both concepts is key to determining its operational efficiency.
Energy is measured in Joules (J) and refers to the total amount of work that can be performed. Power, measured in Watts (W), is the rate at which energy is used or transferred.
To calculate how long the bus can operate using the flywheel, the formula used is:
  • \[\text{Time} = \frac{\text{Energy}}{\text{Power}} \]
Given that the flywheel holds 1.80 x 10^7 J of energy, and the bus requires an average power of 1.86 x 10^4 W, the bus can operate for approximately 968 seconds between stops.
These calculations demonstrate how effective flywheel energy storage can be in managing the energy needs of real-world applications, such as public transportation.

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Most popular questions from this chapter

A metal sign for a car dealership is a thin, uniform right triangle with base length \(b\) and height \(h .\) The sign has mass \(M .\) (a) What is the moment of inertia of the sign for rotation about the side of length \(h ?\) (b) If \(M=5.40 \mathrm{kg}, b=1.60 \mathrm{m},\) and \(h=1.20 \mathrm{m},\) what is the kinetic energy of the sign when it is rotating about an axis along the \(1.20-\mathrm{m}\) side at 2.00 \(\mathrm{rev} / \mathrm{s} ?\)

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A hollow spherical shell has mass 8.20 \(\mathrm{kg}\) and radius 0.220 \(\mathrm{m} .\) It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.890 \(\mathrm{rad} / \mathrm{s}^{2} .\) What is the kinetic energy of the shell after it has turned through 6.00 \(\mathrm{rev} ?\)

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