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Energy is to be stored in a 70.0-kg flywheel in the shape of a uniform solid disk with radius \(R =\) 1.20 m. To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 m/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 is approximately 73,476 joules.

Step by step solution

01

Determine the Maximum Angular Velocity

The maximum radial acceleration of a point is given as 3500 m/sĀ². Since the radial acceleration (a) at the rim of the flywheel is given by \( a = \omega^2 R \), solve for \( \omega \) (angular velocity): \( 3500 = \omega^2 \times 1.20 \). Therefore, \( \omega = \sqrt{\frac{3500}{1.20}} \). Calculating this gives \( \omega \approx 54\, \text{rad/s} \).
02

Calculate the Moment of Inertia

The moment of inertia (I) for a solid disk is \( I = \frac{1}{2} m R^2 \), where \( m = 70.0 \) kg and \( R = 1.20 \) m. Substitute the values to find \( I = \frac{1}{2} \times 70.0 \times (1.20)^2 \). Thus, \( I = 50.4 \text{ kg} \cdot \text{m}^2 \).
03

Determine the Maximum Kinetic Energy

The kinetic energy (KE) stored in a rotating object is given by \( KE = \frac{1}{2} I \omega^2 \). Using the moment of inertia \( I = 50.4 \) kgĀ·mĀ² and angular velocity \( \omega = 54 \) rad/s found previously, substitute these values to get \( KE = \frac{1}{2} \times 50.4 \times 54^2 \).\ Calculating this results in \( KE \approx 73476 \, \text{joules} \).

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Key Concepts

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

Angular Velocity
Angular velocity is a measure of how fast something is rotating and is often represented by the symbol \( \omega \). In the context of a flywheel, it describes how quickly the wheel spins around its axis. It is an important factor when calculating kinetic energy since rotational motion can store energy too.
When finding angular velocity in relation to radial acceleration, the formula \( a = \omega^2 R \) is used, where \( a \) is the radial acceleration and \( R \) is the radius of the circle. Given the radial acceleration and radius, you can solve for \( \omega \):
  • The radial acceleration \( a \) tells us how quickly a point on the rim is accelerating as the flywheel spins.
  • The radius \( R \) is the distance from the center to the rim of the flywheel.
When you plug the given radial acceleration and radius into this formula, you can calculate the angular velocity necessary to prevent structural failure of the flywheel.
Understanding angular velocity helps in determining how energy is distributed and stored in rotational systems.
Moment of Inertia
The moment of inertia, symbolized as \( I \), is a property that measures an object's resistance to changes in its rotational motion. It's similar to mass in linear motion, but applies to rotation. The larger the moment of inertia, the harder it is to change the object's rotation state.
For a solid disk like the flywheel in this exercise, the moment of inertia is calculated using the formula:
  • \( I = \frac{1}{2} m R^2 \)
  • Here, \( m \) is the mass, and \( R \) is the radius of the flywheel.
By substituting the mass and radius values of the flywheel, you find the moment of inertia. This value is crucial for calculating the kinetic energy of the flywheel because it appears in the kinetic energy formula for rotating bodies. Essentially, it quantifies how the mass is distributed in the flywheel and affects how much energy it can store.
Radial Acceleration
Radial acceleration is the rate of change of angular velocity as experienced at a specific point from the axis of rotation, pointing towards the center of rotation. For a rotating object like a flywheel, radial or centripetal acceleration is given by the equation:
  • \( a = \omega^2 R \)
Here, \( a \) is the radial acceleration, \( \omega \) represents the angular velocity, and \( R \) is the radius of the rotation.
The radial acceleration applies to any point on the rim of the flywheel, and it must not exceed certain limits to avoid breakage or structural failure.
This concept is essential because, in calculating the maximum kinetic energy that a flywheel can safely store, knowing the maximum permissible radial acceleration helps to determine the maximum angular velocity the wheel can handle without risk.

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Most popular questions from this chapter

The flywheel of a gasoline engine is required to give up 500 J of kinetic energy while its angular velocity decreases from 650 rev/min to 520 rev/min. What moment of inertia is required?

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

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 cm. Starting from rest at \(t =\) 0, the flywheel rotates with constant angular acceleration 3.00 rad/s\(^2\) about an axis perpendicular to the flywheel at its center. If the flywheel has a density (mass per unit volume) of 8600 kg/m\(^3\), what thickness must it have to store 800 J of kinetic energy at \(t =\) 8.00 s?

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 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-kg block of wood from the free end of the rope, as in Fig. E9.45. You release the system from rest and find that the block descends 12.0 m in 4.00 s. What is the moment of inertia of the wheel for this axis?

At \(t =\) 0 the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t) =\) (250 rad/s)\(t -\) (20.0 rad/s\(^2\))\(t^2 -\) (1.50 rad/s\(^3\))\(t^3\). (a) At what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t =\) 0, when the current was reversed? (e) Calculate the average angular velocity for the time period from \(t =\) 0 to the time calculated in part (a).

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