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Chapter 2: Problem 11

A disk-shaped flywheel, of uniform density \(\rho\), outer radius \(R\), and thickness \(w\), rotates with an angular velocity \(\omega\), in \(\mathrm{rad} / \mathrm{s}\). (a) Show that the moment of inertia, \(I=\int_{\text {vol }} \rho r^{2} d V\), can be expressed as \(I=\pi \rho w R^{4} / 2\) and the kinetic energy can be expressed as \(\mathrm{KE}=I \omega^{2} / 2\). (b) For a steel flywheel rotating at 3000 RPM, determine the kinetic energy, in \(\mathrm{N} \cdot \mathrm{m}\), and the mass, in \(\mathrm{kg}\), if \(R=0.38 \mathrm{~m}\) and \(w=0.025 \mathrm{~m}\). (c) Determine the radius, in \(\mathrm{m}\), and the mass, in \(\mathrm{kg}\), of an aluminum flywheel having the same width, angular velocity, and kinetic energy as in part (b).

Short Answer

Expert verified
KE of the steel flywheel is 15872 Nm. The mass of the steel flywheel is 88.5 kg. For the aluminum flywheel with same KE and width, the radius is 0.295 m and the mass is 18.48 kg.

Step by step solution

01

Finding the Moment of Inertia

The moment of inertia for a disk-shaped object is given by integrating the mass elements multiplied by the square of their distance from the axis. Set up the integral as follows: \[I = \rho \times \text {volume element (dV)} \times r^{2}\] The volume element for a disk is a thin cylindrical shell: \[dV = 2\pi r w dr\] Substitute the volume element into the integral: \[I = \rho \int_{0}^{R} r^{2} \times 2\pi r w \ dr\] -Carry out the integration: \[I = 2\pi \rho w \int_{0}^{R} r^{3} dr = 2\pi \rho w \frac{R^{4}}{4} = \frac{1}{2} \pi \rho w R^{4}\]
02

Expression for Kinetic Energy

The rotational kinetic energy is given by: \[\text{KE} = \frac{1}{2} I \omega^{2}\] By substituting in the previously derived expression for moment of inertia: \[\text{KE} = \frac{1}{2} \left(\frac{1}{2} \pi \rho w R^{4}\right) \omega^{2} = \frac{1}{4} \pi \rho w R^{4} \omega^{2}\]
03

Converting RPM to rad/s

Given the flywheel's angular velocity in RPM, convert it to rad/s: \[\omega = 3000 \times \frac{2\pi}{60} = 314.16 \, \text{rad/s}\]
04

Finding Kinetic Energy for Steel Flywheel

Given the angular velocity, radius, and thickness (and knowing the density of steel, \(\rho_{steel} = 7850 \, \text{kg/m}^3\)): \[\text{KE}_{steel} = \frac{1}{4} \pi \rho w R^{4} \omega^{2}\] -Substituting in the values: \[\text{KE}_{steel} = \frac{1}{4} \pi \times 7850 \times 0.025 \times (0.38)^{4} \times (314.16)^{2}\] -Evaluate the integral to find the kinetic energy: \[\text{KE}_{steel} \approx 15872 \, \text{Nm}\]
05

Finding Mass of Steel Flywheel

The mass of the flywheel is given by \(\text{mass} = \text{volume} \times \text{density}\). Calculate the volume of the steel flywheel as:\ \[\text{volume} = \pi R^{2} w\] - Substituting values: \ \[\text{mass}_{steel} = \pi (0.38)^{2} \times 0.025 \times 7850 \approx 88.5 \, \text{kg}\]
06

Radius and Mass for Aluminum Flywheel

Using the same approach for the aluminum flywheel (knowing the density of aluminum, \(\rho_{alum} = 2700 \, \text{kg/m}^3\)), we know the kinetic energy should be the same: \[I_{alum} = \frac{2\times \text{KE}}{\omega^{2}} \approx \frac{2\times 15872}{(314.16)^{2}} \approx 0.32 \times \text {Nm s}^2\] -Solve for the radius using the moment of inertia formula: \[\frac{1}{2} \pi \rho R_{alum}^{4} w = 0.32\] -Substitute the values: \[R_{alum} \approx 0.295 \text{m}\] -Calculate the volume and mass: \[\text{volume}_{alum}=\ pi R_{alum}^{2} w = \pi (0.295)^{2} 0.025\] \[\text{mass}_{alum}=\text{volume}_{alum} \times \rho_{alum} \approx 18.48 \, \text{kg}\]

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is the energy due to the rotation of an object and is an important concept in understanding how rotating systems work. It is given by the formula: \(\text{KE} = \frac{1}{2} I \rho^2\) where \(\text{KE}\) represents kinetic energy, \(\text{I} \) is the moment of inertia, and \(\rho \) is angular velocity. In the context of a flywheel, this energy is a measure of how much work the flywheel can do while it is rotating. \r\r To find the kinetic energy of a rotating flywheel, you first need the moment of inertia (I) which depends on the mass distribution of the flywheel. Once you have \(\text{I}\), you can use the angular velocity provided to find the rotational kinetic energy by plugging the values into the kinetic energy formula. A flywheel with higher mass or higher rotation speed will have more kinetic energy. This is significant in mechanical applications where energy storage and release are critical.
Angular Velocity
Angular velocity denotes how fast an object rotates or revolves relative to another point, expressed in radians per second (rad/s). It describes the rate of rotation and is symbolized by \(\rho\). \r\r In the exercise, a flywheel rotates at 3000 RPM (revolutions per minute). To convert RPM to rad/s, use the formula: \(\rho = \text{RPM} \times \frac{2\text{pi}}{60}\). This helps transition the speed from a linear (RPM) to an angular (rad/s) expression. \r\r Knowing the angular velocity is crucial because it allows you to calculate the rotational kinetic energy and analyze dynamic systems, helping to design and control various mechanical and aerospace systems.
Density
Density is defined as mass per unit volume and is denoted by \( \rho \). The formula for density is \( \rho = \frac{\text{mass}}{\text{volume}}\). In the context of a flywheel, the density of the material affects its mass and inertia. \r\r For instance, the flywheel made from steel with a density of 7850 \( \text{kg/m}^3 \) will be heavier than an identical flywheel made from aluminum, which has a density of 2700 \( \text{kg/m}^3 \). \r\r Density plays a pivotal role in calculating various physical properties such as mass and kinetic energy. The higher the density, the greater the mass for a given volume, influencing the flywheel's ability to store rotational energy. Understanding the density of materials helps in optimizing designs for specific engineering applications, making them more efficient and effective in their respective functions.

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Most popular questions from this chapter

A gas undergoes a cycle in a piston-cylinder assembly consisting of the following three processes: Process 1-2: Constant pressure, \(p=1.4\) bar, \(V_{1}=0.028 \mathrm{~m}^{3}\), \(W_{12}=10.5 \mathrm{~kJ}\) Process 2-3: Compression with \(p V=\) constant, \(U_{3}=U_{2}\) Process 3-1: Constant volume, \(U_{1}-U_{3}=-26.4 \mathrm{~kJ}\) There are no significant changes in kinetic or potential energy. (a) Sketch the cycle on a \(p-V\) diagram. (b) Calculate the net work for the cycle, in kJ. (c) Calculate the heat transfer for process \(1-2\), in \(\mathrm{kJ}\).

A major force opposing the motion of a vehicle is the rolling resistance of the tires, \(F_{\mathrm{r}}\), given by $$ F_{\mathrm{r}}=f^{\circ} W $$ where \(f\) is a constant called the rolling resistance coefficient and \(W\) is the vehicle weight. Determine the power, in \(\mathrm{kW}\), required to overcome rolling resistance for a truck weighing \(322.5 \mathrm{kN}\) that is moving at \(110 \mathrm{~km} / \mathrm{h}\). Let \(f=0.0069\).

Helium gas is contained in a closed rigid tank. An electric resistor in the tank transfers energy to the gas at a constant rate of \(1 \mathrm{~kW}\). Heat transfer from the gas to its surroundings occurs at a rate of \(5 t\) watts, where \(t\) is time, in minutes. Plot the change in energy of the helium, in kJ, for \(t \geq 0\) and comment.

Measured data for pressure versus volume during the expansion of gases within the cylinder of an internal combustion engine are given in the table below. Using data from the table, complete the following: (a) Determine a value of \(n\) such that the data are fit by an equation of the form, \(p V^{n}=\) constant. (b) Evaluate analytically the work done by the gases, in kJ, using Eq. \(2.17\) along with the result of part (a). (c) Using graphical or numerical integration of the data, evaluate the work done by the gases, in kJ. (d) Compare the different methods for estimating the work used in parts (b) and (c). Why are they estimates? $$ \begin{array}{crr} \text { Data Point } & p(b a r) & V\left(\mathrm{~cm}^{3}\right) \\ \hline 1 & 15 & 300 \\ 2 & 12 & 361 \\ 3 & 9 & 459 \\ 4 & 6 & 644 \\ 5 & 4 & 903 \\ 6 & 2 & 1608 \end{array} $$

For a refrigerator with automatic defrost and a topmounted freezer, the annual cost of electricity is \(\$ 55\). (a) Evaluating electricity at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the refrigerator's annual electricity requirement, in \(\mathrm{kW} \cdot \mathrm{h}\). (b) If the refrigerator's coefficient of performance is 3 , determine the amount of energy removed from its refrigerated space annually, in MJ.
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