How Long Can an Electric Flywheel Delivery Truck Operate Between Chargings?

AI Thread Summary
Electric flywheel delivery trucks in Europe utilize stored rotational energy for operation, reaching a top speed of 198π rad/s. The flywheel's mass is 530 kg with a radius of 1.0 m, and the average power requirement for the truck is 8.5 kW. Participants in the discussion focus on calculating the rotational kinetic energy stored in the flywheel using the formula for rotational inertia. The conversation emphasizes the need to understand the relationship between energy, power, and time to determine operational duration between charges. Overall, the discussion revolves around applying physics principles to solve for the truck's operational time based on its energy storage capabilities.
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Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are charged by using an electric motor to get the flywheel up to its top speed of 198π rad/s. One such flywheel is a solid, homogeneous cylinder with a mass of 530 kg and a radius of 1.0 m. If the truck operates with an average power requirement of 8.5 kW, how long can it operate between chargings?

:confused:

Just a push in the right direction, I've written out so many formulas relating to rotation and Power/Work. And i haven't found a starting point.
 
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Start by figuring how much rotational KE is stored in the flywheel when it rotates at top speed.
 
(edit again)
I solved for I with I=(1/2)mr^2
I plugged I and w into K= Iw^2 and solved for K.
Whew..

I'll keep workin from here, thanks.
 
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Rotational KE

I'm not sure what that equation is supposed to be. Here's another hint: The rotational inertia of a cylinder about its center is I = 1/2 M R2. You'll need it.
 
I=(1/2)mr^2
I=265
(edited)

I'm thinkin my next step is to find alpha or torque, i don't see how to do that but there is
a_r=w^2r
What is a_r?
 
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Oh, I didn't know what a fly wheel was really. : )
 
I get the "I" as being something completely different.

I = \frac{1}{2}mr^2

I = \frac{1}{2}(530)(1)^2

I = 265


Once you get the energy

E = Pt
 
Ya shawn, you're right.
 
Am so out of it. disregard this reply :)
 
  • #10
Power is 8500. Time is what you are supposed to find.
 
  • #11
Got it, Thanks Shawn and Doc Al.

Brain moving slow today.
 
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