## Flywheel in Trucks

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 990 rad/s. One such flywheel is a solid homogeneous cylinder, rotating about its central axis, with a mass of 850 kg and a radius of 0.55 m.

What is the kinetic energy of the flywheel after charging?

What is the kinetic energy of the flywheel after charging?

**6.30×107 J**

If the truck operates with an average power requirement of 7.9 kW, for how many minutes can it operate between charging?**1.33×102 min**## Discussion

**NEW**

*Anonymous 1*

I thought KErot= 1/2*I*w^2?

its not working for prt 1

**NEW**

just make sure to use the I for a cylinder which is given

in the notes as I=.5*m*r^2

**NEW**

**Re:**

Isn't w=v/r? when I plug in the numbers, the final KE is way

too large! Am I using the right formula for w?

**NEW**

**Re: Re:**

*Anonymous 3*

your w is given. Remember that this is angular velocity= w

which is given in rad/s

**NEW**

**part 1**

find I, I=.5MR^2

then

KE=.5Iω^2

**NEW**

**part2**

time= KE/P

multiply power by 1000

divide KE/P

THEN divide by 60

hope this helps!

**NEW**

**help**

*Anonymous 5*

I need some help. I am doing the equation for the first part

and can't seem to get the right answer.

I=.5*510kg*0.65m^2

I=107.7

KE= .5*107.7*760^2rad/s

KE=31103760 J

If anyone could tell me where I'm going wrong that would be

great!

**NEW**

**Re: help**

For mine I did: I=.5*m*r^2 (.5*990*.70^2) = 242.55

then, KE=.5Iw^2 (.5*242.55*870^2) = 9.17e7 J. Hope that helps!