Chapter 10, Problem 20P

To determine

The shape and mass of the flywheel.

Answer to Problem 20P

The flywheel can be in the shape of a cup or open barrel, with mass $\underset{\_}{7.27\text{kg}}$.

Explanation of Solution

To have large energy storage the flywheel should have large moment of inertia. For this the flywheel should have small mass. Consider the flywheel as a hollow cylinder with diameter $18.0\text{cm}$, and $8.00\text{m}$ long.

To support the rim place the disk at the center. The total moment of inertia is the sum of moment of inertia of the disk and hollow cylinder.

Write the expression for total moment of inertia.

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

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

Substitute, $\rho V_{\text{disk}}$ for $M_{\text{disk}}$, and $\rho V_{\text{wall}}$ for $M_{\text{wall}}$ in equation (I).

  $I_{\text{disk}}+I_{\text{hollow cylinder}}=\frac{1}{2}\rho V_{\text{disk}}{R}_{\text{disk}}^2+\frac{1}{2}\rho V_{\text{wall}}\left(R_{\text{outer}}^2+R_{\text{inner}}^2\right)$        (II)

Here, $\rho$ is the density, $V_{\text{diskl}}$ is the volume of the disk, $V_{\text{wall}}$ is the volume of the wall.

Substitute, $\pi R_{\text{outer}}^2\left(2.00\text{cm}\right)$ for $V_{\text{diskl}}$, $\left[\pi R_{\text{outer}}^2-\pi R_{\text{inner}}^2\right]\left(6.00\text{cm}\right)$ for $V_{\text{wall}}$ in equation (II).

  $I_{\text{disk}}+I_{\text{hollow cylinder}}=\frac{1}{2}\rho \pi R_{\text{outer}}^2\left(2.00\text{cm}\right)R_{\text{disk}}^2+\frac{1}{2}\rho \left[\pi R_{\text{outer}}^2-\pi R_{\text{inner}}^2\right]\left(6.00\text{cm}\right)\left(R_{\text{outer}}^2+R_{\text{inner}}^2\right)$        (III)

Write the expression for the total energy stored.

  $W_{\text{out}}=\frac{1}{2}I{\omega }_1^2-\frac{1}{2}{\omega }_2^2$        (IV)

Here, $I$ is the moment of inertia, ${\omega }_1$ is the initial angular velocity, and ${\omega }_2$ is the final angular velocity.

Conclusion:

Substitute, $7.85\times {10}^3\text{kg}/{\text{m}}^3$ for $\rho$, $9.00\text{cm}$ for $R_{\text{outer}}$ in equation (III).

  $\begin{array}{c}{I}_{\text{disk}}+I_{\text{hollow cylinder}}=\frac{1}{2}\left(7.85\times {10}^3\text{kg}/{\text{m}}^3\right)\pi {\left(9.00\text{cm}\right)}^4\left(2.00\text{cm}\right)+\\ \frac{1}{2}\left(7.85\times {10}^3\text{kg}/{\text{m}}^3\right)\left[\pi {\left(9.00\text{cm}\right)}^2-\pi R_{\text{inner}}^2\right]\left(6.00\text{cm}\right)\left({\left(9.00\text{cm}\right)}^2+R_{\text{inner}}^2\right)\\ =7.85\times {10}^3\text{kg}/{\text{m}}^3\pi \left[26244{\text{cm}}^{45}-\left(3.00\text{cm}\right)R_{\text{inner}}^4\right]\end{array}$

Substitute, $600\text{J}$ for $W_{\text{out}}$, $800\text{rev}/\text{min}\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)$ for ${\omega }_1$, $\left(600\text{rev}/\text{min}\right)\left(\frac{2\pi \text{rad}}{60\text{s}}\right)$ for ${\omega }_2$ in equation (IV).

  $\begin{array}{c}600\text{J}=\frac{1}{2}I{\left(800\text{rev}/\text{min}\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)\right)}^2-\frac{1}{2}{\left(\left(600\text{rev}/\text{min}\right)\left(\frac{2\pi \text{rad}}{60\text{s}}\right)\right)}^2\\ I=\frac{600\text{J}}{1535/{\text{s}}^2}\end{array}$        (V)

Compare equation (V) and $I_{\text{disk}}+I_{\text{hollow cylinder}}$.

  $\begin{array}{c}\frac{600\text{J}}{1535/{\text{s}}^2}=7.85\times {10}^3\text{kg}/{\text{m}}^3\pi \left[26244{\text{cm}}^5-\left(3.00\text{cm}\right)R_{\text{inner}}^4\right]\\ R_{\text{inner}}={\left(\frac{26244{\text{cm}}^4-15827{\text{cm}}^4}{3.00}\right)}^{1/4}\\ =7.68\text{cm}\end{array}$

The total mass is the mass of the disk and hollow cylinder.

Write the expression for total mass.

  $M_{\text{disk}}+M_{\text{wall}}=\rho \pi R_{\text{outer}}^2\left(2.00\text{cm}\right)+\rho \left[\pi R_{\text{outer}}^2-\pi R_{\text{inner}}^2\right]\left(6.00\text{cm}\right)$        (VI)

Substitute, $7.85\times {10}^3\text{kg}/{\text{m}}^3$ for $\rho$, $0.090\text{m}$ for $R_{\text{outer}}$, and $0.0768\text{m}$ for $R_{\text{inner}}$ in equation (VI).

  $\begin{array}{c}{M}_{\text{disk}}+M_{\text{wall}}=7.85\times {10}^3\text{kg}/{\text{m}}^3\pi {\left(0.090\text{m}\right)}^2\left(2.00\text{cm}\right)+7.85\times {10}^3\text{kg}/{\text{m}}^3\left[\pi {\left(0.090\text{m}\right)}^2-\pi {\left(0.0768\text{m}\right)}^2\right]\left(6.00\text{cm}\right)\\ =7.27\text{kg}\end{array}$

If the thickness of the disk is less than $2.00\text{cm}$, and the radius of the cylindrical wall is less than $7.68\text{cm}$, hence the mass will be less than $7.27\text{kg}$.

Conclusion:

The flywheel can have a shape of cup or open barrel, with inner radius $7.68\text{cm}$, and outer radius $9.00\text{cm}$, and the height of the cylindrical wall be $6\text{cm}$. The cylinder should be mounted at the center of the disk and turned about the axis of symmetry. If the disk made somewhat thinner and barrel wall thicker, the mass will be less than $7.27\text{kg}$.