A wheel with rotational inertia \(1.27 \mathrm{~kg} \cdot \mathrm{m}^{2}\) is rotating with an angular speed of 824 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with rotational inertia \(4.85 \mathrm{~kg} \cdot \mathrm{m}^{2}\), is suddenly coupled to the same shaft. What is the angular speed of the resultant combination of the shaft and two wheels?

Firstly, the original angular speed of the first wheel needs to be converted from revolutions per minute (rev/min) to radians per second (rad/s), using the conversion factor of \( 2\pi \) radians per revolution and \( \frac{1}{60} \) minutes per second. So if the angular speed is 824 rev/min, the conversion to rad/s is done as follows: \( 824 \times 2\pi \times \frac{1}{60} = 86.32 \) rad/s.

The initial total angular momentum of the system can be calculated since the angular momentum (L) of a rotating object is given by the product of its moment of inertia (I) and its angular speed (\(\omega\)). The second wheel was at rest, thus has zero angular momentum. Hence, the initial total angular momentum of the system (before coupling) is the angular momentum of the first wheel, calculated as follows: \[ L_{\mathrm{initial}} = I_{1} \times \omega_{1} = 1.27 \, \mathrm{kg} \cdot \mathrm{m^{2}} \times 86.32 \, \mathrm{rad/sec} = 109.62 \, \mathrm{kg \cdot m^2/sec}\]

After coupling the second wheel, the total moment of inertia of the system is the sum of the moments of inertia of the two wheels. Let \(\omega_{\mathrm{final}}\) be the final angular speed after the two wheels are coupled. Since angular momentum is conserved when no external torques are present, the total angular momentum after coupling (Lfinal) will be equal to the total initial angular momentum (Linitial). This can be written as following: \[ L_{\mathrm{initial}} = L_{\mathrm{final}} \] \[ I_{1} \times \omega_{1} = \left(I_{1} + I_{2}\right) \times \omega_{\mathrm{final}} \] Solving for \(\omega_{\mathrm{final}}\) we find: \[ \omega_{\mathrm{final}} = \frac{I_{1} \times \omega_{1}}{I_{1} + I_{2}} \] Inserting the given values and solving, we get: \[ \omega_{\mathrm{final}}= \frac{109.62 \, \mathrm{kg \cdot m^2/sec}}{1.27 \, \mathrm{kg} \cdot \mathrm{m^{2}} + 4.85 \, \mathrm{kg} \cdot \mathrm{m^{2}}} = 17.70 \, \mathrm{rad/sec} \]

Finally, to have the final angular speed in a more commonly used unit, convert it from rad/s to rev/min using the conversion factors: 1 revolution = \(2\pi\) rad and 60 s = 1 min. The final angular speed becomes: \[ \omega_{\mathrm{final}} = 17.70 \, \mathrm{rad/sec} \times \frac{1 \, \mathrm{rev}} {2\pi \, \mathrm{rad}} \times \frac{60 \, \mathrm{s}} {1 \, \mathrm{min}} = 169.0 \, \mathrm{rev/min} \]