91Ó°ÊÓ

The number of complete revolutions is\(\theta = 1250\;{\rm{rev}}\).

The cooling fan is running at \({\omega _1} = 850\;{\rm{rev/min}}\).

The equation of kinematics for rotational motion can be used for solving problems related to rotational and linear kinematics, where acceleration or angular acceleration does not change with time.

The relation to find the angular acceleration is given by:

\(\alpha = \frac{{{\omega _2} - {\omega _1}}}{{2\theta }}\)

Here, \({\omega _2}\) is the final angular speed, whose value is zero and \(\alpha \) is the angular acceleration.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\alpha &= \left( {\frac{{0 - {{\left( {850\;{\rm{rev/min}}} \right)}^2}}}{{2\left( {1250\;{\rm{rev}}} \right)}}} \right)\\\alpha &= \left( { - 289\;{\rm{rev/mi}}{{\rm{n}}^2} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}} \times {{\left( {\frac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right)}^2}} \right)\\\alpha &= - 0.5\;{\rm{rad/}}{{\rm{s}}^2}\end{aligned}\)

Thus, \(\alpha = - 0.5\;{\rm{rad/}}{{\rm{s}}^2}\) is the required angular acceleration.

The relation to find the time required is given by:

\(\theta = \frac{1}{2}\left( {{\omega _1} + {\omega _2}} \right)t\)

Here, tis the required time.

On plugging the values in the above relation, you get:

\(\begin{aligned}{c}1250\;{\rm{rev}} &= \frac{1}{2}\left( {850\;{\rm{rev/min}} + 0} \right)\left( t \right)\\t &= \left( {2.94\;{\rm{min}} \times \frac{{60\;{\rm{s}}}}{{1\;{\rm{min}}}}} \right)\\t &= 176.4\;{\rm{s}}\end{aligned}\)

Thus, \(t = 176.4\;{\rm{s}}\) is the required time.