Question

A spinning flywheel has rotational inertia I = 474.0 kg·m2. Its angular velocity decreases from 26.2 rad/s to zero in 204.0 s due to friction. What is the frictional torque acting?

Answer 1

`\tau = 60.88 N.m`

Step-by-step 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 = ?

`\tau = I \alpha`

`\alpha = (\omega_1-\omega_2)/(t)`

`\alpha = (26.2-0)/(204)`

α = 0.128 rad/s²

`\tau = I \alpha`

`\tau = 474* 0.128`

`\tau = 60.88 N.m`

frictional torque acting by the flywheel is equal to 60.88 N.m