91Ó°ÊÓ

We have the given data:

The radius of the wheel =0.120 m.

Force applied to the rim of the wheel =80.0 N.

Angular speed of the wheel$\left({\mathrm{Ó\neg }}_z\right)$= 12.0 rev/s =75.40 rad/s.

The system is initially at rest.

Time interval during which the revolutions take place is,

$∆t=2.00s$s.

If a rigid object free to rotate about a fixed axis has a net external torque acting on it, the object undergoes an angular acceleration , where,

$∑{\mathrm{Ï„}}_{\text{ext}}=l\mathrm{Î\pm }⋯⋯\left(1\right)$

First, we will calculate angular acceleration which is given by,

${\mathrm{Ó\neg }}_z={\mathrm{Ó\neg }}_{0z}+{\mathrm{Î\pm }}_{z}t$

Since the system starts from the rest, we have,${\mathrm{Ó\neg }}_{0z}$= 0 .

Therefore,

$\begin{array}{r}{\mathrm{Ó\neg }}_z={\mathrm{Î\pm }}_{z}t\\ ⇒{\mathrm{Î\pm }}_z=\frac{{\mathrm{Ó\neg }}_z}{t}\\ ⇒{\mathrm{Î\pm }}_z=\frac{75.40}{2.00}\\ ⇒{\mathrm{Î\pm }}_z=37.70\mathrm{rad}/{\mathrm{s}}^2\end{array}$

From , the net torque is given by,

$\begin{array}{r}∑{\mathrm{Ï„}}_z=I{\mathrm{Î\pm }}_z\\ ⇒I=\frac{∑{\mathrm{Ï„}}_z}{{\mathrm{Î\pm }}_z}\end{array}$

$\begin{array}{r}⇒I=\frac{Fr}{{\mathrm{Î\pm }}_z}\\ ⇒I=\frac{\left(80.0\right)\left(0.120\right)}{\left(37.70\right)}\\ ⇒I=0.255\mathrm{kg}â‹\dots {\mathrm{m}}^2\end{array}$

Hence, the moment of inertia of the wheel is,$I=0.255kg.m^2$ .