It has been argued that power plants should make useof off-peak hours (such as late at night) to generate mechanicalenergy and store it until it is needed during peak load times, suchas the middle of the day. One suggestion has been to store theenergy in large flywheels spinning on nearly frictionless ballbearings. Consider a flywheel made of iron (density \(7800\,\,{{kg} \mathord{\left/ {\vphantom {{kg} {{m^3}}}} \right.} {{m^3}}}\))in the shape of a 10.0-cm-thick uniform disk.

(a) What would thediameter of such a disk need to be if it is to store 10.0 megajoulesof kinetic energy when spinning at 90.0 rpm about an axis perpendicularto the disk at its center?

(b) What would be the centripetalacceleration of a point on its rim when spinning at this rate?

Given data in the question

The density of the flywheel, \(\rho = \,\,7800\,\,{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.} {{{\rm{m}}^{\rm{3}}}}}\)

The thickness of the flywheel,\(t = 10.0\,\,{\rm{cm}}\)

The kinetic energy of the flywheel,\(KE = 10\,\,{\rm{MJ}}\)

Angular velocity of the flywheel, \(\omega = 90\,\,{\rm{rpm}}\)

Converting angular velocity into \({{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\) \(\omega = 90\,\,\left( {\frac{{2\pi }}{{60}}} \right)\,{{\,{\rm{rad}}} \mathord{\left/ {\vphantom {{\,{\rm{rad}}} {\rm{m}}}} \right.} {\rm{m}}}\)

Rotational energy

Rotational kinetic energy is also known as angular kinetic energy, this iskinetic energydue to the rotation of any object or body and is part of thetotal kinetic energy of the system.

Since the flywheel is rotating it has rotational kinetic energy.

Where r is radius and\(\omega \) is the angular velocity

\(a = r{\omega ^2}\)

Where r is radius and \(\omega \)is angular velocity.

First, divide the flywheel disk into circular rings of width dr

So, the mass of the circular ring of radius r and thickness t can be given as,

\(\begin{aligned}{\rm{mass = density}}\,{\rm{ \times }}\,{\rm{volume}}\\dm = \rho \left( {2\pi rtdr} \right)\end{aligned}\)

Therefore, the kinetic energy of a circular ring can be given as,

\(\begin{aligned}K = \frac{1}{2}m{v^2}\\dK = \frac{1}{2}dm{\left( {r\omega } \right)^2}\end{aligned}\)

\(dK = \frac{1}{2}\left( {\rho \left( {2\pi rtdr} \right)} \right){\left( {r\omega } \right)^2}\)

The total kinetic energy of the flywheel disc is the sum of the energy of each circular ring.

\(K = \int\limits_0^r {\frac{1}{2}\left( {\rho \left( {2\pi rtdr} \right)} \right){{\left( {r\omega } \right)}^2}} \)

Substituting the values

\(\begin{aligned}K = \int\limits_0^r {\frac{1}{2}\left( {\rho \left( {2\pi rtdr} \right)} \right){{\left( {r\omega } \right)}^2}} \\10 \times {10^6}\,{\rm{J}} = \int\limits_0^r {\frac{1}{2}\left( {\,\,7800\,\,{{{\rm{kg}}} \mathord{\left/ {\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.} {{{\rm{m}}^{\rm{3}}}}}} \right)\left( {2\pi r\left( {10.0 \times {{10}^{ - 2}}\,\,{\rm{m}}} \right)dr} \right){{\left( {90\,\,\left( {\frac{{2\pi }}{{60}}} \right)\,{{\,{\rm{rad}}} \mathord{\left/ {\vphantom {{\,{\rm{rad}}} {\rm{m}}}} \right.} {\rm{m}}}\,\,r\,} \right)}^2}} \\10 \times {10^6}\,{\rm{J}} = \frac{{4{\pi ^3}\left( {7800} \right)\left( {0.1} \right){{\left( {90} \right)}^2}}}{{{{\left( {60} \right)}^2}}}\int\limits_0^r {{r^3}} dr\\10 \times {10^6}\,{\rm{J}} = \frac{{4{\pi ^3}\left( {7800} \right)\left( {0.1} \right){{\left( {90} \right)}^2}}}{{{{\left( {60} \right)}^2}}}\left( {\frac{{{r^4}}}{4}} \right)\\r = 3.65\,\,{\rm{m}}\end{aligned}\)

The radius of the flywheel dick is \(3.65\,\,{\rm{m}}\)

Therefore

The diameter of the flywheel disk is

\(\begin{aligned}{c}d = 2r\\ = 2 \times 3.65\,\,{\rm{m}}\\{\rm{ = 7}}{\rm{.36}}\,\,{\rm{m}}\end{aligned}\)

Hence the diameter of the flywheel disk is \({\rm{7}}{\rm{.36}}\,\,{\rm{m}}\)

The centripetal acceleration of the flywheel can be given as

\(a = r{\omega ^2}\)

Substituting the values

\(\begin{aligned}a = \left( {{\rm{7}}{\rm{.36}}\,\,{\rm{m}}} \right){\left( {90\,\,\left( {\frac{{2\pi }}{{60}}} \right)\,{{\,{\rm{rad}}} \mathord{\left/ {\vphantom {{\,{\rm{rad}}} {\rm{m}}}} \right.} {\rm{m}}}} \right)^2}\\ = 327\,\,{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\end{aligned}\)

Hence the centripetal acceleration of flywheel discs is