A disk, with a radius of$0.25 \text{m}$ , is to be rotated like a merrygo-round through , starting from rest, gaining angular speed at the constant rate ${\mathit{\alpha }}_{\mathbf{1}}$through the first$\mathbf{400}\mathbf{}\mathbf{ }\mathbf{\text{rad}}$ and then losing angular speed at the constant rate$-{\alpha }_1$ until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed$\mathbf{400}\mathbf{ }\raisebox{1ex}{${\displaystyle \mathbf{\text{m}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\mathbf{\text{s}}}^{\mathbf{\text{2}}}}$}\right.$ .

(a) What is the least time required for the rotation?

(b) What is the corresponding value of${\mathit{\alpha }}_{\mathbf{1}}$ ?

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, $\theta =400 \text{rad}$.

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 ${\alpha }_1$. The formula are given below.

1. Angular velocity is${\omega }_{max}=\sqrt{\frac{a}{r}}$
2. The kinematic equation for $\theta$ is$\theta -{\theta }_0=\frac{1}{2}\left({\omega }_0+\omega \right)t$
3. The kinematic equation for$\omega$is$\omega ={\omega }_0+{\alpha }_{1}t$

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

${\omega }_{\max }=\sqrt{\frac{a}{r}}$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{\omega }_{\max }& =& \sqrt{\frac{{\displaystyle 400\raisebox{1ex}{$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

$\theta -{\theta }_0=\frac{1}{2}\left({\omega }_0+\omega \right)t$

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$}\right.\right)t\\ t& =& \frac{{\displaystyle 400 \text{rad}\times 2}}{{\displaystyle 40\raisebox{1ex}{${\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 ${\alpha }_1$

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

$$\begin{gathered} \omega ={\omega }_0+{\alpha }_{1}t \\ {\alpha }_1=\frac{\omega -{\omega }_0}{t} \end{gathered}$$

Substitute the all the value in the above equation.

$\begin{array}{rcl}{\alpha }_1& =& \frac{{\displaystyle 40\raisebox{1ex}{${\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 ${\alpha }_1$ is,$2.0\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$ .