Chapter 10, Problem 10.88CP

As a gasoline engine operates, a flywheel turning with the crankshaft stores energy after each fuel explosion, providing the energy required to compress the next charge of fuel and air. For the engine of a certain lawn tractor, suppose a flywheel must be no more than 18.0 cm in diameter. Its thickness, measured along its axis of rotation, must be no larger than 8.00 cm. The flywheel must release energy 60.0 J when its angular speed drops from 800 rev/min to 600 rev/min. Design a sturdy steel (density 7.85 × 10³ kg/m³) flywheel to meet these requirements with the smallest mass you can reasonably attain. Specify the shape and mass of the flywheel.

To determine

The design of sturdy steel flywheel with the small mass and specify it shape.

Answer to Problem 10.88CP

The shape of the flywheel is hollow cylinder of inner radius $0.08\text{m}$ and outer radius $0.09\text{m}$ with the mass $7.34\text{kg}$.

Explanation of Solution

The diameter of the flywheel must be no more than $18.0\text{cm}$ and the length is not larger than $8.00\text{cm}$. The energy release by the flywheel is $60.0\text{J}$ and drops in angular speed is $800\text{rev}/\text{min}$ to $600\text{rev}/\text{min}$. The density of the steel is $7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$.

From the law of energy conservation,

    $K_{\text{rf}}-K_{\text{ri}}=W_{\text{out}}$

Here, $K_{\text{ri}}$ is the initial rotational energy of the flywheel, $K_{\text{ri}}$ is the initial rotational energy of the flywheel and $W_{\text{out}}$ is the output work produce by the flywheel.

Write the expression for the initial rotational energy of the flywheel is,

    $K_{\text{ri}}=\frac{1}{2}I{\omega }_{\text{i}}^2$

Here, $I$ is the moment of inertia of the flywheel and ${\omega }_{\text{i}}$ is the initial angular speed of the flywheel.

Write the expression for the final rotational energy of the flywheel is,

    $K_{\text{rf}}=\frac{1}{2}I{\omega }_{\text{f}}^2$

Here, ${\omega }_{\text{i}}$ is the final angular speed of the flywheel.

Substitute $\frac{1}{2}I{\omega }_{\text{i}}^2$ for $K_{\text{ri}}$ and $\frac{1}{2}I{\omega }_{\text{f}}^2$ for $K_{\text{rf}}$ in equation (1) and rearrange the equation for $I$.

    $\begin{array}{c}\frac{1}{2}I{\omega }_{\text{f}}^2-\frac{1}{2}I{\omega }_{\text{i}}^2=W\\ \frac{1}{2}I\left({\omega }_{\text{f}}^2-{\omega }_{\text{f}}^2\right)=W\\ I=\frac{2W}{\left({\omega }_{\text{f}}^2-{\omega }_{\text{f}}^2\right)}\end{array}$

Substitute $60.0\text{J}$ for $W$, $600\text{rev}/\text{min}$ for ${\omega }_{\text{i}}$ and $800\text{rev}/\text{min}$ for ${\omega }_{\text{f}}$ in above equation to find $I$.

    $\begin{array}{c}I=\frac{2\times 60.0\text{J}}{\left({\left(800\text{rev}/\text{min}\times \left(\frac{1\min }{60\text{s}}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\right)}^2-{\left(600\text{rev}/\text{min}\times \left(\frac{1\min }{60\text{s}}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\right)}^2\right)}\\ =\frac{120\text{J}\times \frac{\text{1}\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{1\text{J}}}{3070.54{\text{s}}^{-\text{2}}}\\ =0.03908\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

Formula to calculate the outer radius of the hollow cylinder is,

  $R_{\text{outer}}=\frac{D}{2}$

Here, $D$ is the diameter of the cylinder.

Substitute $18.0\text{cm}$ for $D$ in above equation to find $R_{\text{outer}}$.

  $\begin{array}{c}{R}_{\text{outer}}=\frac{18.0\text{cm}}{2}\\ =9\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\\ =0.09\text{m}\end{array}$

For large energy storage by the moment of inertia of flywheel must be large but the mass should be small as much as possible for design to fulfill this requirement the mass should place as far away from the axis as possible to increase the moment of inertia.

Let choose a hollow cylinder to make the flywheel of $18.0\text{cm}$ in diameter and $8.00\text{cm}$ long. To support this rim, place a disk across its center. Assume the disk is $2.00\text{cm}$ thick will be enough to support the hollow cylinder securely.

Formula to calculate the moment of inertia of the flywheel is,

    $I=I_{\text{disk}}+I_{\text{hollow}\text{cylinder}}$        (1)

Here, $I$ is the total moment of inertia of the flywheel, $I_{\text{disk}}$ is the moment of inertia of the sturdy disk and $I_{\text{hollow}\text{cylinder}}$ is the moment of inertia of the hollow cylinder.

Write the expression for the moment of inertia of the disk is,

    $I_{\text{disk}}=\frac{1}{2}{M}_{\text{disk}}{R}_{\text{disk}}^2$

Here, $M_{\text{disk}}$ is the mass of the disk and $R_{\text{disk}}$ is the radius of the disk.

Assume the radius of the disk is equal to the outer radius of the hollow cylinder.

Write the expression for the moment of inertia of the hollow cylinder is,

    $I_{\text{hollow}\text{cylinder}}=\frac{1}{2}{M}_{\text{wall}}\left(R_{\text{outer}}^2+R_{\text{inner}}^2\right)$

Here, $M_{\text{wall}}$ is the mass of the hollow cylinder wall, $R_{\text{outer}}$ is the outer radius of the hollow cylinder and $R_{\text{inner}}$ is the inner radius of the hollow cylinder.

Substitute $\frac{1}{2}{M}_{\text{disk}}{R}_{\text{disk}}^2$ for $I_{\text{disk}}$ and $\frac{1}{2}{M}_{\text{wall}}\left(R_{\text{outer}}^2+R_{\text{inner}}^2\right)$ for $I_{\text{hollow}\text{cylinder}}$ in equation (1).

    $I=\frac{1}{2}{M}_{\text{disk}}{R}_{\text{disk}}^2+\frac{1}{2}{M}_{\text{wall}}\left(R_{\text{outer}}^2+R_{\text{inner}}^2\right)$        (2)

Write the expression for the mass of the hollow cylinder wall is,

    $M_{\text{wall}}=\rho \pi \left(R_{\text{outer}}^2-R_{\text{inner}}^2\right)L$

Here, $\rho$ is the density of the disk and $L$ is the length of the cylinder wall.

Write the expression for the mass of the disk is,

    $M_{\text{disk}}=\rho \left(\pi R_{\text{disk}}^2{t}_{\text{disk}}\right)$

Here, $t_{\text{disk}}$ is the thickness of the disk.

Substitute $\rho \pi \left(R_{\text{outer}}^2-R_{\text{inner}}^2\right)L$ for $M_{\text{wall}}$ and $\rho \left(\pi R_{\text{disk}}{t}_{\text{disk}}\right)$ for $M_{\text{disk}}$ in equation (2).

    $\begin{array}{c}I=\frac{1}{2}\left(\rho \left(\pi R_{\text{disk}}^2{t}_{\text{disk}}\right)\right)R_{\text{disk}}^2+\frac{1}{2}\left(\rho \pi \left(R_{\text{outer}}^2-R_{\text{inner}}^2\right)L\right)\left(R_{\text{outer}}^2+R_{\text{inner}}^2\right)\\ =\frac{\rho \pi }{2}\left[R_{\text{disk}}^4{t}_{\text{disk}}+\left(R_{\text{outer}}^4-R_{\text{inner}}^4\right)L\right]\end{array}$

Substitute $0.03908\text{kg}\cdot {\text{m}}^{\text{2}}$ for $I$, $2.00\text{cm}$ for $t_{\text{disk}}$, $0.09\text{m}$ for $R_{\text{outer}}$ and $R_{\text{disk}}$ , $8.00\text{cm}$ for $L$ and $7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ to find $R_{\text{inner}}$.

    $\begin{array}{c}\frac{\left(7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\pi }{2}\left[\begin{array}{l}{\left(0.09\text{m}\right)}^4\left(2.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)+\\ \left({\left(0.09\text{m}\right)}^4-R_{\text{inner}}^4\right)\\ \left(8.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\end{array}\right]=0.03908\text{kg}\cdot {\text{m}}^{\text{2}}\\ \left[\begin{array}{l}1.31\times {10}^{-6}{\text{m}}^5-\\ \left(6.5\times {10}^{-5}{\text{m}}^4-R_{\text{inner}}^4\right)\left(0.08\text{m}\right)\end{array}\right]=\frac{0.03908\text{kg}\cdot {\text{m}}^{\text{2}}}{12330.75\text{kg}/{\text{m}}^{\text{3}}}\\ \left(6.5\times {10}^{-5}{\text{m}}^4-R_{\text{inner}}^4\right)\left(0.08\text{m}\right)=\left(\begin{array}{l}1.31\times {10}^{-6}{\text{m}}^5-\\ 3.17\times {10}^{-6}{\text{m}}^5\end{array}\right)\\ 6.5\times {10}^{-5}{\text{m}}^4-R_{\text{inner}}^4=\frac{-1.86\times {10}^{-6}{\text{m}}^5}{0.08\text{m}}\end{array}$

Solve the equation further,

    $\begin{array}{c}6.5\times {10}^{-5}{\text{m}}^4-R_{\text{inner}}^4=2.325\times {10}^{-5}{\text{m}}^4\\ R_{\text{inner}}^4=6.5\times {10}^{-5}{\text{m}}^4-2.325\times {10}^{-5}{\text{m}}^4\\ R_{\text{inner}}^4=4.175\times {10}^{-5}{\text{m}}^4\\ R_{\text{inner}}=0.080\text{m}\end{array}$

Formula to calculate the mass of the flywheel is,

    $M=M_{\text{disk}}+M_{\text{wall}}$

Substitute $\rho \left(\pi R_{\text{disk}}^2{t}_{\text{disk}}\right)$ for $M_{\text{disk}}$ and $\rho \pi \left(R_{\text{outer}}^2-R_{\text{inner}}^2\right)L$ for $M_{\text{wall}}$ to find $M$.

    $\begin{array}{c}M=\rho \left(\pi R_{\text{disk}}^2{t}_{\text{disk}}\right)+\rho \pi \left(R_{\text{outer}}^2-R_{\text{inner}}^2\right)L\\ =\rho \pi \left[R_{\text{disk}}^2{t}_{\text{disk}}+\left(R_{\text{outer}}^2-R_{\text{inner}}^2\right)L\right]\end{array}$

Conclusion:

Substitute $7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2.00\text{cm}$ for $t_{\text{disk}}$, $0.09\text{m}$ for $R_{\text{outer}}$ and $R_{\text{disk}}$ , $8.00\text{cm}$ for $L$ and $0.080\text{m}$ for $R_{\text{inner}}$ to find $M$.

    $\begin{array}{c}M=\left(7.85\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\pi \left[\begin{array}{l}{\left(0.09\text{m}\right)}^2\left(2.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)+\\ \left({\left(0.09\text{m}\right)}^2-{\left(0.08\text{m}\right)}^2\right)\left(8.00\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\end{array}\right]\\ =2.466\times {10}^4\text{kg}/{\text{m}}^{\text{3}}\times 2.98\times {10}^{-4}{\text{m}}^{\text{3}}\\ =7.34\text{kg}\end{array}$

Therefore, shape of the flywheel is hollow cylinder of inner radius $0.08\text{m}$ and outer radius $0.09\text{m}$ with the mass $7.34\text{kg}$.