A rope unwounded from a cylindrical wheel

Question:
A cylindrical wheel with radius of 2-m rotates on a fixed frictionless horizontal axis. The moment of inertia of the wheel is 10.0 kg m$^2$. A constant tension on the rope exerted around the rim is 40.0 N. If the wheel starts from rest at $t=0$ s, calculate the length of the rope unwounded after 3.00 seconds.

Answer:
The equation of motion for rotation is
\[
\sum \tau = rF = I\alpha
\]
where $\tau$ is the torque, $F$ is the tension, $I$ is the moment of inertia, and $\alpha$ is the angular acceleration. Now, solve for the angular acceleration.
\begin{equation}
\alpha = \frac{rT}{I} = \frac{2\cdot 40}{10} = 8.00 \ \mathrm{rad/s^2}
\end{equation}
According to angular kinematics, we can calculate the angular displacement.
\begin{equation}
\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 = 0+0+\frac{1}{2}\alpha t^2 =\frac{1}{2}\cdot 8.00 \cdot 3^2 = 36.0 \ \mathrm{rad}
\end{equation}
The linear displacement is
\[
d = r\theta = 2.0 \cdot 36.0 = 72.0 \ \mathrm{m}
\]

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