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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

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} \\)\(M_0R_0^2\). 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 bicycle wheel has an initial angular velocity of 1.50 rad/s. (a) If its angular acceleration is constant and equal to 0.200 rad/s\(^2\), what is its angular velocity at \(t =\) 2.50 s? (b) Through what angle has the wheel turned between \(t =\) 0 and \(t =\) 2.50 s?

A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of 3.00 rad/s\(^2\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{rad} = \omega^2r\) and (b) \(a_{rad} = v^2/r\)

You must design a device for shooting a small marble vertically upward. The marble is in a small cup that is attached to the rim of a wheel of radius 0.260 m; the cup is covered by a lid. The wheel starts from rest and rotates about a horizontal axis that is perpendicular to the wheel at its center. After the wheel has turned through 20.0 rev, the cup is the same height as the center of the wheel. At this point in the motion, the lid opens and the marble travels vertically upward to a maximum height \(h\) above the center of the wheel. If the wheel rotates with a constant angular acceleration \(\alpha\), what value of a is required for the marble to reach a height of \(h =\) 12.0 m?

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