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 known as the torque, \(F\) is the tension, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration. Then, solve for the angular acceleration.

\[  \alpha = \frac{rT}{I} = \frac{2\cdot 40}{10} = 8.00 \ \mathrm{rad/s^2}  \]

According to angular kinematics, we can derive the angular displacement.

\[  \theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2 = 0+0+\frac{1}{2}\alpha t^2 =\frac{1}{2}\cdot 8.00 \cdot 3^2 = 36.0 \ \mathrm{rad} \]

The linear displacement is

\[  d = r\theta = 2.0 \cdot 36.0 = 72.0 \ \mathrm{m}  \]