/*! 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 62 You are designing a rotating met... [FREE SOLUTION] | 91影视

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You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius \(25.0 \mathrm{~cm} .\) Starting from rest at \(t=0,\) the flywheel rotates with constant angular acceleration \(3.00 \mathrm{rad} / \mathrm{s}^{2}\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of \(8600 \mathrm{~kg} / \mathrm{m}^{3},\) what thickness must it have to store \(800 \mathrm{~J}\) of kinetic energy at \(t=8.00 \mathrm{~s} ?\)

Short Answer

Expert verified
The thickness needed for the flywheel to store 800 joules of kinetic energy, given a constant angular acceleration of 3 rad/s虏, a radius of 0.25 m, and a density of 8600 kg/m鲁, can be calculated by first determining the angular speed after 8 seconds, then the kinetic energy, the mass of the flywheel and finally, the thickness. Using the equations described in the solution steps, one finds that all this leads to the solution for the thickness \(h\).

Step by step solution

01

Calculate the Angular Speed

The angular speed \(\omega\) at a certain time \(t\) can be calculated using the formula \(\omega = \alpha \cdot t\), where \(\alpha\) is the constant angular acceleration. Substituting the given values: \(\omega = 3.00 \,rad/s^2 * 8.00 \, s = 24 \, rad/s\)
02

Calculate the Kinetic Energy

To store 800 Joules of energy, we will use the kinetic energy formula for rotational motion, \(\frac{1}{2}I\omega^2\), where \(I\) is the moment of inertia. The moment of inertia for a uniform disk is \(0.5 m R^2\), where \(m\) is the mass of the disk and \(R\) is its radius. Substituting this into the kinetic energy formula gives \(\frac{1}{2}*0.5mR^2蠅^2 = 800\). This leads to a quadratic equation for \(m\), \(0.25m(0.25)^2(24)^2 = 800\).
03

Solve for the Disk's Mass

Solve the equation for \(m\) which will give the mass in kg. The resulting equation is \(m = \frac{800}{0.25*(0.25)^2*(24)^2}\).
04

Calculate the Flywheel's Thickness

We know that the volume of the disk \(V = \pi R^2h\), where \(R\) is the radius and \(h\) is the thickness, we are trying to find. Also, \(m = 蟻V\), where \(蟻\) is the given mass density. Hence, we have \(h = \frac{m}{\pi R^2 蟻}\), substitute for \(m\) and solve for \(h\).

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Key Concepts

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

Angular Speed
Angular speed, often represented by the Greek letter \(\omega\), is a measure of how fast an object rotates or revolves around a fixed axis. It is typically expressed in radians per second (rad/s). In the context of our exercise, angular speed can be calculated using the relationship \(\omega = \alpha \cdot t\), with \(\alpha\) being the constant angular acceleration and \(t\) the time during which the object has been accelerating.

In the given problem, the flywheel starts from rest and accelerates uniformly to an angular speed. This is akin to a linear object starting with zero velocity and gaining speed over time due to constant acceleration. The formula used indicates a direct proportionality; as time \(t\) increases, the angular speed \(\omega\) also increases linearly. This concept is crucial in understanding how the flywheel's speed of rotation builds up over time and is foundational to calculating the energy stored in its rotational motion.
Moment of Inertia
The moment of inertia, symbolized by \(I\), is like the rotational equivalent of mass for linear motion. It quantifies an object's resistance to changes in its rotational speed and depends on the mass distribution relative to the rotation axis. For a uniform disk such as a flywheel, the moment of inertia is given by the formula \(I = 0.5 m R^2\), where \(m\) is the disk's mass and \(R\) is its radius.

Understanding the Moment of Inertia

When we look at the equation for rotational kinetic energy, \(\frac{1}{2}I\omega^2\), we see that the moment of inertia plays a significant role. If the mass is distributed further from the axis, the same object will have a larger moment of inertia and will store more energy for a given angular speed. In the context of our problem, calculating the moment of inertia is critical for determining how much mass is needed for the flywheel to store the specified amount of kinetic energy. The resistance to rotational acceleration due to moment of inertia is also the reason why heavier flywheels can store more energy鈥攁t the cost of needing more energy to speed up.

Flywheel Energy Storage
Flywheel energy storage systems harness the principle of conservation of angular momentum to store energy. The energy is stored in the form of rotational kinetic energy of the flywheel鈥攁s the flywheel spins faster, it stores more energy. The relationship between the rotational kinetic energy (\(E_k\)) and the angular speed is given by the equation \(E_k = \frac{1}{2}I\omega^2\), where \(I\) is the moment of inertia and \(\omega\) the angular speed.

In our exercise, we're asked to determine the thickness of the flywheel needed to store a specific amount of energy (800 J). This energy storage depends on both the moment of inertia and the angular speed of the flywheel. The design of a flywheel for energy storage is a delicate balance鈥攊ncreasing its mass or its radius can increase its energy storage capacity, but it will also make accelerating the flywheel to the desired speed more demanding. Consequently, materials with high strength and density, such as those specified in the exercise, are generally favored in the design of flywheels to maximize energy storage while minimizing size and rotational speed.

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Most popular questions from this chapter

You are to design a rotating cylindrical axle to lift \(800 \mathrm{~N}\) buckets of cement from the ground to a rooftop \(78.0 \mathrm{~m}\) above the ground. The buckets will be attached to a hook on the free end of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise. (a) What should the diameter of the axle be in order to raise the buckets at a steady \(2.00 \mathrm{~cm} / \mathrm{s}\) when it is turning at \(7.5 \mathrm{rpm} ?\) (b) If instead the axle must give the buckets an upward acceleration of \(0.400 \mathrm{~m} / \mathrm{s}^{2},\) what should the angular acceleration of the axle be?

A rotating wheel with diameter \(0.600 \mathrm{~m}\) is speeding up with constant angular acceleration. The speed of a point on the rim of the wheel increases from \(3.00 \mathrm{~m} / \mathrm{s}\) to \(6.00 \mathrm{~m} / \mathrm{s}\) while the wheel turns through 4.00 revolutions. What is the angular acceleration of the wheel?

An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

A uniform disk has radius \(R_{0}\) and mass \(M_{0}\). Its moment of inertia for an axis perpendicular to the plane of the disk at the disk's center is \(\frac{1}{2} M_{0} R_{0}^{2}\). You have been asked to halve the disk's moment of inertia by cutting out a circular piece at the center of the disk. In terms of \(R_{0}\), what should be the radius of the circular piece that you remove?

A meter stick with a mass of \(0.180 \mathrm{~kg}\) is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate (a) the change in gravitational potential energy that has occurred; (b) the angular speed of the stick; (c) the linear speed of the end of the stick opposite the axis. (d) Compare the answer in part (c) to the speed of a particle that has fallen \(1.00 \mathrm{~m}\), starting from rest.

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