B

Step 1: Given parameters: - The larger wheel has an initial angular velocity, \( \omega_1 = 2000 \text{ rev/min} \). - The smaller wheel starts from rest, so \( \omega_2 = 0 \text{ rev/min} \). - The radius of the larger wheel is \( R_1 = 2R_2 \). - The moment of inertia of the larger wheel is \( I_1 = 16 I_2 \).
Step 2: Convert angular velocity from rev/min to rad/s: \[ \omega_1 = 2000 \times \frac{2\pi}{60} = \frac{2000\pi}{30} = \frac{200\pi}{3} \text{ rad/s} \].
Step 3: The conservation of angular momentum states that \( I_1 \omega_1 + I_2 \omega_2 = I_1 \omega_1' + I_2 \omega_2' \), where \( \omega_1' \) and \( \omega_2' \) are the angular velocities after they are in contact.
Step 4: Since they will roll without slipping after they come into contact, \( R_1 \omega_1' = R_2 \omega_2' \)
Substituting \( R_1 = 2R_2 \): \[ 2R_2 \omega_1' = R_2 \omega_2' \quad \Rightarrow \quad \omega_2' = 2 \omega_1' \].
Step 5: Substitute \( I_1 = 16 I_2 \) into the conservation of momentum equation: \[ 16 I_2 \left(\frac{200\pi}{3}\right) + I_2 (0) = 16 I_2 \omega_1' + I_2(2 \omega_1') \].
\[ \Rightarrow 16 \left(\frac{200\pi}{3}\right) = 18 \omega_1' \quad \Rightarrow \quad \omega_1' = \frac{16 \times \frac{200\pi}{3}}{18} = \frac{1600\pi}{27} \text{ rad/s} \].
Step 6: Since \( \omega_2' = 2 \omega_1' \): \[ \omega_2' = 2 \times \frac{1600\pi}{27} = \frac{3200\pi}{27} \text{ rad/s} \].
Step 7: Convert back to rev/min: \[ \omega_2' = \frac{3200\pi}{27} \times \frac{60}{2\pi} = \frac{3200 \times 30}{27} = \frac{96000}{27} \approx 3555.56 \text{ rev/min} \].
Therefore, the final angular velocity of the smaller wheel after equilibrium occurs is approximately 3555.56 rev/min.