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.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-07-03T04:48:22+02:00

Answer:

a

[tex]KE = 7.17 *10^{7} \ J[/tex]

b

[tex]t = 6411.09 \ s[/tex]

Explanation:

From the question we are told that

The radius of the flywheel is [tex]r = 1.50 \ m[/tex]

The mass of the flywheel is [tex]m = 430 \ kg[/tex]

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

The power supplied by the motor is [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]

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

[tex]I = \frac{1}{2} mr^2[/tex]

substituting values

[tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]

[tex]I = 483.75 \ kgm^2[/tex]

The kinetic energy that is been stored is

[tex]KE = \frac{1}{2} * I * w^2[/tex]

substituting values

[tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]

[tex]KE = 7.17 *10^{7} \ J[/tex]

Generally power is mathematically represented as

[tex]P = \frac{KE}{t}[/tex]

=> [tex]t = \frac{KE}{P}[/tex]

substituting the value

[tex]t = \frac{7.17 *10^{7}}{11190}[/tex]

[tex]t = 6411.09 \ s[/tex]