Answer:
Approximately [tex]0.293\; {\rm kg \cdot m^{2}}[/tex].
Explanation:
Since the wheel started from rest, initial angular velocity will be [tex]\omega_{0} = 0\; {\rm rad \cdot s^{-1}}[/tex]. It is given that the angular velocity [tex]\omega_{1}[/tex] is [tex]14.9\; {\rm rev \cdot s^{-1}}[/tex] after [tex]t = 3.41\; {\rm s}[/tex]. Apply unit conversion and ensure that all angular velocity are measured in radians-per-second:
[tex]\begin{aligned} \omega_{1} &= 14.9\; {\rm rev \cdot s^{-1}} \times \frac{2\, \pi \; {\rm rad}}{1\; {\rm rev}} \\ &\approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}[/tex].
Change in angular velocity:
[tex]\begin{aligned} \Delta \omega = \omega_{1} - \omega_{0} \approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}[/tex].
Since the tangential force is constant and there is no friction on the wheel, the angular acceleration [tex]\alpha[/tex] of this wheel will be constant. Since the change in velocity [tex]\Delta \omega \approx 93.620\; {\rm rad \cdot s^{-1}}[/tex] was achieved within [tex]t = 3.41\; {\rm s}[/tex], the average angular acceleration will be:
[tex]\begin{aligned} \alpha &= \frac{\Delta \omega}{t} \\ &\approx \frac{93.620\; {\rm rad \cdot s^{-1}}}{3.41\; {\rm s}} \\ &\approx 27.45\; {\rm rad \cdot s^{-2}}\end{aligned}[/tex].
At a distance of [tex]r = 0.110\; {\rm m}[/tex] from the axis of rotation, the tangential force [tex]F = 73.0\; {\rm N}[/tex] will exert on the wheel a torque [tex]\tau[/tex] of magnitude:
[tex]\begin{aligned} \tau &= F\, r \\ &= (73.0\; {\rm N})\, (0.110\; {\rm m}) \\ &\approx 8.030\; {\rm N \cdot m}\end{aligned}[/tex].
Let [tex]I[/tex] denote the moment of inertia of this wheel. The equation [tex]\alpha = (\tau / I)[/tex] relates angular acceleration [tex]\alpha[/tex] to moment of inertia [tex]I\![/tex] and net torque [tex]\tau[/tex]. Rearrange this equation to find the moment of inertia:
[tex]\begin{aligned}I &= \frac{\tau}{\alpha} \\ &\approx \frac{8.030\; {\rm N\cdot m}}{27.45\; {\rm rad \cdot s^{-2}}} \\ &\approx 0.293\; {\rm N \cdot m \cdot s^{2}} \\ &= 0.293 \; {\rm kg \cdot m^{2}}\end{aligned}[/tex].
Note that the unit "radians" is typically ignored. Additionally, [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].
Hence, the moment of inertia of this wheel is approximately [tex]0.293\; {\rm kg \cdot m^{2}}[/tex].
