A flywheel with a moment of inertia of 5.0 kg-m^2 is rotated at a...

Initial angular velocity, $\omega_0 = 60 \text{ rad/s}$

Time taken to come to rest, $t = 5.0 \text{ minutes} = 300 \text{ s}$

Final angular velocity, $\omega = 0$

Using the first equation of rotational motion, $\omega = \omega_0 + \alpha t$

$0 = 60 + \alpha (300)$

$\alpha = -\dfrac{60}{300} = -0.2 \text{ rad/s}^2$

We need to find the angular momentum $1 \text{ minute}$ ($60 \text{ s}$) before it stops. The time elapsed from the start is $t' = 5.0 - 1.0 = 4.0 \text{ minutes} = 240 \text{ s}$.

Angular velocity at $t' = 240 \text{ s}$ is given by:

$\omega' = \omega_0 + \alpha t'$

$\omega' = 60 + (-0.2)(240) = 60 - 48 = 12 \text{ rad/s}$

Angular momentum at this instant is:

$L = I\omega'$

$L = 5.0 \times 12 = 60 \text{ kg-m}^2/\text{s}$