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Q31P

Expert-verifiedFound in: Page 289

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A disk, with a radius of $0.25\u200a\text{m}$** **, 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 ${\mathbf{400}}{\mathbf{}}{\mathbf{\u200a}}{\mathbf{\text{rad}}}$** **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${\mathbf{400}}{\mathbf{\u200a}}\raisebox{1ex}{$\mathbf{\text{m}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{s}}}^{\mathbf{\text{2}}}$}\right.$ ** **.**

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

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

a) The least time required for rotation t is $40\u200a\text{s}$.

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

- The acceleration a is $400\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$ .
- The radius r is $r=0.25\u200a\text{m}$
- The angular displacement is, $\theta =400\u200a\text{rad}$ .

**By using****the ****formula 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.**

**Angular velocity is ${{\omega}}_{max}{=}\sqrt{\frac{a}{r}}$****The kinematic equation for ${\theta}$ is ${\theta}{-}{{\theta}}_{{0}}{=}\frac{1}{2}\left({\omega}_{0}+\omega \right){t}$****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,

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

Substitute the all the value in the above equation.

$\begin{array}{rcl}{\omega}_{\mathrm{max}}& =& \sqrt{\frac{400\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}{0.25\u200a\text{m}}}\\ & =& 40\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Now, by applying kinematic equation to the first 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.

$\begin{array}{rcl}400\u200a\text{rad}& =& \frac{1}{2}\left(0+40\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)t\\ t& =& \frac{400\u200a\text{rad}\times 2}{40\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}\\ & =& 20\u200a\text{s}\end{array}$

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

Hence, the total time is $40\u200a\text{s}$.

**Step 3: (b) Calculation for the corresponding value of ****${{\alpha}}_{{1}}$**

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

**$\omega ={\omega}_{0}+{\alpha}_{1}t\phantom{\rule{0ex}{0ex}}{\alpha}_{1}=\frac{\omega -{\omega}_{0}}{t}\phantom{\rule{0ex}{0ex}}$**

**Substitute the all the value in the above equation.**

** $\begin{array}{rcl}{\alpha}_{1}& =& \frac{40\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}{20\u200a\text{s}}\\ & =& 2.0\u200a\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\end{array}$**

**Hence the value of ${\alpha}_{1}$ is,$2.0\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$ .**

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