Chapter 10, Problem 88CP

To determine

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

Answer to Problem 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

Given information: 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.

$W_{\text{out}}$ 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.

${\omega }_{\text{i}}$ 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 $I$.

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

Thus, the moment of inertia of the flywheel is $0.03908\text{kg}\cdot {\text{m}}^{\text{2}}$.

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 $R_{\text{outer}}$.

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

Thus, the outer radius of the hollow cylinder is $0.09\text{m}$.

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.

$I_{\text{hollow}\text{cylinder}}$ 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.

$R_{\text{disk}}$ 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.

$R_{\text{inner}}$ 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.

$L$ 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,

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

Thus, the inner radius of the hollow cylinder flywheel is $0.080\text{m}$.

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 $M$.

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

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 $M$.

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

Thus, the mass of the flywheel is $7.34\text{kg}$.

Conclusion:

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 $7.34\text{kg}$.