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

Answer:

$v = \sqrt{\frac{1.1g}{(1 + \frac{1}{r^2})}}$

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:

$gh = 0.5v^2 + \frac{v^2}{2r^2}$

$gh = (\frac{v^2}{2})(1 + \frac{1}{r^2})$

$v^2 = \frac{2gh}{(1 + \frac{1}{r^2}})$

$v = \sqrt{\frac{2gh}{(1 + \frac{1}{r^2})}}$

$v = \sqrt{\frac{1.1g}{(1 + \frac{1}{r^2})}}$