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?
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
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
NEWAnonymous 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!
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!