A steel flywheel in the form of a solid disk has a radius of 0.2 m and a thickness of $0.1 \mathrm {~m}$.
(a) What is the maximum rotational frequency that is compatible with the material characteristics of this flywheel?
(b) What is its energy storage capacity at maximum rotational frequency?
(c) If the flywheel is energized by a motor with a power output of $1 \mathrm {~kW}$ (at $100 \%$ efficiency) how long would it take to increase the rotational frequency from zero to the maximum?

Question

Quizplus · Accepted Answer

The Answer is (a) From equation (18.10)

\sigma = ho r ^ { 2 } \omega ^ { 2 } = 4 ho r ^ { 2 } \pi ^ { 2 } f ^ { 2 }

so

f = \frac { 1 } { 2 \pi } \sqrt { \frac { \sigma } { ho r ^ { 2 } } }

So for

\begin{array} { l } r = 0.2 \mathrm {~m} \ ho = 7.8 imes 10 ^ { 3 } \mathrm {~kg} / \mathrm { m } ^ { 3 } \\sigma = 1.72 imes 10 ^ { 9 } \mathrm {~N} / \mathrm { m } ^ { 2 }\end{array}

then we find a value of the maximum rotational frequency

f = [ 1 / ( 2 imes 3.14 ) ] imes \left[ \left( 1.72 imes 10 ^ { 9 } \mathrm {~N} / \mathrm { m } ^ { 2 } ight) / \left[ \left( 7.8 imes 10 ^ { 3 } \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) imes ( 0.2 \mathrm {~m} ) ^ { 2 } ight] = 373 \mathrm {~s} ^ { - 1 } ight.

that is about

\left( 373 \mathrm {~s} ^ { - 1 } ight) imes ( 60 \mathrm {~s} / \mathrm { m } ) = 22,400 \mathrm { rpm }

.
(b) The rotational energy is

E = ( 1 / 2 ) I \omega ^ { 2 }

where the angular frequency is

\omega = 2 \pi f

and for a disk the moment of inertia is

I = k m r ^ { 2 }

with

k = 1 / 2

. Substituting into the equation for the energy gives

E = \pi ^ { 2 } m r ^ { 2 } f ^ { 2 }

The mass is found to be

m = ( 3.14 ) imes ( 0.2 \mathrm {~m} ) ^ { 2 } imes ( 0.1 \mathrm {~m} ) imes \left( 7.8 imes 10 ^ { 3 } \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) = 98.0 \mathrm {~kg}

Using the values from the problem gives

E = ( 3.14 ) ^ { 2 } imes ( 98.0 \mathrm {~kg} ) imes ( 0.2 \mathrm {~m} ) ^ { 2 } imes \left( 373 \mathrm {~s} ^ { - 1 } ight) ^ { 2 } = 5.38 \mathrm { MJ }

(c) If the input power is

1 \mathrm {~kW}

then the energy stored after a time

t

is

( 1 \mathrm {~kW} ) t

so the time is

\left( 5.38 imes 10 ^ { 6 } \mathrm {~J} ight) / ( 1000 \mathrm {~W} ) = 5.38 imes 10 ^ { 3 } \mathrm {~s}

or about an hour and a half.

A Steel Flywheel in the Form of a Solid Disk 0.1 m0.1 \mathrm {~m}0.1 m

A Steel Flywheel in the Form of a Solid Disk $0.1 \mathrm {~m}$