Question

Special delivery trucks, which operate by making use of the energy stored in a rotating flywheel, have been in use for some time. The trucks are “charged up” before leaving by using an electric motor to get the flywheel up to its top speed of 7000 revolutions per minute. If one such flywheel is a solid cylinder (I = 1/2 MR2) of weight 5000 N and a diameter of 2 m, how long can the truck operate before returning to base for “recharging,” if its average power requirement is 8000 Watts? (Watt = Joule/second)

Answer 1

we are given final angular velocity =7000 revolution per minute = 7000*2$\pi$/60 = 732.66 rad/s

Total kinetic energy = rotational kinetic energy + translational kinetic energy

Rotational kinetic energy = $(1/2)*I*\omega^2$

translational kinetic energy =$(1/2)*m*v^2=(1/2)m*(\omega r)^2$

mass of the wheel is 500 kg

radius of the wheel is 1 m

moment of inertia of the wheel =$(1/2)mR^2=(1/2)*500*1^2 = 250$

now calculating total kinetic energy = $(1/2)*250*(732.66)^2+(1/2)*500*(732.66)^2$

=$201,296,503.35 \ J$

energy= power *time

time = energy/power =$201,296,503.35 \ /8000 = 25162 seconds =7hours$