/*! 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 60 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. Whenever the bus was stopped at a station, the wheel was brought up to speed with the use of an electric motor that could then be attached to the electric power lines. The flywheel was a solid cylinder with a mass of 1000 \(\mathrm{kg}\) and a diameter of \(1.80 \mathrm{m} ;\) its top angular speed was 3000 \(\mathrm{rev} / \mathrm{min.}\) At this angular speed, what was the kinetic energy of the flywheel?

Short Answer

Expert verified
The kinetic energy of the flywheel is approximately 19,987,980 joules.

Step by step solution

01

Convert Angular Speed from RPM to Radians per Second

First, you need to convert the angular velocity from revolutions per minute (rpm) to radians per second. Use the conversion factor, knowing that one revolution is \(2\pi\) radians, and there are 60 seconds per minute.\[\omega = 3000 \times \frac{2\pi}{60} = 100\pi \, \text{rad/s}\]
02

Calculate the Moment of Inertia for the Flywheel

The moment of inertia \(I\) for a solid cylinder is given by the formula:\[I = \frac{1}{2} m r^2\]Where:- \(m\) is the mass of the flywheel: 1000 kg,- \(r\) is the radius of the flywheel: \( \frac{1.80}{2} = 0.90 \, \text{m} \).Substitute the values:\[I = \frac{1}{2} \times 1000 \times (0.90)^2 = 405 \, \text{kg}\cdot\text{m}^2\]
03

Calculate the Kinetic Energy of the Flywheel

The kinetic energy \(K\) of a rotating object is given by:\[K = \frac{1}{2} I \omega^2\]Substitute \(I = 405\, \text{kg}\cdot\text{m}^2\) and \(\omega = 100\pi \, \text{rad/s}\):\[K = \frac{1}{2} \times 405 \times (100\pi)^2\]Simplify and calculate:\[K = 0.5 \times 405 \times 10000 \times \pi^2\]\[K = 2,025,000\pi^2 \, \text{Joules}\]Approximating \(\pi^2 \approx 9.8696\), we find:\[K \approx 19,987,980 \, \text{J}\]
04

Summarize the Result

The final kinetic energy of the flywheel is approximately 19,987,980 joules. This energy represents the amount of rotational kinetic energy stored in the flywheel at its peak angular speed.

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Key Concepts

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

Angular Speed
Angular speed is how fast something spins around an axis. It's similar to how velocity is the speed of how fast something moves forward. When describing angular speed, we often use revolutions per minute (RPM) or radians per second (rad/s).
For calculations, rad/s is preferred because it connects so beautifully with other formulas in physics. In rotational motion, one full revolution equals 2Ï€ radians.
So, to switch from RPM to rad/s, you multiply the RPM by 2Ï€ and then divide by 60, since there are 60 seconds in a minute. In the given problem, the angular speed of the flywheel was converted into rad/s, which allowed us to find how quickly the flywheel is spinning in a way that's useful for further calculations, like finding kinetic energy.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotation. For linear motion, you think about mass, but for rotational motion, you think about the moment of inertia. The farther the mass is from the axis of rotation, the larger the inertia.
For a solid cylinder like the flywheel in the exercise, the formula to calculate moment of inertia is \[I = \frac{1}{2} m r^2\] where \(m\) is the mass and \(r\) is the radius. This formula highlights that not only the mass but also the distribution of this mass (how far it is from the axis) matters a lot in rotational dynamics.
The moment of inertia was calculated as 405 kg·m² for the given flywheel, indicating how its mass and shape influence its rotational properties.
Rotational Motion
Rotational motion involves objects spinning around a central point or axis. Unlike linear motion, rotational motion takes into account angular quantities like angular speed, angular acceleration, and angular displacement. Just like in linear motion, where Newton's laws can be applied, rotational motion has analogous principles.
The key aspects of rotational motion include the relationship between torque, which is a force that causes rotation, and the moment of inertia which we discussed earlier. In the context of the exercise, the flywheel's rotation allows it to store energy in the form of rotational kinetic energy, which is later used to drive the bus.
This stored energy is what makes flywheels valuable in applications where consistent power over a stretched time is appreciated.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed. In the context of the flywheel, the energy from the electric motor converts into kinetic energy as the flywheel spins faster.
The equation for rotational kinetic energy is \[K = \frac{1}{2} I \omega^2\] In this formula, the kinetic energy \(K\) increases with both the moment of inertia \(I\) and the square of the angular speed \(\omega\). The stored energy in the flywheel can then be used to power the bus when the electric current isn't available, making use of energy already stored without wasting it. This aspect of energy conservation is a fundamental concept in physics and engineering, optimizing resources by transforming energy from one form to another efficiently.

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Most popular questions from this chapter

\(\bullet\) A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle \(\theta\) above the horizontal. Both spheres start from rest at the same vertical height \(h\) . (a) How fast is each sphere moving when it reaches the bottom of the hill? (b) Which sphere will reach the bottom first, the hollow one or the solid one?

\(\cdot\) A wheel turns with a constant angular acceleration of 0.640 \(\mathrm{rad} / \mathrm{s}^{2} .\) (a) How much time does it take to reach an angular velocity of \(8.00 \mathrm{rad} / \mathrm{s},\) starting from rest? (b) Through how many revolutions does the wheel turn in this interval?

Compound objects. Moment of inertia is a scalar. There- fore, if several objects are connected together, the moment of inertia of this compound object is simply the scalar (algebraic) sum of the moments, of inertia of each of the component. objects. Use this principle to answer each of the following questions about the moment of inertia of compound objects: (a) A thin uniform 2.50 \(\mathrm{kg}\) bar 1.50 \(\mathrm{m}\) long has a small 1.25 \(\mathrm{kg}\) mass glued to each end. What is the moment of inertia of this object about an axis perpendicular to the bar through its center? (b) What is the moment of inertia of the object in part (a) about an axis perpendicular to the bar at one end? (c) \(A 725\) g metal wire is bent into the shape of a hoop 60.0 \(\mathrm{cm}\) in diameter. Six wire spokes, each of mass 112 g, are added from the center of the hoop to the rim. What is the moment of inertia of this object about an axis perpendicular to it through its center?

A A flywheel having constant angular acceleration requires 4.00 s to rotate through 162 rad. Its angular velocity at the end of this time is 108 rad/s. Find (a) the angular velocity at the beginning of the 4.00 s interval; (b) the angular acceleration of the flywheel.

A thin uniform rod 50.0 \(\mathrm{cm}\) long with mass 0.320 \(\mathrm{kg}\) is bent at its center into a \(\mathrm{V}\) shape, with a \(70.0^{\circ}\) angle at its vertex. Find the moment of inertia of this V-shaped object about an axis perpendicular to the plane of the \(\mathrm{V}\) at its vertex.

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