Question

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${\mathit{\alpha }}_{\mathbf{1}}$ ?

Short answer

a) The least time required for rotation t is $40 \text{s}$.

b) The corresponding value of ${\alpha }_1$ is ${\alpha }_1$,$2.0\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$ .

Step-by-step solution

Understanding the given information

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

Concept and Formula used

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$\omega ={\omega }_0+{\alpha }_{1}t$

(a) Calculation for the least time required for the rotation

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}}}$}\right.$ .