Answer :
Explanation:
Given that,
Mass of the bus, m = 1500 kg
Mass of the flywheel, M = 10,000 kg
Radius of flywheel, r = 0.6 m
Speed of the bus, v = 24.5 m/s
(a) Let [tex]\omega[/tex] is the angular velocity of the flywheel. It is assumed that 85.0% of the rotational kinetic energy can be transformed into translational energy. Using the conservation of energy as :
[tex]85\%\ of\ K_r=K_t[/tex]
[tex]0.85\times \dfrac{1}{2}I\omega^2=\dfrac{1}{2}Mv^2[/tex]
[tex]0.85\times I\omega^2=Mv^2[/tex]
[tex]0.85\times \dfrac{1}{2}mr^2\omega^2=Mv^2[/tex]
[tex]\omega^2=\dfrac{Mv^2}{0.85\times mr^2}[/tex]
[tex]\omega^2=\dfrac{10000 \times (24.5)^2}{0.85\times 1500\times (0.6)^2}[/tex]
[tex]\omega=114.35\ rad/s[/tex]
(b) Let h is the height climb with this stored energy. Again using the conservation of energy as :
[tex]0.85K_r=K_t+mgh[/tex]
[tex]0.85\times \dfrac{1}{2}I\omega^2=\dfrac{1}{2}Mv^2+Mgh[/tex]
[tex]Mgh=0.85\times \dfrac{1}{4}mr^2\omega^2-\dfrac{1}{2}Mv^2[/tex]
[tex]h=\dfrac{0.85\times \dfrac{1}{4}mr^2\omega^2-\dfrac{1}{2}Mv^2}{Mg}[/tex]
[tex]h=\dfrac{0.85\times \dfrac{1}{4}1500\times (0.6)^2\times (114.35)^2-\dfrac{1}{2}\times 10000\times 24.5^2}{10000\times 9.8}[/tex]
h = 15.31 meters
Hence, this is the required solution.