A flywheel of moment of inertia 50 kg m^2 starts from rest. If it...

Given, initial angular velocity $\omega_i = 0$
Final angular velocity $\omega_f = 20 \text{ rad/s}$
Time $t = 4 \text{ s}$

Using the first equation of rotational kinematics:
$\omega_f = \omega_i + \alpha t$
$20 = 0 + \alpha (4)$
$\alpha = 5 \text{ rad/s}^2$

The required constant torque is given by:
$\tau = I \alpha$
$\tau = 50 \times 5 = 250 \text{ N m}$

Answer: $250 \text{ N m}$