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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.

User Huseyint
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1 Answer

4 votes

Answer:

a


KE = 7.17 *10^(7) \ J

b


t = 6411.09 \ s

Step-by-step 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 * (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 = (1)/(2) mr^2

substituting values


I = (1)/(2) ( 430)(1.50)^2


I = 483.75 \ kgm^2

The kinetic energy that is been stored is


KE = (1)/(2) * I * w^2

substituting values


KE = (1)/(2) * 483.75 * (544.61)^2


KE = 7.17 *10^(7) \ J

Generally power is mathematically represented as


P = (KE)/(t)

=>
t = (KE)/(P)

substituting the value


t = (7.17 *10^(7))/(11190)


t = 6411.09 \ s

User Parchambeau
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