physics please help
posted by roni on .
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.59 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed v0, so that the rod begins to rotate upward about the pivot. What must be the value of v0, such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?
The initial kinetic energy of the bar, which is
(1/2) I wo^2 = (1/2)*M (L^2/3)*(vo/L)^2
= (1/6)*M vo^2
Must equal the increase in potential energy in the upside-down position, which is
M g L
since the center of mass raises by an amount L.
I is the moment of inertia of the rod about the pivot at the top.
M cancels out and you are left with
(1/6)vo^2 = g*L
Solve for vo