Can a Flywheel Efficiently Power a Low-Pollution Car for Long Distances?

  • Thread starter Thread starter LBloom
  • Start date Start date
  • Tags Tags
    Car Flywheel
AI Thread Summary
The discussion focuses on the feasibility of using a flywheel to power a low-pollution car over long distances. Key calculations involve determining the energy needed to be stored in the flywheel, which is estimated at 1.65 x 10^8 J, and the angular velocity when fully charged. Participants are also exploring how to calculate the average velocity and acceleration required for the trip, with one user estimating an average velocity of approximately 12.92 m/s and an acceleration of 0.443 m/s². The significance of the 20 acceleration periods is debated, with suggestions to analyze the work done over each segment. The conversation highlights the complexities of energy storage and motion dynamics in this innovative automotive concept.
LBloom
Messages
169
Reaction score
0

Homework Statement



"One possibility for a low-pollution automobile is for it to use energy stored in a heavy rotating flywheel. Suppose such a car has a total mass of 1400 kg, uses a uniform cylindrical flywheel of diameter 1.90 m and mass 210 kg, and should be able to travel 330 km without needing a flywheel "spinup."

a)Make reasonable assumptions (average frictional retarding force = 500 N, twenty acceleration periods from rest to 93 km/h, equal uphill and downhill, and that energy can be put back into the flywheel as the car goes downhill), and estimate what total energy needs to be stored in the flywheel.
b)What is the angular velocity of the flywheel when it has a full "energy charge"?
c)About how long would it take a 150-\rm hp motor to give the flywheel a full energy charge before a trip?

Homework Equations



Kfw=1/2*I*\omega^2
I=1/2*Mfw*R^2
Frictional Work: W=Ffd

The Attempt at a Solution



I converted the km/h to m/s, to get that the car maxes out at (155/6) m/s
I converted km to m, to get that the total trip was 330000m

I found that I=94.7625 and that the Wf=1.65*10^8
I assume that I calculate the acceleration to find the average velocity for the entire trip, although I'm not quite sure how to do that yet, but I'll get started on that. Honestly, I'm just confused what the significance of the fact that there are 20 acceleration periods. Could I divide the trip in 20, and figure out the work required for that distance and multiply it by 20, or is that unneeded?

Edit: I think I found the average velocity of the trip to be 12.9167 m/s. Can anyone tell me if I'm right or on the right track?
 
Last edited:
Physics news on Phys.org
Edit: So I think I figured out the acceleration. I took the difference between the max velocity and the average velocity and divided it by 20, to get that the acceleration was 0.443 m/s^2. Does this sound right?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top