(II) A rotating uniform cylindrical platform of mass 220 kg and radius 5.5 m slows down from to rest in 16 s when the driving motor is disconnected. Estimate the power output of the motor (hp) required to maintain a steady speed of\({\bf{3}}{\bf{.8 }}rev/s\).

The given data can be listed below as:

• The mass of the cylindrical platform is\(M = 220{\bf{ }}{\rm{kg}}\).
• The radius of thecylindrical platformis \(R = 5.5{\rm{ m}}\).
• The initial angular velocity of the rotating platform is\({\omega _i} = 3.8{\rm{ rev/s}}\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right) = 23.88{\rm{ rad/s}}\).
• The final angular velocity of the rotating platform is\({\omega _f} = 0{\rm{ rad/s}}\).
• The time taken by the rotating platform in the motion is\(\Delta t = 16{\rm{ s}}\).
• The steady speed of the motor is\({\omega _s} = 3.8{\rm{ rev/s}}\left( {\frac{{2\pi {\rm{ rad}}}}{{1{\rm{ rev}}}}} \right) = 23.88{\rm{ rad/s}}\).

The motor’s power is the product of the torque developed by the motor and the steady angular speed of the motor.In this problem, the torque developed by that motor is equal to the negative torque developed due to friction.

The angular acceleration of the motor is the rate of change of angular velocity of the motor.

The torque developed due to friction can be expressed as:

\({T_{fr}} = I\alpha \) … (i)

The angular acceleration of the motor can be expressed as:

\(\alpha = \frac{{{\omega _f} - {\omega _i}}}{{\Delta t}}\)

Substitute the values in the above equation.

\(\begin{align}\alpha &= \frac{{0{\rm{ rad/s}} - 23.88{\rm{ rad/s}}}}{{16{\rm{ s}}}}\\ &= - 1.49{\rm{ rad/}}{{\rm{s}}^2}\end{align}\)

The moment of inertia of the cylindrical platform can be expressed as:

\(I = \frac{1}{2}M{R^2}\)

Substitute the values in the above equation.

\(\begin{align}I &= \frac{1}{2} \times 220{\bf{ }}{\rm{kg}} \times {\left( {5.5{\rm{ m}}} \right)^2}\\ &= 3327.5{\rm{ kg}} \cdot {{\rm{m}}^2}\end{align}\)

Substitute the values of\(\left( I \right)\)and\(\left( \alpha \right)\)in equation (i).

\(\begin{align}{T_{fr}} &= 3327.5{\rm{ kg}} \cdot {{\rm{m}}^2} \times - 1.49{\rm{ rad/}}{{\rm{s}}^2}\\ &= - 4957.97{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}} \cdot {\rm{m}}}}{{1{\bf{ }}{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^2}}}} \right)\\ &= - 4957.97{\rm{ N}} \cdot {\rm{m}}{\bf{ }}\end{align}\)

The power developed by the motor can be expressed as:

\(\begin{align}P &= {T_m}{\omega _s}\\ &= - {T_{fr}}{\omega _s}\end{align}\)

Here,\({T_{fr}}\)is the torque developed due to friction,\({\omega _s}\)is the angular speed of the motor, and\({T_m}\)is the torque developed by the motor.

Substitute the values in the above equation.

\(\begin{align}P &= - \left( { - 4957.97{\rm{ N}} \cdot {\rm{m}}} \right) \times {\bf{ }}23.88{\rm{ rad/s}}\\ &= 118396.32{\rm{ N}} \cdot {\rm{m/s}}\left( {\frac{{1{\rm{ W}}}}{{1{\rm{ N}} \cdot {\rm{m/s}}}}} \right)\left( {\frac{{1{\rm{ hp}}}}{{746{\rm{ W}}}}} \right)\\ &= 158.71{\rm{ hp}}\end{align}\)

Thus, the power output of the motor is \(158.71{\rm{ hp}}\).