Other Materials (In Addition to Those Given in Table 18$18.3$

Question

Examlex · Accepted Answer

The Answer is The maximum energy storage capability is

E = ( 1 / 2 ) k m \sigma / ho

Results of this calculation with

m = 1 \mathrm {~kg}

and

k = 1 / 2

for a solid disk are given in the table below.

\begin{array}{llll}\hline ext { material } & \begin{array}{l} ext { density } \\left(10^{3} \mathrm{~kg} / \mathrm{m}^{3} ight)\end{array} & \begin{array}{l} ext { tensile strength } \\left(10^{9} \mathrm{~N} / \mathrm{m}^{2} ight)\end{array} & E_{\max }(\mathrm{MJ}) \\hline ext { bamboo } & 0.4 & 0.5 & 0.31 \ ext { graphene } & 1.0 & 130.0 & 32.5 \ ext { kevlar } & 1.44 & 3.76 & 0.65 \ ext { zylon } & 5.8 & 1.56 & 0.067 \ ext { boron } & 2.46 & 3.1 & 0.31\end{array}

These results may be compared with the maximum energy capacity of flywheels made from some commonly used materials using the densities and tensile strengths as given in the text

\begin{array}{llll}\hline ext { material } & \begin{array}{l} ext { density } \\left(10^{3} \mathrm{~kg} / \mathrm{m}^{3} ight)\end{array} & \begin{array}{l} ext { tensile strength } \\left(10^{9} \mathrm{~N} / \mathrm{m}^{2} ight)\end{array} & E_{\max }(\mathrm{MJ}) \\hline ext { steel } & 7.2 & 1.71 & 0.055 \ ext { aluminum } & 2.7 & 0.59 & 0.055 \ ext { titanium } & 4.5 & 1.22 & 0.0675 \ ext { fibreglass } & 2.0 & 1.60 & 0.20 \ ext { carbon fibre } & 1.5 & 2.4 & 0.40 \\hline\end{array}

It is seen that even some common materials, e.g. bamboo and boron, are comparable to the best materials given in the table in the text. Graphene is two orders of magnitude better than the best common materials but is, at present, not suitable for the manufacture of bulk materials.

Other Materials (In Addition to Those Given in Table 18 $18.3$

Definitions:

Other Materials (In Addition to Those Given in Table 18 18.318.318.3

Definitions:

Other Materials (In Addition to Those Given in Table 18 $18.3$