A thin uniform rod of length 0.810 m and mass M rotates horizontally at angular speed 21.0 rad/s about an axis through its center. A particle of mass M/3.00 attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed vp is 6.00 m/s greater than the speed of the end of the rod just after ejection, what is the value of vp?
Li = Lf
L = IW
1/12 * m * r^2 = I for a rod rotating about its center
mvr* sin(angle between r and v) = I for particle
The Attempt at a Solution
(1/12) * M * (.81^2) * 21rad/s = [1/12 * M * (.81^2)]W + [1/3*M * r * v]
so the problem is I have one equation with two unknowns, and no other equations....
(I tried to use that v = Wr, but I think that only works for rolling things and the answer didn't work out...)
Any help would be great! Thanks