Question

A uniform, thin rod of length h and mass M is held vertically with its lower end resting on a frictionless horizontal surface. The rod is then released to fall freely. (a) What is the speed of its center of mass just before it hits the horizontal surface? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)

Answer 1

Answer:

$v = \sqrt{gh}$

Explanation:

Since there's no external force beside gravity acting on the rod, we can use the law of energy conservation to calculate for the speed of rod center of gravity before hitting the surface.

Also as the rod is uniform and this, its center of gravity is at distance h/2 from the surface.

$E_k = E_p$

$mv^2/2 = mgh/2$

$v^2 = gh$

$v = \sqrt{gh}$