A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 600 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 4000 rev/min.

(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.

Model the flywheel as a sold disk, look up the moment of inertial

KE= 1/2 I w^2 where w=4000^2PI/60 rad/sec

change 15hp to kwatts
time= power/KEabove

Can i have more help i tried that every time in part a and get the wrong answer on my webassign i tried it 5 times and got five different answers. the answer would be nice so i can get part b.

To find the kinetic energy stored in the flywheel, we can use the formula for rotational kinetic energy:

K = (1/2) I ω^2

Where:
K is the kinetic energy,
I is the moment of inertia of the flywheel, and
ω is the angular velocity of the flywheel.

(a) First, let's find the moment of inertia of the flywheel. The moment of inertia for a solid cylinder rotating around its axis is given by the formula:

I = (1/2) m r^2

Where:
m is the mass of the flywheel, and
r is the radius of the flywheel.

Plugging in the given values, we get:

I = (1/2) (600 kg) (1.50 m)^2 = 450 kg.m^2

Next, we need to convert the angular velocity from rev/min to rad/s. There are 2π radians in one revolution, and 60 seconds in one minute, so:

ω = (4000 rev/min) (2π rad/1 rev) (1 min/60 s) = 418.88 rad/s

Substituting these values into the kinetic energy formula:

K = (1/2) (450 kg.m^2) (418.88 rad/s)^2 ≈ 19890720 J

Therefore, the kinetic energy stored in the flywheel is approximately 19,890,720 J.

(b) To determine the length of time the car could run before the flywheel needs to be brought back up to speed, we can use the power-time relationship:

P = ΔE/Δt

Where:
P is the power supplied by the flywheel,
ΔE is the change in energy of the flywheel, and
Δt is the time interval.

We are told that the flywheel is supplying energy to the car as a 15.0 hp motor. One horsepower (hp) is equal to 745.7 watts, so:

P = (15.0 hp) (745.7 W/hp) = 11185.5 W

The change in energy of the flywheel is equal to the kinetic energy stored in it, which we found to be approximately 19,890,720 J.

Substituting these values into the power-time relationship:

11185.5 W = 19890720 J / Δt

Solving for Δt:

Δt = 19890720 J / 11185.5 W ≈ 1777 s

Therefore, the car could run for approximately 1777 seconds or 29.62 minutes before the flywheel needs to be brought back up to speed.