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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 \(R=1.20 \mathrm{m.}\) 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 stored is approximately 73411.2 J.

Step by step solution

01

Understanding the Problem

We need to find the maximum kinetic energy stored in a flywheel, modeled as a solid disk, given its mass (70.0 kg), radius (1.20 m), and maximum allowable radial acceleration (3500 m/s²).
02

Known Equations

We will use the formulas for the moment of inertia of a disk, the relationship between angular velocity and radial acceleration, and the expression for rotational kinetic energy. - Moment of Inertia (I) for a solid disk: \[ I = \frac{1}{2} m R^2 \]- Radial acceleration (a) in terms of angular velocity (\(\omega\)): \[ a = \omega^2 R \]- Rotational kinetic energy (K):\[ K = \frac{1}{2} I \omega^2 \]
03

Calculate the Moment of Inertia

Using the moment of inertia formula for a solid disk, we calculate:\[ I = \frac{1}{2} \times 70.0 \text{ kg} \times (1.20 \text{ m})^2 \]\[ I = \frac{1}{2} \times 70.0 \times 1.44 = 50.4 \text{ kg m}^2 \]
04

Relate Angular Velocity to Radial Acceleration

Solve for angular velocity (\(\omega\)) using the radial acceleration formula:\[ \omega^2 = \frac{a}{R} = \frac{3500 \text{ m/s}^2}{1.20 \text{ m}} = 2916.67 \text{ s}^{-2} \]\[ \omega = \sqrt{2916.67} \approx 54.00 \text{ rad/s} \]
05

Calculate Rotational Kinetic Energy

Substitute \(I\) and \(\omega\) into the kinetic energy formula:\[ K = \frac{1}{2} \times 50.4 \text{ kg m}^2 \times (54.00 \text{ rad/s})^2 \]\[ K = 25.2 \times 2916.00 \]\[ K \approx 73411.2 \text{ J} \]
06

Interpret the Result

The maximum kinetic energy that can be stored in the flywheel, without exceeding the maximum radial acceleration, is approximately 73411.2 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
When talking about rotational motion, the moment of inertia is a key concept. It essentially describes how mass is distributed in an object and affects its ability to rotate. For a solid disk like the flywheel in this exercise, the moment of inertia is calculated using the formula:
  • \[ I = \frac{1}{2} m R^2 \]
In this equation,
  • \(m\) stands for the mass of the object, which is 70 kg for our flywheel,
  • \(R\) represents the radius of the object, which is 1.20 meters.
By plugging in these values, we determine that the moment of inertia \(I\) is \(50.4 \, \text{kg} \, \text{m}^2\). This value indicates how the mass is spread out relative to the axis of rotation. The larger the moment of inertia, the more energy is required to change the object's rotation. It's similar to inertia in linear motion, which depends on mass, but here we also consider how the mass is distributed relative to the axis.
Angular Velocity
Angular velocity (\(\omega\)) defines how fast an object spins or rotates. It's measured in radians per second (\(\text{rad/s}\)). In our exercise, we need to find the angular velocity that would result in a given radial acceleration without surpassing the limits.To relate angular velocity to radial acceleration, use the formula:
  • \[ a = \omega^2 R \]
Here,
  • \(a\) is the radial acceleration, given as 3500 \(\text{m/s}^2\),
  • \(R\) is 1.20 m.
Rearranging this formula helps us find \(\omega\):
  • \[ \omega = \sqrt{\frac{a}{R}} \approx 54.00 \, \text{rad/s} \]
So, at \(54.00 \, \text{rad/s}\), the flywheel achieves its maximum radial acceleration without structural failure. Angles measured in radians provide a convenient way to express rotations, frequently helping in calculations related to rotation and oscillation.
Radial Acceleration
Radial acceleration is a measure of how quickly an object changes its velocity when moving in a circular path. Unlike linear acceleration, radial acceleration is directed towards the center of the circle and can be thought of as what keeps an object moving in a circle rather than flying off in a straight line.For the flywheel, the radial acceleration is crucial because it is limited to 3500 \(\text{m/s}^2\) to prevent any structural damage. It's calculated from angular velocity using the formula:
  • \[ a = \omega^2 R \]
This tells us that for larger angular velocities, the radial acceleration becomes significant, and care must be taken to ensure that it doesn't surpass safe levels.Knowing this, we can adjust design specifications, like dimensions and material choice, to stay within safe limits under expected operating conditions. Radial acceleration is partly what makes high-speed rotations challenging in engineering applications, demanding precision and durability from the materials used.

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Most popular questions from this chapter

9.75\(\cdots .\) It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. Consider a flywheel made of iron (density 7800 \(\mathrm{kg} / \mathrm{m}^{3}\) ) in the shape of a 10.0 -cm-thick uniform disk. (a) What would the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm about an axis perpendicular to the disk at its center? (b) What would be the centripetal acceleration of a point on its rim when spinning at this rate?

A light, flexible rope is wrapped several times around a hollow cylinder, with a weight of 40.0 \(\mathrm{N}\) and a radius of 0.25 \(\mathrm{m}\), that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force \(P\) for a distance of \(5.00 \mathrm{m},\) at which point the end of the rope is moving at 6.00 \(\mathrm{m} / \mathrm{s} .\) If the rope does not slip on the cylinder, what is the value of \(P ?\)

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm. The power is off for 30.0 s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 200 complete revolutions. (a) At what rate is the flywheel spinning when the power comes back on? (b) How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

(a) Derive an equation for the radial acceleration that includes \(v\) and \(\omega,\) but not \(r .\) (b) You are designing a merry-go-roundfor which a point on the rim will have a radial acceleration of 0.500 \(\mathrm{m} / \mathrm{s}^{2}\) when the tangential velocity of that point has magnitude 2.00 \(\mathrm{m} / \mathrm{s} .\) What angular velocity is required to achieve these values?

Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7}\) m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 \(\mathrm{mm}\) and \(58.0 \mathrm{mm},\) respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 \(\mathrm{m} / \mathrm{s}\) . (a) What is the angular speed of the CD when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration \(C D\) if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum-duration CD during its 74.0 -min playing time? Take the direction of rotation of the disc to be positive.

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