Answer:
[tex]\tau = 60.88 N.m[/tex]
Explanation:
given,
rotational inertia, I = 474.0 kg·m²
decrease in angular velocity
ω₁ = 26.2 rad/s ω₂ = 0 rad/s
time = 204 s
torque = ?
[tex]\tau = I \alpha[/tex]
[tex]\alpha = \dfrac{\omega_1-\omega_2}{t}[/tex]
[tex]\alpha = \dfrac{26.2-0}{204}[/tex]
α = 0.128 rad/s²
[tex]\tau = I \alpha[/tex]
[tex]\tau = 474\times 0.128[/tex]
[tex]\tau = 60.88 N.m[/tex]
frictional torque acting by the flywheel is equal to 60.88 N.m
