A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 N is applied to the rim of the wheel. The wheel has radius 0.120 m. Starting from rest, the wheel has an angular speed of 12 rev/s after 2.00 s. What is the moment of inertia of the wheel?

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$\left({\omega }_z\right)$= 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,

$\sum {\tau }_{\text{ext}}=l\alpha \cdots \cdots \left(1\right)$

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

${\omega }_z={\omega }_{0z}+{\alpha }_{z}t$

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

Therefore,

$\begin{array}{r}{\omega }_z={\alpha }_{z}t\\ \Rightarrow {\alpha }_z=\frac{{\omega }_z}{t}\\ \Rightarrow {\alpha }_z=\frac{75.40}{2.00}\\ \Rightarrow {\alpha }_z=37.70\mathrm{rad}/{\mathrm{s}}^2\end{array}$

From , the net torque is given by,

$\begin{array}{r}\sum {\tau }_z=I{\alpha }_z\\ \Rightarrow I=\frac{\sum {\tau }_z}{{\alpha }_z}\end{array}$

$\begin{array}{r}\Rightarrow I=\frac{Fr}{{\alpha }_z}\\ \Rightarrow I=\frac{\left(80.0\right)\left(0.120\right)}{\left(37.70\right)}\\ \Rightarrow I=0.255\mathrm{kg}\cdot {\mathrm{m}}^2\end{array}$

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