Question

A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.

Answer 1

Answer:

a

  $KE = 7.17 *10^{7} \ J$

b

 $t = 6411.09 \ s$

Explanation:

From the question we are told that

    The radius of the flywheel is  $r = 1.50 \ m$

      The mass of the flywheel is $m = 430 \ kg$

          The rotational speed of the flywheel is $w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec$

      The power supplied by the motor is  $P = 15.0 hp = 15 * 746 = 11190 \ W$

         

     Generally the moment of inertia of the flywheel is  mathematically represented as

       $I = \frac{1}{2} mr^2$

substituting values

       $I = \frac{1}{2} ( 430)(1.50)^2$

       $I = 483.75 \ kgm^2$

The kinetic energy that is been stored is  

       $KE = \frac{1}{2} * I * w^2$

substituting values

        $KE = \frac{1}{2} * 483.75 * (544.61)^2$

        $KE = 7.17 *10^{7} \ J$

Generally power is mathematically represented as

          $P = \frac{KE}{t}$

=>      $t = \frac{KE}{P}$

substituting the value

        $t = \frac{7.17 *10^{7}}{11190}$

        $t = 6411.09 \ s$