Question

A disk and a hoop roll without slipping down the incline plane, starting from rest at height h=0.55m above the level surface. Using the laws of energy conservation find the speeds of the objects at the bottom of the incline.

Answer 1

`v = \sqrt{(1.1g)/((1 + (1)/(r^2)))}`

Step-by-step explanation:

By the law of energy conservation, the potential energy can be converted to kinetic energy and rotational energy as it rolling downhill

`E_p = E_k + E_r`

`mgh = 0.5mv^2 + 0.5m\omega^2`

Also velocity of the disk is its angular velocity times its radius:

`gh = 0.5v^2 + (v^2)/(2r^2)`

`gh = ((v^2)/(2))(1 + (1)/(r^2))`

`v^2 = (2gh)/((1 + (1)/(r^2)))`

`v = \sqrt{(2gh)/((1 + (1)/(r^2)))}`

`v = \sqrt{(1.1g)/((1 + (1)/(r^2)))}`