Q9-67P - University Physics with Modern Physics

Question

It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball bearings. 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 the diameter of such a disk need to be if it is to store 10.0 megajoules of kinetic energy when spinning at 90.0 rpm about an axis perpendicular to the disk at its center? 

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

Step-by-Step Solution

Answer

1. The diameter of the flywheel disk is \({\rm{7}}{\rm{.36}}\,\,{\rm{m}}\) 
2. The centripetal acceleration of a point on the flywheel rim is 

Step 1: Identification of the given data.

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

Step 2: Concept used to solve the questions

Rotational energy 

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

Since the flywheel is rotating it has rotational kinetic energy.

Step 3: formula used to solve the problem

• The kinetic energy 
  \(K = \frac{1}{2}m{v^2}\) 
  Where m is the mass and v is the velocity

• Tangential Velocity 
  \(v = r\omega \) 

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

• Centripetal acceleration

           \(a = r{\omega ^2}\) 

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

Step 4: (a) Finding the diameter of the flywheel disk

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 \({\rm{7}}{\rm{.36}}\,\,{\rm{m}}\)

Step 4: (b) Finding the centripetal acceleration

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