Flywheel Kinetic Energy in Delivery Trucks

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The discussion focuses on calculating the kinetic energy of a flywheel used in delivery trucks, specifically a solid homogeneous cylinder with a mass of 550 kg and a radius of 0.65 m, rotating at 960 rad/s. The kinetic energy formula K = 1/2 Iω^2 is applied, where the moment of inertia I is calculated as I = 1/2 mr^2. The resulting kinetic energy is found to be approximately 5.35 x 10^7 J. Additionally, the average power requirement of the truck is 9.3 kW, prompting a calculation of operational time between charges. The discussion emphasizes the importance of verifying calculations and converting units as necessary.
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Homework Statement


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 960 rad/s. One such flywheel is a solid homogenous cylinder, rotating about its central axis, with a mass of 550 kg and a radius of 0.65 m. What is the kinetic energy of the flywheel after charging?

If the truck operates with an average power requirement of 9.3 kW, for how many minutes can it operate between charging?



Homework Equations


K = 1/2Iw^2
I = 1/2mr^2

The Attempt at a Solution


K = 1/2 * (.5 * 550 * .65^2) * 960^2 = 5.35 x 10^7
I a, getting a really huge number I wanted to make sure I am doing it right.
 
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Well convert radians per second to metres per second.

Then K = 0.5 (m) (v)^2
 
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