A thin rod having a mass of 4 kg is balanced vertically as shown. Determine the height h at which it can be struck with a horizontal force F so that it rotates but does not slip on the floor. Assume the coefficient of friction is zero.

Force Diagram and Velocities

The horizontal linear impulse and momentum is,
     ΣF Δt = mv₂ - mv₁
     FΔt = mv_G - 0

The rod will not slip, so it will rotate about its bottom point where it touches the ground,
     FΔt = m(0.4ω)  

Since the rod will also rotate, the angular impulse and momentum equation must also apply about its center of mass.
     ΣM Δt = I_Gω₂ - I_Gω₁
     F(h - 0.4) Δt = [m(0.8²)/12] ω - 0

The term, "F Δt" can be eliminated by combining the two equations, giving,
     (h - 0.4) m(0.4ω) = [m(0.8²)/12] ω
     0.4 (h - 0.4) = 0.64/12

Solving for h gives,

h = 0.5333 m