Suggested languages for you:

Americas

Europe

Q107P

Expert-verifiedFound in: Page 294

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A pulley wheel that is ${\text{8.0 \hspace{0.17em}cm}}$ ****in diameter has a ${\text{5.6 m}}$ ****long cord wrapped around its periphery. Starting from rest, the wheel is given a constant angular acceleration of ${\mathbf{1}}{\text{.}}{\mathbf{5}}{\mathbf{}}{\text{\hspace{0.17em}}}{\mathbf{rad}}{\mathbf{/}}{{\mathbf{s}}}^{{2}}{\mathbf{}}$****. (a) Through what angle must the wheel turn for the cord to unwind completely? (b) How long will this take?**

a) The angle that the wheel must turn for the cord to unwind completely is .$140\text{\hspace{0.17em}}\text{rad}$

b) Time taken by wheel to turn for the cord to unwind completelyis.$14\text{sec}$

i) Diameter of pulley is $D=8.0\text{\hspace{0.17em}}\text{cm}$,

ii) Length of cord is $L=5.6\text{\hspace{0.17em}}\text{m}$,

iii) Angular acceleration is $\alpha =1.5\text{\hspace{0.17em}}\text{rad}/{s}^{2}$,

**Find the circumference of the wheel. Using circumference and length of cord, find the number of revolutions. Find the angle using the revolutions. Then, use rotational kinematic equations to find the time.**

**The circumference of the pulley, **

${\mathbf{c}}{\mathbf{=}}{\mathbf{\pi d}}$

**The number of revolutions made,**

${\mathbf{\theta}}{\mathbf{=}}{\mathbf{2}}{\mathbf{\pi n}}$

**Total angular distance travelled, **

${\mathbf{\theta}}{\mathbf{=}}{\mathbf{2}}{\mathbf{\pi n}}$

**Relation between angular distance and time taken,**

${\mathbf{\theta}}{\mathbf{=}}{{\mathbf{\omega}}}_{{0}}{\mathbf{t}}{\mathbf{+}}\frac{1}{2}{{\mathbf{\alpha t}}}^{{2}}$

**where, Iisthe moment of inertia, M, m are masses, r is radius, t is time taken.**

Circumference of wheel is c,

$c=\pi d$

$c=\pi \times 0.08\text{\hspace{0.17em}m}$

$c=0.25\text{\hspace{0.17em}}\text{m}$

Now, number of revolutions *n* as follows:

$n=\frac{\text{Length of cord}}{\text{circumference of wheel}}$

$n=\frac{5.6\text{\hspace{0.17em}m}}{0.25\text{\hspace{0.17em}m}}$

$n=22.3\text{\hspace{0.17em}}\text{revolutions}$

Now, angle is as follows:

$\theta =2\pi n$

$\theta =2\pi \times 22.3$

$\theta =140\text{\hspace{0.17em}}\text{rad}$

Hence,the angle that the wheel must turn for the cord to unwind completely is $140\text{\hspace{0.17em}}\text{rad}$.

Now, time can be calculated using the equation-

$\theta ={\omega}_{0}\times t+\frac{1}{2}\alpha {t}^{2}$

$140\text{\hspace{0.17em}rad}=0\times t+\frac{1}{2}\times 1.5\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\times {t}^{2}$

$t=13.7\text{sec}$

In two significant figures,

$t=14\text{s}$

Hence, time taken by wheel to turn for the cord to unwind completelyis $14\text{\hspace{0.17em}sec}$.

94% of StudySmarter users get better grades.

Sign up for free