Question

A solid cylinder of radius 0:500[m] is wrapped with a massless string to simulate a

yo-yo. Holding the free end of the string stationary, the cylinder is dropped from a height

of 2:50 [m] from the ground. Right before it hits the ground, the cylinder was found to have

an angular velocity of 5:00 [rad=s]. Assuming that the string does not slip or stretch as the

cylinder descends and rotates, what is the angular acceleration of the cylinder?

Answer 1

The angular acceleration can be found with the following equation:

$$\alpha=\frac{\omega_f^2-\omega_i^2}{2\phi}=\frac{\omega_f^2}{2\phi}.$$

Find the angle $\phi$ the cylinder turned while rotating down 2.5 m: it is $2\pi$ times the number of complete rotations N:

$$\phi=2\pi N=2\pi \frac{H}{C}=2\pi \frac{H}{2\pi r}=\frac Hr.\\\space\\ \alpha=\frac{\omega_f^2r}{2H}=2.5\text{ s}^{-2}.$$