Answer to Question #113841 in Classical Mechanics for Caylin

Question

A 5.0kg, 60cm diameter disk rotates on a axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle,then released.

1) what's the cylinders initial angular acceleration?

2) What is the cylinders angular velocity when it is directly below the axle?

Please assist

Answer 1

The moment of inertia of a disk rotating this way (around the edge):

$$I=\frac{3}{2}mr^2.$$

The torque around the same axis when the center of the disk was at equal height with the axis:

$$\tau=mgr.$$

Newton's second law for rotational motion:

$$\tau=I\alpha,\\ mgr=\frac{3}{2}mr^2\alpha,\\ \space\\ \alpha=\frac{2g}{3r}.$$

When the disk is below the axle, it became one radius lower. Conservation of energy:

$$mgr=\frac{1}{2}I\omega^2=\frac{3}{4}mr^2\omega^2.\\ \space\\ \omega=\sqrt{\frac{4g}{3r}}=6.6\text{ rad/s}.$$