A computer disk drive is turned on starting from rest

and has constant angular acceleration. If it took 0.0865 s for the

drive to make its secondcomplete revolution, (a) how long did it

take to make the first complete revolution, and (b) what is its

angular acceleration, in rad/s sq

The time to complete the second revolution is\({t_2} = 0.0865\,\,s\)

Angular acceleration is constant

Since the disk drive starts from rest, therefore, the initial angular speed\({\omega _0} = 0\)

And initial angular displacement\({\theta _0} = 0\).

If angular acceleration is constant, we can use the relation,

\(\theta = {\theta _0} + {\omega _0}t + \frac{1}{2}\alpha {t^2}\)

Where

\(\theta \)is the final angular deviation.

\({\theta _0}\)is the initial angular deviation.

\({\omega _0}\)is the initial angular velocity.

\(t\)is the time.

\(\alpha \)the angular acceleration.

Since angular acceleration is constant, we can use the equation,

\(\theta = {\theta _0} + {\omega _0}t + \frac{1}{2}\alpha {t^2}\)

We know in one revolution disk will cover\(2\pi \,\,rad\)

Suppose the time to complete the first revolution is\({t_1}\)

Then, substituting the values for the first revolution in the equation.

\(\begin{aligned}{c}2\pi \,rad = \left( 0 \right) + \left( 0 \right)\left( {{t_1}} \right) + \frac{1}{2}\alpha {\left( {{t_1}} \right)^2}\\2\pi \,\,rad = \frac{1}{2}\alpha {\left( {{t_1}} \right)^2}...\left( i \right)\end{aligned}\)

For second complete revolution \(\theta = 4\pi \,\,rad\) and time to will be

\(\begin{aligned}{}t = \,{t_1} + {t_2}\\ = {t_1} + 0.086\,\,s\end{aligned}\)

Substituting the values into the equation

\(\begin{aligned}{}4\pi \,\,rad = \left( 0 \right) + \left( 0 \right)\left( {{t_1} + 0.0865} \right) + \frac{1}{2}\alpha {\left( {{t_1} + 0.0865} \right)^2}\\4\pi \,\,rad = \frac{1}{2}\alpha {\left( {{t_1} + 0.0865} \right)^2}...\left( {ii} \right)\end{aligned}\)

Dividing equation (i) by (ii)

\(\begin{aligned}{c}\frac{{4\pi }}{{2\pi }} = \frac{{{1 \mathord{\left/{\vphantom {1 2}} \right.} 2}\alpha {{\left( {{t_1} + 0.0865\,s} \right)}^2}}}{{{1 \mathord{\left/{\vphantom {1 2}} \right.} 2}\alpha {{\left( {{t_1}} \right)}^2}}}\\t_1^2 = \frac{{{{\left( {{t_1} + 0.0865\,\,s} \right)}^2}}}{2}\\2t_1^2 = {t_1}^2 + 2\left( {0.0865\,} \right){t_1} + {\left( {0.0865} \right)^2}\\{t_1}^2 - 0.173t + 0.00748 = 0\end{aligned}\)\({t_1} = 0.209\,s,\,\, - 0.0358\,s\)

Since time cannot have negative value

Therefore, the time to cover first complete revolution is \(0.209\,\,s\)

We know,

\({t_1} = 0.209\,\,s\)

Substituting this value into equation (i)

\(\begin{aligned}{c}2\pi \,\,rad = \frac{1}{2}\alpha {\left( {{t_1}} \right)^2}\\2\left( {3.14} \right)\,\,rad = \frac{1}{2}\alpha {\left( {0.209\,\,s} \right)^2}\\\alpha = 287.69\,\,{{rad} \mathord{\left/{\vphantom {{rad} {{s^2}}}} \right.} {{s^2}}}\end{aligned}\)

Hence the angular acceleration is \(287.69\,\,{{{rad} \mathord{\left/{\vphantom {{rad} s}} \right.} s}^2}\)