Some European trucks run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 120 ð rad/s. One such flywheel is a solid, uniform cylinder with a mass of 270 kg and a radius of 1.10 m. (a) What is the kinetic energy of the flywheel after charging? (b) If the truck uses an average power of 7.3 kW, for how many minutes can it operate between chargings?

What is your question here? a) is straightforeqrd, KE=1/2 I*w^2

To find the kinetic energy of the flywheel after charging, we can use the formula for kinetic energy:

Kinetic Energy (K.E.) = 1/2 * I * ω²

where:
I is the moment of inertia
ω (omega) is the angular velocity

(a) Finding the kinetic energy of the flywheel after charging:
First, we need to find the moment of inertia (I) of the solid, uniform cylinder.

The moment of inertia of a solid, uniform cylinder is given by the formula:

I = (1/2) * m * r²

where:
m is the mass of the cylinder
r is the radius of the cylinder

Given:
m = 270 kg
r = 1.10 m

Plugging in the values, we can calculate the moment of inertia (I):

I = (1/2) * 270 kg * (1.10 m)²

Now, we can find the angular velocity (ω). The top speed of the flywheel is given as 120 ð rad/s. So, ω = 120 ð rad/s.

Now, we can use the formula for kinetic energy:

K.E. = 1/2 * I * ω²

Plugging in the calculated values:

K.E. = 1/2 * (1/2) * 270 kg * (1.10 m)² * (120 ð rad/s)²

(b) To find the operating time between chargings, we can use the formula for power:

Power (P) = Work Done (W) / Time (t)

Given:
Power (P) = 7.3 kW (kilowatts)
1 kW = 1000 W
Time (t) = ? (in minutes)

We know that Power (P) = Work Done (W) / Time (t), and Work Done (W) is equal to the change in kinetic energy (ΔK.E.) of the flywheel.

Therefore:

P = ΔK.E. / t

We already know the power (P) value, and we can find the energy change (ΔK.E.) by subtracting the initial kinetic energy from the final kinetic energy.

Finally, solving for time (t):

t = ΔK.E. / P

We can plug in the values and calculate the time (t) in minutes.