Answer:[tex]\omega =93.51 rad/s[/tex]
Explanation:
Given
mass of Flywheel [tex]m_1=1440 kg[/tex]
mass of bus [tex]m_b=10200 kg[/tex]
radius of Flywheel [tex]r=0.63 m[/tex]
final speed of bus [tex]v=21 m/s[/tex]
Conserving Energy i.e.
0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus
Let [tex]\omega [/tex]be the angular velocity of Flywheel
[tex]0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}[/tex]
[tex]I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2[/tex]
[tex]0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}[/tex]
[tex]\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}[/tex]
[tex]\omega =21\times 4.45=93.51 rad/s[/tex]