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In the
Curvilinear Motion: Rectilinear Coordinates section, it was shown that velocity is always tangent to the path of motion, and acceleration is generally not.
If the component of acceleration along the path of motion is known, motion in terms of normal and tangential components can be analyzed. |
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Consider a point P moving along a curved path.
The position vector r specifies the position of P with respect to the reference point O, and s measures the position of P along the path relative to the reference point O'. The unit tangent vector, et, is tangent to the path at point P. |
Velocity of a Particle in a Plane |
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The speed of the point along the path is found by taking the derivative
of the position,
v = ds/dt
Since the velocity is tangent to the path, it can be expressed in terms
of et,
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Relationship Between Normal
and Tangent Direction
Relationship between et and en
Radius of Curvature Definition
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Similar to rectilinear coordinates, acceleration is obtained by differentiating
the velocity (two parts) as
a = dv/dt = dv/dt et + v det/dt
Since the acceleration is not, in general, tangent to the path, it is useful to express it in terms of components that are normal and tangent to the path. To do so, the time derivative of the unit tangent vector, et, will be found.
Let et(t) be the unit tangent vector at time t, and et(t + Δt) be the unit tangent vector at time t + Δt. If et(t) and et(t + Δt) are drawn from the same origin, they form two radii of length 1 on the unit circle. The magnitude of Δet is given by the equation
Now, consider the vector Δ>et/Δθ.
As Δθ approaches 0, Δet/Δθ becomes tangent to the unit circle (perpendicular to the path of P). The
magnitude of Δet/Δθ< approaches 1.
Next, define the unit normal vector as
The chain rule can be used with the time derivative of the unit tangent vector to give
But ds/dt = v, det/dθ = en, and dθ/ds = 1/ρ (where ρ is the radius of curvature). After substituting, the time derivative of the unit tangent vector becomes |