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Chapter 9: Problem 45

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

Short Answer

Expert verified
The required moment of inertia is approximately 0.60 kg路m虏.

Step by step solution

01

Understand the given problem

The problem involves finding the moment of inertia of a flywheel. We know that it must lose 500 J of kinetic energy as its angular velocity decreases from 650 revolutions per minute (RPM) to 520 RPM.
02

Convert angular velocities to rad/s

First, we need to convert the angular velocities from RPM to radians per second (rad/s). The conversion factor is \(\frac{2\pi}{60}\). - Initial angular velocity: \[ \omega_i = 650 \, \text{RPM} \times \frac{2\pi}{60} = \frac{1300\pi}{60} \, \text{rad/s} \approx 68.1 \, \text{rad/s} \]- Final angular velocity:\[ \omega_f = 520 \, \text{RPM} \times \frac{2\pi}{60} = \frac{1040\pi}{60} \, \text{rad/s} \approx 54.5 \, \text{rad/s} \]
03

Use the kinetic energy formula

The kinetic energy (KE) of a rotating object is given by:\[ KE = \frac{1}{2}I\omega^2 \]where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.
04

Apply energy conservation

The change in kinetic energy can be described by:\[ \Delta KE = KE_{initial} - KE_{final} = \frac{1}{2}I\omega_i^2 - \frac{1}{2}I\omega_f^2 = 500 \, \text{J} \]
05

Rearrange the equation for moment of inertia

Rearrange the equation to solve for \(I\):\[ 500 = \frac{1}{2}I(\omega_i^2 - \omega_f^2) \]\[ I = \frac{1000}{\omega_i^2 - \omega_f^2} \]
06

Calculate \(\omega_i^2\) and \(\omega_f^2\)

Calculate the squared values using converted angular velocities:\[ \omega_i^2 = (68.1)^2 = 4637.61 \, \text{rad}^2/ ext{s}^2 \]\[ \omega_f^2 = (54.5)^2 = 2970.25 \, \text{rad}^2/ ext{s}^2 \]
07

Solve for moment of inertia \(I\)

Substitute the squared values into the equation from Step 5:\[ I = \frac{1000}{4637.61 - 2970.25} \approx \frac{1000}{1667.36} \approx 0.60 \, \text{kg} \cdot \text{m}^2 \]

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Key Concepts

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

Flywheel
A flywheel is an important mechanical device used to store or release energy in a system. It plays a vital role in engines and various machines by smoothing out the fluctuations in angular velocity and ensuring consistent energy supply.

Flywheels achieve this through their rotational motion, which means they can store energy when spinning and release that energy when slowing down. This balance is especially useful in situations like vehicles or energy-generating machines, where the power output needs to be stable.
  • In the context of a gasoline engine, the flywheel helps balance the energy cycles, making the engine run more smoothly.
  • The efficiency of a flywheel depends on its size and the moment of inertia, which allows it to store more energy during its motion.
By understanding flywheels, we can appreciate how they play a key role in managing and storing kinetic energy across different systems.
Kinetic Energy
Kinetic energy is the energy of motion. For rotating objects, such as a flywheel, we often talk specifically about rotational kinetic energy.

The formula for the kinetic energy (KE) of a rotating object is \[ KE = \frac{1}{2} I \omega^2 \] where \(I\) is the moment of inertia鈥攁n object's resistance to changes in its rotation鈥攁nd \(\omega\) is the angular velocity, or how quickly something is spinning.
  • Rotational kinetic energy is greater in objects that have a larger moment of inertia and higher angular velocity.
  • When a flywheel accelerates, its kinetic energy increases, storing energy within the system.
In practical terms, managing kinetic energy is about controlling how much energy is stored or released from the flywheel during operation. This principle is why flywheels are used in engines, aiding in stabilizing the energy they emit or absorb.
Angular Velocity
Angular velocity is the rate at which an object rotates or spins. It's a critical aspect when studying the dynamics of rotating systems like flywheels.

When we talk about angular velocity, we refer to how fast something rotates, usually measured in radians per second (rad/s).
  • For a flywheel, higher angular velocity means more kinetic energy is stored, while a lower angular velocity means energy is being released.
  • Conversions from revolutions per minute (RPM) to radians per second are often needed to calculate angular velocity accurately.
In systems where precision and timing are essential, controlling angular velocity ensures consistent performance. Understanding this concept helps us in recalibrating and optimizing systems, whether in vehicles or other mechanical settings.

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Most popular questions from this chapter

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 \(120.0^{\circ} .\)

Calculate the moment of inertia of a uniform solid cone about an axis through its center (Fig. 9.40). The cone has mass \(M\) and altitude \(h\) . The radius of its circular base is \(R\) . figure can't copy

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 computer disk drive is turned on starting from rest and has constant angular acceleration. If took 0.750 s for the drive to make its second complete revolution, (a) how long did it take to make the first complete revolution, and (b) what is its angular acceleration, in \(\operatorname{rad} / s^{2} ?\)

The angular velocity of a fywheel obeys the equation \(\omega_{z}(t)=A+B t^{2},\) where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75\((\text { for } A)\) and 1.50\((\text { for } B)\) . (a) What are the units of \(A\) and \(B\) if \(\omega\) is in rad/s? \((b)\) What is the angular acceleration of the wheel at (i) \(t=0.00\) and (ii) \(t=5.00 \mathrm{s} ?\) (c) Through what angle does the flywheel turn during the first 2.00 \(\mathrm{s} ?(\text { Hint: See Section } 2.6 .)\)
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