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

Answer:

$\tau = 60.88 N.m$

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 = ?

$\tau = I \alpha$

$\alpha = \dfrac{\omega_1-\omega_2}{t}$

$\alpha = \dfrac{26.2-0}{204}$

   α = 0.128 rad/s²

$\tau = I \alpha$

$\tau = 474\times 0.128$

$\tau = 60.88 N.m$

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