posted by John on .
One end of a thin uniform rod, 1.00m in length, is attached to a pivot. The rod is free to rotate about the pivot without friction or air resistance. Initially, it is hanging straight down (the "6 o' clock position). Then the lower tip is given an initial horizontal speed of Vti, so that the rod rotates upward, about its pivot. Find the value of Vti so that the rod just reaches the "12 o' clock" position (straight up), coming to a momentary halt there.
The initial kinetic energy must be enough to increase the potential energy enough to let it reach the 12 O'clock position.
The initial kinetic energy is:
KE = (1/2) I w^2 = (1/2)(1/3)M L^2*(Vti/L)^2
= (1/6)M Vti^2
(I is the moment of inertia and w is angular velocity)
The potential energy change (with the CM in the middle of the rod) is
deltaPE = M g [L/2 - (-L/2)] = M g L
Therefore the requirement is
(1/6)Vti^2 = gL
Vti = sqrt(6 g L)