/*! 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 45 Energy is to be stored in a \(70... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 45

Energy is to be stored in a \(70.0 \mathrm{~kg}\) flywheel in the shape of a uniform solid disk with radius \(R=1.20 \mathrm{~m}\). To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is \(3500 \mathrm{~m} / \mathrm{s}^{2}\). What is the maximum kinetic energy that can be stored in the flywheel?

Short Answer

Expert verified
The maximum kinetic energy that can be stored in the flywheel is 72930.51 Joules.

Step by step solution

01

Calculate Angular Velocity

Given that the maximum radial acceleration \(a_r = 3500 \mathrm{m/s}^2\) and radius \(r=1.20 \mathrm{m}\), the angular velocity \(\omega\) can be found using the formula for radial acceleration: \(a_r = \omega^2 r\). Solving for \(\omega\) gives: \(\omega = \sqrt{\frac{a_r}{r}} = \sqrt{\frac{3500}{1.2}} = \sqrt{2916.67} = 53.98 \mathrm{rad/s}\)
02

Calculate the Moment of Inertia

The moment of inertia for a uniform solid disc is given by \(I = \frac{1}{2} m r^2\). Substituting the given mass \(m = 70.0 \mathrm{kg}\) and radius \(r=1.20 \mathrm{m}\) into this formula gives: \(I = \frac{1}{2} * 70.0 * (1.2)^2 = 50.4 \mathrm{kg\cdot m^2}\)
03

Calculate Maximum Kinetic Energy

The kinetic energy of a rotating object is given by \(K = \frac{1}{2} I \omega^2\). Substituting the calculated angular velocity and moment of inertia into this formula gives: \(K = \frac{1}{2} * 50.4 * (53.98)^2 = 72930.51 \mathrm{J}\)

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Key Concepts

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

Radial Acceleration
Radial acceleration is a component of acceleration experienced by an object moving in a circular path, directed towards the center of curvature of the path. It plays a pivotal role in determining the structural integrity of rotating objects, like flywheels, when they store kinetic energy.

When calculating radial acceleration, we use the formula a_r = \( \(omega\) \)^2 r, where \(omega\) is the angular velocity and r is the radius of the circular path. For a point on the rim of the flywheel, with a known maximum safe radial acceleration, we can determine the maximum angular velocity at which the flywheel can rotate without failure.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates around a specific point or axis. It's an essential quantity in physics when dealing with rotating bodies as it gives us an idea of the rate at which an angle is swept out.

The unit of angular velocity is radians per second (rad/s). In our case, the calculation of angular velocity is derived from the flywheel's radial acceleration and radius, using the aforementioned radial acceleration formula. Knowing the angular velocity of a rotating flywheel is crucial to understanding its kinetic energy storage capabilities.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It is determined by both the mass of the body and the distribution of that mass relative to the axis of rotation.

In formulas, the moment of inertia for a uniform solid disc, like our flywheel, is given by I = \( \frac{1}{2} m r^2\), where m is the mass and r is the radius. This property directly influences the kinetic energy that a rotating object can hold, making it a fundamental concept in rotational dynamics.
Kinetic Energy of Rotating Objects
Kinetic energy in the context of rotating objects is termed rotational kinetic energy, and it's the energy due to an object's rotation. It's given by the formula K = \( \frac{1}{2} I \omega^2\), where I is the moment of inertia and \(omega\) is the angular velocity.

The kinetic energy of a flywheel is a practical concern, as it represents the amount of energy that can be stored and later retrieved for use. The maximum kinetic energy the flywheel can store depends directly on the limits imposed by its maximum safe radial acceleration.
Physics of Rigid Bodies
The physics of rigid bodies involves analyzing the motion and forces applied to objects that do not deform under external forces or torques. It encompasses concepts like moment of inertia and kinetic energy for rotating bodies.

A rigid body like a flywheel is subject to these principles, which determine how it can absorb and distribute energy through rotation. Understanding the physics of rigid bodies allows us to calculate properties such as radial acceleration and the resulting kinetic energy, critical for flywheel function and design.

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Most popular questions from this chapter

About what axis will a uniform, balsa-wood sphere have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius, with the axis along a diameter?

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?

\(9.88^{\circ}\) DATA You are analyzing the motion of a large flywheel that has radius \(0.800 \mathrm{~m}\). In one test run, the wheel starts from rest and turns in a horizontal plane with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration \(a\) of a point on the rim of the flywheel as a function of the angle \(\theta-\theta_{0}\) through which the wheel has turned. You collect these results: $$ \begin{array}{l|cccccccc} \boldsymbol{\theta}-\boldsymbol{\theta}_{\mathbf{0}}(\mathbf{r a d}) & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 \\ \hline \boldsymbol{a}\left(\mathbf{m} / \mathbf{s}^{\mathbf{2}}\right) & 0.678 & 1.07 & 1.52 & 1.98 & 2.45 & 2.92 & 3.39 & 3.87 \end{array} $$ Construct a graph of \(a^{2}\left(\right.\) in \(\left.\mathrm{m}^{2} / \mathrm{s}^{4}\right)\) versus \(\left(\theta-\theta_{0}\right)^{2}\) (in rad \(^{2}\) ). (a) What are the slope and \(y\) -intercept of the straight line that gives the best fit to the data? (b) Use the slope from part (a) to find the angular acceleration of the flywheel. (c) What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of \(135^{\circ} ?\) (d) When the flywheel has turned through an angle of \(90.0^{\circ},\) what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

\(\mathrm{At} t=0\) a grinding wheel has an angular velocity of \(24.0 \mathrm{rad} / \mathrm{s}\) It has a constant angular acceleration of \(30.0 \mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{~s}\). From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?
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