91Ó°ÊÓ

1. The acceleration a is $400\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.
2. The radius r is$r=0.25 \text{m}$
3. The angular displacement is, $\mathrm{θ}=400 \text{rad}$.

By usingtheformula for angular velocity and applying kinematic equation considering first half of the motion, we can find the least time required for rotation and corresponding value of ${\mathrm{Î\pm }}_1$. The formula are given below.

1. Angular velocity is${\mathrm{Ó\neg }}_{max}=\sqrt{\frac{a}{r}}$
2. The kinematic equation for $\mathrm{θ}$ is$\mathrm{θ}-{\mathrm{θ}}_0=\frac{1}{2}\left({\mathrm{Ó\neg }}_0+\mathrm{Ó\neg }\right)t$
3. The kinematic equation for$\mathrm{Ó\neg }$is$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+{\mathrm{Î\pm }}_{1}t$

The upper limit for centripetal acceleration places an upper limit of the spin by considering a point at the rim. Thus,

${\mathrm{Ó\neg }}_{\max }=\sqrt{\frac{a}{r}}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{\mathrm{Ó\neg }}_{\max }& =& \sqrt{\frac{{\displaystyle 400\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.}}{{\displaystyle 0.25 \text{m}}}}\\ & =& 40\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Now, by applying kinematic equation tothefirst half of the motion, we get

$\mathrm{θ}-{\mathrm{θ}}_0=\frac{1}{2}\left({\mathrm{Ó\neg }}_0+\mathrm{Ó\neg }\right)t$

Substitute the all the value in the above equation.

$\begin{array}{rcl}400 \text{rad}& =& \frac{1}{2}\left(0+40\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)t\\ t& =& \frac{{\displaystyle 400 \text{rad}×2}}{{\displaystyle 40\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}\\ & =& 20 \text{s}\end{array}$

The second half of the motion takes the same amount of time.

Hence, the total time is $40 \text{s}$.

Step 3: (b) Calculation for the corresponding value of ${\mathrm{Î\pm }}_1$

Considering the first half of the motion and applying kinematic equation, we get

$$\begin{gathered} \mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+{\mathrm{Î\pm }}_{1}t \\ {\mathrm{Î\pm }}_1=\frac{\mathrm{Ó\neg }-{\mathrm{Ó\neg }}_0}{t} \end{gathered}$$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{\mathrm{Î\pm }}_1& =& \frac{{\displaystyle 40\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{{\displaystyle 20 \text{s}}}\\ & =& 2.0 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

Hence the value of ${\mathrm{Î\pm }}_1$ is,$2.0\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$ .