Flywheels are large, massive wheels used to store energy. they can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. an industrial flywheel has a 1.5 m diameter and a mass of 250 kg. its maximum angular velocity is 12000 rpm.

a) a motor spins up the flywheel with a constant torque of 50 nm. how long does it take the flywheel to reach top speed?

b) how much energy is stored in the flywheel?

c) the flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. half of the energy stored in the flywheel is delivered in 2.0 s. what is the average power delivered to the machine?

d) how much torque does the flywheel exert on the machine?

a) 1766 sec. b) 55.5 MJ c) 13.9 MW d) -12,944 Nm

Explanation:

a) The torque and  the angular acceleration, are related by the following expression, which resembles very much to the Newton's 2nd Law for point masses:

ζ = I . γ, where ζ=external torque, I = rotational inertia and γ = angular acceleration.

We also know that a flywheel is a solid disk, so the rotational inertia for this type of body is equal to MR² / 2.

By definition, angular acceleration is the rate of change of angular velocity with time, so we can write the following:

γ = ωf -ω₀ / t

Assuming that the flywhel starts from rest, we know that ω₀ = 0, and ωf = 12,000 rpm.

As all the units are given in SI units, it is advisable to convert the rpm to rad/sec, as follows:

12,000 rpm = 12,000 rev. (2π/rev) . (1min/60 sec) = 400 π rad/sec

Returning to the original equation, we have:

ζ = MR² / 2 . (ωf/ t)

Replacing by the values, and solving for t, we have:

t = 250 Kg. (0.75)² m² . 400 π / 2. 50 Nm = 1,766 sec.

b) Due to the flywheel is just rotating, all the stored energy is rotational kinetic energy, which can be written as follows:

K = 1/2 I ωf² = 1/2 (MR²/2) ωf² = 1/4. 250 Kg. (0.75)² m². (400π)²

K= 55.5 MJ

c) Power is defined as energy delivered in a given time.

The energy delivered, is just the half of the originally stored value, i.e. , 55.5 MJ /2, equal to 27.75 MJ.

Dividing this value by 2.0 sec, we have the average power delivered to the machine, that we found to be equal to 27.75 MJ / 2s =  13. 9 MW

d) Using the same relationship than in a), we can write the following:

ζ = I. γ

I remains the same (as the flywheel is the same), so the only unknown is the angular acceleration.

Angular acceleration, by definition, is as follows:

γ = ωf - ω₀ / t

We know the value of ω₀, as it is the top speed value that we have already got,i.e., 400 π rad/sec.

We don't know the value for ωf, but we know the value of the rotational kinetic energy after 2.0 secs, which is equal to the half of the one we obtained in step b).

So, we can write the following:

Kf = 1/2 I ωf² = 1/2 (1/2 I ω₀²) ⇒ 1/ 2 ωf² = 1/4 ω₀² ⇒ωf = ω₀/√2

Replacing in the expression for angular acceleration:

γ = (ω₀/√2 - ω₀) / t = -0.29. 400. π/ 2 rad/sec²= -184.1 rad/sec²

Finally, we can get the torque as follows:

ζ = (250 kg. (0.75)² m² /2) . 184.1 rad/sec² = -12,944 Nm

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be