Question

Short answer

The results for parts (a) and (b) are \(46\;{\rm{rev/mi}}{{\rm{n}}^2}\) and \(46\;{\rm{rev/min}}\), respectively.

Step-by-step solution

Given data

The number of complete revolutions is\(\theta = 23\,{\rm{rev}}\).

The time required is \(t = 1\;{\rm{min}}\).

Understanding angular acceleration

The relation to find the angular acceleration is given by:

\(\theta = {\omega _1}t + \frac{1}{2}\alpha {t^2}\)

Here, \({\omega _1}\) is the initial angular speed, whose value is zero and \(\alpha \) is the angular acceleration.

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

\(\begin{aligned}{c}23\;{\rm{rev}} = 0 + \frac{1}{2}\alpha {\left( {1\;{\rm{min}}} \right)^2}\\\alpha = 46\;{\rm{rev/mi}}{{\rm{n}}^2}\end{aligned}\)

Thus, \(\alpha = 46\;{\rm{rev/mi}}{{\rm{n}}^2}\) is the required angular acceleration.

Determine the angular speed

The relation to find the angular speed is given by:

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

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

\(\begin{aligned}{c}23\;{\rm{rev}} &= \frac{1}{2}\left( {0 + {\omega _2}} \right)\left( {1\;{\rm{min}}} \right)\\{\omega _2} &= 46\;{\rm{rev/min}}\end{aligned}\)

Thus, \({\omega _2} = 46\;{\rm{rev/min}}\) is the required angular speed.