Motion of a Particle in Space  (continued)

 3 It is often convenient to describe motion of a particle in terms of its normal and tangential components.   Let    T  be the unit tangential vector to the path, C, of motion of a particle and let   N  be its unit normal vector.                              Then since   v ,  the velocity of the particle is always tangent to its path                          v    =    v  T          so       dv / dt    =    dv/dt  T   +   v  dT/dt   here   dv/dt   represents the tangential component of acceleration of the particle   Now          dT/dt   =   [dT/ds] [ ds/dt]    using the chain rule of differentiation.  So           v  dT/dt   =    v dT/ds  v   =    v2  dT/ds       since    ds/dt   =  v   Note:  Since      T  ∙ T   =    1    (T   is a unit tangent vector to the path)                   T  ∙  dT/ds    =   0    which shows that  dT/ds    is perpendicular to  T    .   Define the curvature of C as   | dT/ds |    =   κ  and let   N  be the unit normal vector to C                 dT/ds     =   κ  N   so the acceleration of the particle becomes (where κ   = curvature )      a   =   dv / dt   =    dv/dt  T   +   κ v2  N,    κ   =  1 / ρ,   ρ  =  radius of curvature   Note:  The acceleration in general has components tangential and normal to its path.   Note further:  To find:  T  =   v /  | v |                    a ∙ a   =   a2  =  aT 2  +   aN 2   To find:  N         N  =    [dT/ds] / κ   =   [dT/ds] / | dT/ds |   =    [dT/dt] / | dT/dt |     Click here for an example of these calculations.