Problem 47
Question
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?
Step-by-Step Solution
Verified Answer
The maximum kinetic energy is approximately 73,442 J.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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:\[ I = \frac{1}{2} \times 70.0 \times (1.20)^2 = 50.4 \text{ kg} \cdot \text{m}^2 \]
4Step 4: 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/s² and \( R = 1.20 \) m.So, we can solve for \( \omega \):\[ 3500 = \omega^2 \times 1.20 \]\[\omega^2 = \frac{3500}{1.20} \approx 2916.67 \]\[ \omega = \sqrt{2916.67} \approx 54.00 \text{ rad/s} \]
5Step 5: 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:\[ KE = 0.5 \times 50.4 \times 2916.67 \approx 73,442 \text{ J (joules)} \]
Key Concepts
Moment of InertiaAngular VelocityCentripetal Acceleration
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.
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.
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.
Other exercises in this chapter
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