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Pilots can be tested for the stresses of flying high-speed jets in a whirling 鈥渉uman centrifuge,鈥 which takes 1.0 min to turn through 23 complete revolutions before reaching its final speed. (a) What was its angular acceleration (assumed constant), and (b) what was its final angular speed in rpm?

Short Answer

Expert verified

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

01

Given data

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

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

02

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.

03

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.

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