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/m3) 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?
Solution 75P Step 1: The kinetic energy of a rotating disk is K.E = 1/2 Iw2 Where K.E = kinetic energy in joules 2 I = moment of inertia in kg/m w = angular velocity in rad/s The moment of inertia of rotating disk is 2 I = Mr /2 2 Mass of disk M = dr Here d is density,r is radius and h is thickness. So 2 2 2 K.E = 1/4 * * r * *r *w 4 2 K.E = 1/4 * * r * *w 2 1/4 r = (4K.E/(d*hw ) Step 2: w = 90.0 rpm *(2 rad/rev)/(60 s/1 min)] = 9.424 rad/s K.E = 10 × 10 J6 2 h = 11 × 10 m 3 d = 7800 kg/m Then r = [4(10 × 10 J)/(7800 kg/m 3 × 11 × 10 2(9.424 rad/s) ]1/4 * r = 3.595 m Then diameter d = 2r = 2(3.595 m) = 7.190 m