91Ó°ÊÓ

i) Mass of box, $m=30\text{ k²µ}$

ii) Outer radius of the device,$R=0.50\text{ M}$

iii) Inner radius of the device,$r=0.20\text{ }\text{m}$

iv) Applied force, $F_{app}=140\text{ }\text{N}$

v) Acceleration, $a=0.80\text{″¾}/{\text{s}}^2$

Find the tension in the rope using Newton’s second law of motion. This tension can be used to find the moment of inertia of the device about its axis of rotation by applying the equation of the rotational motion to the device.

Formulae are as follow:

$T−mg=ma$

$F_{app}R−Tr=I\mathrm{Î\pm }$

$\mathrm{Î\pm }=a/r$

where,$F_{app}$ is apparent force, m is mass, R, r are radii, I is moment of inertia, T is tension, a is an acceleration,g is an acceleration due to gravity and $\mathrm{Î\pm }$ is angular acceleration.

For the tension in the rope,

$T−mg=ma$

$\begin{array}{c}T=ma+mg\\ =(30\text{ }\text{kg})\left(0.80\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)+(30\text{ k²µ})\left(9.8\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\\ =318\text{ N}\end{array}$

Now, an equation for the rotational motion to the device is,

$F_{app}R−Tr=I\mathrm{Î\pm }$

As,

$\mathrm{Î\pm }=a/r$

$\begin{array}{l}{F}_{ap}R−Tr=I\left(\frac{a}{r}\right)\\ I=\frac{(F_{ap}R−Tr)}{\left(\frac{a}{r}\right)}\\ I=(F_{ap}R−Tr)\left(\frac{r}{a}\right)\end{array}$

For the given values, the above equation becomes-

$\begin{array}{c}I=(140\text{ N}\left(0.50\text{″¾}\right)−\left(318\text{ N}\right)0.20\text{″¾})\left(\frac{0.20\text{″¾}}{0.8\text{″¾}/{\text{s}}^2}\right)\\ =1.6\text{ k²µ}.{\text{m}}^2\end{array}$

Hence, the rotational inertia of the device about its axis rotation is$1.6{\text{ k²µ³¾}}^2$.