B

To calculate the constant torque necessary to stop the flywheel, we'll follow these steps:
Step 1: Calculate the moment of inertia (I) of the flywheel.
The moment of inertia for a solid disk is given by the formula:
$$ I = \frac{1}{2} m r^2 $$
where $m$ is the mass and $r$ is the radius.
First, we convert the weight of the flywheel to mass:
$$ m = \frac{weight}{g} = \frac{600 \text{ lbs}}{32.2 \text{ ft/s}^2} \approx 18.52 \text{ slugs} $$
The radius is half of the diameter:
$$ r = \frac{4}{2} = 2 \text{ ft} $$
Now, substituting into the moment of inertia formula:
$$ I = \frac{1}{2} \times 18.52 \text{ slugs} \times (2 \text{ ft})^2 = \frac{1}{2} \times 18.52 \times 4 = 37.04 \text{ slug ft}^2 $$
Step 2: Calculate the angular velocity (\omega) in rad/s.
To convert rpm to rad/s, use the conversion factor:
$$ \omega = \frac{1200 \text{ rev/min} \times 2\pi ext{ rad/rev}}{60 \text{ s/min}} = 126.0 \text{ rad/s} $$
Step 3: Calculate the angular deceleration (\alpha).
We want to stop the flywheel in 2.0 minutes (which is 120 seconds).
Using the formula:
$$ \alpha = \frac{\Delta \omega}{t} $$
where $\Delta \omega = 0 - 126.0 = -126.0 \text{ rad/s}$ and $t = 120 \text{ s}$.
Therefore:
$$ \alpha = \frac{-126.0}{120} = -1.05 \text{ rad/s}^2 $$
Step 4: Calculate the torque (\tau).
The relationship between torque, moment of inertia, and angular acceleration is given by:
$$ \tau = I \cdot \alpha $$
Substituting the values we found:
$$ \tau = 37.04 \text{ slug ft}^2 \times (-1.05 \text{ rad/s}^2) \approx -38.97 \text{ lb ft} $$
Since we are looking for the required constant torque (which is a positive value), we take the magnitude:
Therefore, the constant torque necessary to stop the flywheel is approximately 39 lb ft.