91Ó°ÊÓ

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

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

iii) Angular acceleration is $\mathrm{Î\pm }=1.5\text{ }\text{rad}/s^2$,

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,

$\mathbf{c}\mathbf{=}\mathbf{Ï€»å}$

The number of revolutions made,

$\mathbf{θ}\mathbf{=}\mathbf{2}\mathbf{Ï€²Ô}$

Total angular distance travelled,

$\mathbf{θ}\mathbf{=}\mathbf{2}\mathbf{Ï€²Ô}$

Relation between angular distance and time taken,

$\mathbf{θ}\mathbf{=}{\mathbf{Ó\neg }}_0\mathbf{t}\mathbf{+}\frac{1}{2}{\mathbf{Î\pm ³Ù}}^2$

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

Circumference of wheel is c,

$c=\mathrm{Ï€}d$

$c=\mathrm{Ï€}×0.08\text{″¾}$

$c=0.25\text{ }\text{m}$

Now, number of revolutions n as follows:

$n=\frac{\text{Lengthofcord}}{\text{circumferenceofwheel}}$

$n=\frac{5.6\text{″¾}}{0.25\text{″¾}}$

$n=22.3\text{ }\text{revolutions}$

Now, angle is as follows:

$\mathrm{θ}=2\mathrm{π}n$

$\mathrm{θ}=2\mathrm{π}×22.3$

$\mathrm{θ}=140\text{ }\text{rad}$

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

Now, time can be calculated using the equation-

$\mathrm{θ}={\mathrm{Ó\neg }}_0×t+\frac{1}{2}\mathrm{Î\pm }{t}^2$

$140\text{ r²¹»å}=0×t+\frac{1}{2}×1.5\text{″¾}/{\text{s}}^{\text{2}}×t^2$

$t=13.7\text{sec}$

In two significant figures,

$t=14\text{s}$

Hence, time taken by wheel to turn for the cord to unwind completelyis $14\text{ s±ð³¦}$.