Question

A solid uniform disk with a mass M and a radius R is pivoted around a horizontal axis through its center, and a small object with mass M is attached to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular velocity of the small object at the bottom.

Answer 1

`\omega = \sqrt{(4g)/(3R)}`

Step-by-step explanation:

By conservation of energy:

The initial energy of the system is:
`M*g*R`

The final energy of the system is:
`1/2*M*R^2*\omega^2+1/2*(1/2*M*R^2)*\omega^2`

`M*g*R=1/2*M*R^2*\omega^2+1/2*(1/2*M*R^2)*\omega^2`

Solving for ω:

`\omega=\sqrt{(4g)/(3R)}`