How Does a Glass Fibre Flywheel Store Kinetic Energy in Buses?

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SUMMARY

The forum discussion centers on the mechanics of a glass fibre flywheel developed by BP for storing kinetic energy in buses. The flywheel can store a maximum kinetic energy of 1.42 MJ, derived from a 16,000 kg bus braking from 48 km/h. Key calculations include the moment of inertia at 1.01 kg and a radius of 0.215 m. Participants also explore the conversion of angular velocity from revolutions per minute to radians per second, and the determination of mean frictional torque during a two-minute stop.

PREREQUISITES
  • Understanding of kinetic energy calculations (E = ½mv²)
  • Familiarity with angular motion equations and torque
  • Knowledge of unit conversions, particularly speed from km/h to m/s
  • Basic principles of rotational dynamics and moment of inertia
NEXT STEPS
  • Learn about the principles of energy conservation in mechanical systems
  • Study the equations of motion for rotational dynamics
  • Explore the concept of torque and its applications in engineering
  • Investigate the design and applications of flywheel energy storage systems
USEFUL FOR

Engineers, physics students, and professionals in the transportation sector interested in energy storage solutions and the mechanics of flywheels.

  • #31
Katy96 said:
2Pi/60?

A poor guess. It's ##2\pi##. You could always have googled that!
 
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  • #32
Katy96 said:
so how do I change 48km/h into m/s? never mind I got it now!
Is editing a response as you type it a lost skill?

I see this so often it's ridiculous !
 
  • #33
PeroK said:
A poor guess. It's ##2\pi##. You could always have googled that!
okay well do I just divide by 2pi? because I thought you would do 16000X2pi/60
 
  • #34
Katy96 said:
okay well do I just divide by 2pi? because I thought you would do 16000X2pi/60

I see what you did. There are ##2\pi## radians in a revolution, so yes it's ##2\pi /60## to convert from revs per minute to radians per second.
 
  • #35
PeroK said:
I see what you did. There are ##2\pi## radians in a revolution, so yes it's ##2\pi /60## to convert from revs per minute to radians per second.
when I did that I got the right answer so in part c do I use vmax=rωmax or do I use I=0.606mr2
 
  • #36
Katy96 said:
when I did that I got the right answer so in part c do I use vmax=rωmax or do I use I=0.606mr2

Does that give the right answer?

I'm going offline now. Good luck with the rest of it.
 
  • #37
PeroK said:
Does that give the right answer?

I'm going offline now. Good luck with the rest of it.
thanks.
 
  • #38
Its just part e I'm struggling on now
 
  • #39
Katy96 said:
e) Determine also the

Katy96 said:
Its just part e I'm struggling on now
Yes. It is quite the conundrum.
 
  • #40
jbriggs444 said:
Yes. It is quite the conundrum.
sorry it meant to say Determine also the mean frictional torque causing the slowing down in the two minute stop.
 
  • #41
You know how much energy it had at the beginning of the two minute stop. You have already used that to calculate its angular velocity at that time.

The question tells you what fraction of the starting energy it has at the end of the two minute stop. How much energy is that? How fast does that mean that it must be rotating at the end of the two minutes?
 
  • #42
this is what I know, i need to use the change in the KE to find the final angular speed, and hence the angular deceleration, using the equations of angular motion and torque Important to show answer is negative, as result of deceleration
The force providing the centripetal acceleration acts towards the axis so cannot have a torque about it.and flywheel does not stop during the 2 mins.
 
  • #43
Katy96 said:
this is what I know, i need to use the change in the KE to find the final angular speed
Then there is a starting point. What is the change in KE?
 
  • #44
jbriggs444 said:
Then there is a starting point. What is the change in KE?
that's where I get stuck
 
  • #45
The problem statement says that four fifths of the flywheel's energy remains after two minutes.
 

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