Motion of a Particle in Space (continued)

In a Nut Shell: 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 a = 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 = a_T ² + a_N ²

To find: N N = [dT/ds] / κ = [dT/ds] / | dT/ds | = [dT/dt] / | dT/dt |

Click here for an example.

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