Example of Motion of a Particle in Space  (continued)                

 

 

 

 

 

 

 

 

 

 

 

 

4.

 

 

 

 

Next find the unit normal vector,  N  ;  recall   N  =   [dT/ds] / κ

 

But  κ  =  |  dT/ds  |

 

Therefore   N  =  dT/ds  /  | dT/ds |   =  [dT/dt] / | dT/dt |  

 

and

 

     dT/dt    =  [- sin t j  -  cos t k ]  / √2

 

     | dT/dt |   =    1 / √2

 

So    [dT/dt] / | dT/dt |   =  [- sin t j  -  cos t k ]   =   N                     (result)

 

 

Check:  The unit tangential and normal vectors are perpendicular.  Therefore their

dot product should be zero.

 

           Is        T  ∙  N   =  0 ?

 

     { [ i   +  cos t j  -  sin t k ]  / √2 }  ∙  {  [- sin t j  -  cos t k ] } 

 

            =  - cos t  sin t  +  sint t  cos t  =  0    (check)

 

 

 

 

 

5.

 

 

 

Next find the curvature,  κ .

 

          κ    =    | dT/ds |     =    | dT/dt |  dt/ds   using the chain rule

 

         Now   dt/ds  =  1/v

 

         κ    =    | dT/ds |     =    | dT/dt |  ( 1 / v )   =  (1 / √2) (1 / √2)  =  ½      (result

 

 

 

 

 

6.

 

 

Finally, the acceleration of the particle is the rate of change of the velocity of the

particle,   dv/dt.   Recall   v   =     i   +  cos t j  -   sin t k   .   So 

 

   a  =  - sin t j  -   cos t k     (result in Cartesian components)

 

Click here to continue with calculation of the tangential  and normal components

of acceleration.

 


Copyright © 2011 Richard C. Coddington

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