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Chapter 9: Problem 47

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 \(\mathrm{rm}\) 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 is approximately 73,442 J.

Step by step solution

01

Understand the Problem

We need to find the maximum kinetic energy that can be stored in a flywheel that has been described as a uniform solid disk. We know the disk's mass, radius, and the maximum radial acceleration allowed. The challenge is to use this information to calculate the kinetic energy.
02

Recall the Necessary Formulas

To find kinetic energy stored in the flywheel, we use the equation for rotational kinetic energy: \[ KE = \frac{1}{2} I \omega^2 \]where \( I \) is the moment of inertia of the flywheel and \( \omega \) is the angular velocity.
03

Calculate the Moment of Inertia

For a solid disk, the moment of inertia \( I \) is given by: \[ I = \frac{1}{2} m R^2 \]Substituting \( m = 70.0 \) kg and \( R = 1.20 \) m, we find:\[ I = \frac{1}{2} \times 70.0 \times (1.20)^2 = 50.4 \text{ kg} \cdot \text{m}^2 \]
04

Relate Radial Acceleration to Angular Velocity

The radial (centripetal) acceleration \( a_c \) is related to angular velocity \( \omega \) by:\[ a_c = \omega^2 R \]We are given \( a_c = 3500 \) m/s虏 and \( R = 1.20 \) m.So, we can solve for \( \omega \):\[ 3500 = \omega^2 \times 1.20 \]\[\omega^2 = \frac{3500}{1.20} \approx 2916.67 \]\[ \omega = \sqrt{2916.67} \approx 54.00 \text{ rad/s} \]
05

Calculate the Maximum Kinetic Energy

Using the values of \( I \) and \( \omega \), substitute into the kinetic energy formula:\[ KE = \frac{1}{2} \times 50.4 \times (54.00)^2 \]Calculate:\[ KE = 0.5 \times 50.4 \times 2916.67 \approx 73,442 \text{ J (joules)} \]

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics. It can be thought of as the rotational equivalent of mass for linear motion. While mass determines how much an object resists linear acceleration, the moment of inertia reflects how much an object resists angular acceleration. For objects that rotate, such as wheels, discs, or flywheels, the moment of inertia is an essential factor in calculating rotational kinetic energy. It depends on the mass of the object and the distribution of that mass relative to the axis of rotation.
  • For a solid disk or flywheel, the moment of inertia, denoted as \( I \), can be calculated using the formula: \[ I = \frac{1}{2} m R^2 \]where \( m \) is the mass, and \( R \) is the radius.
In practical terms, the greater the moment of inertia, the more energy is needed to change the object's rotational speed. In our exercise, the moment of inertia for the flywheel was calculated to be 50.4 kg路m虏 based on a mass of 70 kg and a radius of 1.2 meters.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins around a central point. It's an important element when calculating the rotational kinetic energy, as seen in our flywheel exercise.Angular velocity, typically represented by the symbol \( \omega \), is expressed in radians per second (rad/s). It tells us how many radians the object moves through in a unit of time.
  • To find the angular velocity when given centripetal acceleration and radius, the relationship is:\[ a_c = \omega^2 R \]where \( a_c \) is the centripetal acceleration.
In our example, with a maximum radial acceleration of 3500 m/s虏 and a radius of 1.20 m, the angular velocity was calculated to be approximately 54.00 rad/s. The angular velocity indicates how quickly the flywheel rotates and is directly related to the kinetic energy stored within it.
Centripetal Acceleration
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is crucial in scenarios involving rotating objects, such as wheels or disks.Centripetal acceleration is defined by the equation:
  • \[ a_c = \omega^2 R \]where \( \omega \) is the angular velocity, and \( R \) is the radius of the circular path.
The concept highlights the continuous change in direction of the velocity of the object, which requires a constant inward force and therefore an acceleration towards the center. It's this force that maintains the object's circular motion.In our exercise, the maximum allowed centripetal acceleration was given as 3500 m/s虏. This constraint was important to prevent structural failure of the flywheel, ensuring safety and integrity in practical applications. Understanding centripetal acceleration helps in calculating the limits within which the flywheel can safely and effectively store energy.

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Most popular questions from this chapter

The angle \(\theta\) through which a disk drive turns is given by \(\theta(t)=a+b t-c t^{3},\) where \(a, b,\) and \(c\) are constants \(t\) is in seconds, and \(\theta\) is in radians. When \(t=0, \theta=\pi / 4\) rad and the angular velocity is \(2.00 \mathrm{rad} / \mathrm{s},\) and when \(t=1.50 \mathrm{s},\) the angular acceleration is 1.25 \(\mathrm{rad} / \mathrm{s}^{2}\) , (a) Find \(a, b,\) and \(c,\) including their units. b) What is the angular acceleration when \(\theta=\pi / 4\) rad? (c) What are \(\theta\) and the angular velocity when the angular acceleration is 3.50 \(\mathrm{rad} / \mathrm{s}^{2} ?\)

A roller in a printing press turns through an angle \(\theta(t)\) given by \(\theta(t)=\gamma t^{2}-\beta t^{3},\) where \(\gamma=3.20 \operatorname{rad} / \mathrm{s}^{2}\) and \(\beta=\) 0.500 \(\mathrm{rad} / \mathrm{s}^{3}\) , (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of \(t\) does it occur?

At \(t=3.00 \mathrm{s}\) a point on the rim of a \(0.200-\mathrm{m}\) -radius wheel has a tangential speed of 50.0 \(\mathrm{m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude 10.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00 \mathrm{s}\) and \(t=0\) . (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{s} ?\) (d) At what time will the radial acceleration equal \(g ?\)

A thin, uniform rod is bent into a square of side length \(a\) . If the total mass is \(M\) , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2}\) . Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start; (b) after it has turned through \(60.0^{\circ} ;\) (c) after it has turned through \(120.0^{\circ} .\)
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