Motion of a Particle in Space (continued)
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 |
Copyright © 2011 Richard C. Coddington
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