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DYNAMICS - EXAMPLE


Thin Vertical Rod
  Example

 

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.

   
    Solution


Force Diagram and Velocities

 

The horizontal linear impulse and momentum is,
     ΣF Δt = mv2 - mv1
     FΔt = mvG - 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 = IGω2 - IGω1
     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

     
   
 
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