Question

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 in diameter. Its thickness, measured along its axis of rotation, must be no larger than$8.00cm$ . The flywheel must release energy $60.0J$when its angular speed drops from .role="math" localid="1663659241681" $800rev/min$ Design a sturdy steel (density ) flywheel to meet these requirements with the smallest mass you can reasonably attain. Specify the shape and mass of the flywheel.

Short answer

Reduce the thickness of the disc to less than $2cm$and the inner radius of the cylinder wall to less than $7.68cm$ resulting in a mass of less than$7.27kg$

Step-by-step solution

Given Information

Consider the flywheel as a hollow cylinder with a diameter of $18cm$and a length of $8cm.$A disc is placed across the middle of this rim to provide stability. We assume that a disc $2cm$thick will be sufficient to firmly hold the hollow cylinder.

Moment of Inertia

Using the notion of moment of inertia $I=\sum m{r}^2$rotational kinetic energy $K=\frac{1}{2}I{\omega }^2$, this problem is a substitution problem.

Analyzing the given information

The fly wheel has an outside radius of $2.00cm$

$\begin{array}{l}=2.00cm\left(\frac{1m}{100cm}\right)\\ =0.02cm\end{array}$

Outer radius of the fly wheel is$R_{outer}=\frac{d}{2}$

$\begin{array}{l}=\frac{18cm}{2}\\ =9cm\left(\frac{1m}{100cm}\right)\\ =0.09m\end{array}$

We need a huge moment of inertia for large energy storage at a specific rotation rate. We situate the mass as far away from the axis as possible to meet this criteria while maintaining a low mass. The thickness of the hollow cylinder's wall is an adjustable parameter.

The inertia moment can be expressed as

$\begin{array}{l}{I}_{disk}+I_{hollowcylinder}=\frac{1}{2}{M}_{disk}{R}_{disk}^2+\frac{1}{2}{M}_{wall}(R_{outer}^2+R_{inner}^2)\\ =\frac{1}{2}\left(\rho V_{disk}\right)R_{disk}^2+\frac{1}{2}\left(\rho V_{wall}\right)(R_{outer}^2+R_{inner}^2)\end{array}$

$\begin{array}{l}=\frac{1}{2}\rho \left[\left(\pi R_{outer}^2\right)(2cm)\right]R_{outer}^2+\frac{1}{2}\rho \left[(\pi R_{outer}^2-\pi R_{inner}^2)(6cm)\right](R_{outer}^2+R_{inner}^2)\\ =\frac{\rho \pi }{2}\left[{(9cm)}^4(2cm)+(6cm)({(9cm)}^2-R_{inner}^2)({(9cm)}^2+R_{inner}^2)\right]\\ =\rho \pi \left[{(9cm)}^4(1cm)+(3cm)({(9cm)}^4-R_{inner}^4)\right]\end{array}$

$\begin{array}{l}=\rho \pi \left[6561c{m}^5+(3cm)({(9cm)}^4-R_{inner}^4)\right]\\ =\rho \pi \left[26244c{m}^5-(3cm)R_{inner}^4\right]\end{array}$

Use of the law of conservation of energy

Energy conservation is based on the law of conservation of energy.

$\frac{1}{2}I{\omega }_1^2=\frac{1}{2}I{\omega }_2^2+W_{out}$

$\begin{array}{l}\frac{1}{2}I{\left(7011.27rad/s\right)}^2=\frac{1}{2}I(3943.84rad/s)+60J\\ I{\left(7.11.27rad/s\right)}^2-I(3943.84rad/s)=2(60J)\\ I(3067.43)=120\end{array}$

$\begin{array}{l}\rho \pi \left[26244c{m}^5-(3cm)R_{inner}^4\right](3067.43)=120\\ (7.85\times {10}^{3}kg/m^3)\pi \left[26244c{m}^5-(3cm)R_{inner}^4\right]=\frac{120}{3067.43}\\ \left[26244c{m}^5-(3cm)R_{inner}^4\right]=\frac{120}{(3067.43)\pi (7.85\times {10}^{3}kg/m^3)}\end{array}$

$\begin{array}{l}\left[26244c{m}^5-(3cm)R_{inner}^4\right]=1.587\times {10}^{-6}\\ R_{inner}={\left[\frac{26244\times {10}^{-10}-1.587\times {10}^{-6}}{0.03m}\right]}^{1/4}\\ =0.0768m\left(\frac{100cm}{1m}\right)\end{array}$

$=7.68cm$

The flywheel has a mass of

$\begin{array}{l}{M}_{disk}+M_{wall}=\rho \pi R_{outer}^2(2cm)+\rho [\pi R_{outer}^2-\pi R_{inner}^2](6cm)\\ =(7.85\times {10}^{3}kg/m^3)\pi \left[{(0.09m)}^2(0.02m)+\{{(0.09m)}^2-{(0.0768m)}^2\}(0.06m)\right]\\ =7.27kg\end{array}$

To compensate, we might reduce the thickness of the disc to less than $2cm$and the inner radius of the cylinder wall to less than $7.68cm$, resulting in a mass of less than$7.27kg$