A passenger bus in Zurich, Switzerland derived its motive power from the energy stored in a large flywheel. The wheel was brought up to speed periodically, when the bus stopped at a station, by an electric motor, which could then be attached to the electric power lines. The flywheel was a solid cylinder with a mass of 1010 kg and a diameter of 1.80 m; its top angular speed was 3080 rev/min.
At this angular speed, what is the kinetic energy of the flywheel?
PART B
If the average power required to operate the bus is 1.81×104 , how long could it operate between stops?
I am confused in this question.
For part A
I=mr^2/2 = (1010)(0.81)/2 = 409.05
For KE= 1/2 I w ^2
= 1/2 (409.05)(19352.21075rev/min)
= 3958010.903
I think i did something wrong, can someone please direct me if i have.
Here's a step-by-step explanation of how to solve Part A of the question:
1. Calculate the moment of inertia (I) of the flywheel using the formula:
I = (mr^2)/2, where
m = mass of the flywheel in kg (1010 kg in this case),
r = radius of the flywheel in meters (diameter is given as 1.80 m).
I = (1010 kg)(0.9 m)^2 / 2
I = 408.15 kg⋅m²
2. Convert the angular speed from revolutions per minute (rev/min) to radians per second (rad/s):
We know that 1 revolution is equal to 2π radians.
So, the angular speed in rad/s can be calculated as:
angular speed in rad/s = (3080 rev/min) * (2π rad/1 rev) * (1 min/60 s)
angular speed in rad/s ≈ 323.27 rad/s
3. Finally, calculate the kinetic energy of the flywheel using the formula:
KE = (1/2) * I * (angular speed)^2
KE = (1/2) * (408.15 kg⋅m²) * (323.27 rad/s)^2
KE ≈ 21,049,797 J
Therefore, the kinetic energy of the flywheel at an angular speed of 3080 rev/min is approximately 21,049,797 Joules.
Now, let's move on to Part B:
To calculate the time the bus could operate between stops, we need to determine the time it takes for the kinetic energy of the flywheel to decrease to zero, given the average power required to operate the bus.
The average power (P) is given as 1.81 × 10^4 W.
The power (P) can be expressed as the rate of change of kinetic energy (dKE/dt):
P = dKE/dt
Since KE = (1/2) * I * (angular speed)^2, we can differentiate it with respect to time (t):
dKE/dt = (1/2) * I * (d/dt)(angular speed)^2
We know that (d/dt)(angular speed)^2 is equal to twice the angular speed multiplied by the angular acceleration (α). So:
dKE/dt = (1/2) * I * (2Ωα)
Now, α can be expressed as the change in angular speed (Δω) divided by the change in time (Δt):
α = Δω / Δt
Rearranging the equation, we get:
dKE/dt = I * Ω * (Δω / Δt)
Replacing Δω with the change in angular speed between stops and Δt with the time between stops, we get:
dKE/dt = I * Ω * (dω / dt)
Now, we can substitute the average power (P) with dKE/dt:
P = I * Ω * (dω / dt)
Rearranging the equation, we get:
Δt = I * Ω * (dω / P)
Plugging in the values we have:
Δt = (408.15 kg⋅m²) * 323.27 rad/s * (dω / 1.81 × 10^4 W)
Therefore, to calculate the time the bus can operate between stops, you need to know the rate of change of angular speed (dω), which is not provided in the given information. Once you have that information, you can substitute it into the equation and solve for Δt.