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DYNAMICS - EXAMPLE


Particle Motion along Path
  Example

 

A particle moves along a wire that follows the curve y = 4(e2x + e-2x) at a constant velocity of 4 m/s. What is the normal acceleration of the particle when x = 0?

 

   
    Solution


Partical at x = 0

 

Since the velocity is constant, there is no acceleration tangent to its path. However, there is acceleration normal (perpendicular) to the path as given by,

      an = v2

Need to determine, ρ
     

     dy/dx = d[ 4(e2x + e-2x) ] / dx
              = 8 (e2x - e-2x )

     d2y/dx2 = d[ 8(e2x - e-2x ) ] / dx
                 = 16 (e2x + e-2x )

     (dy/dx)x=0 = 8(1 - 1) = 0
     (d2y/dx2)x=0 = 16(1 + 1) = 32

     

Substituting back into the acceleration equation,
     an = v2/ρ = 42 (32)

an = 512 ft/s2

     
   
 
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