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DYNAMICS - THEORY

   

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.

     

Motion of a Particle in a Plane
  Plane Motion of a Particle

 

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


Velocity of a Particle in a Plane
 

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,

 
v = ds/dt et = v et
 
     
    Acceleration


Relationship Between Normal
and Tangent Direction

 


Relationship between et and en

 


Radius of Curvature Definition

 

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

     

Normal and Tangential Acceleration
 

     

The final equation for the acceleration in terms of both normal and tangential components is

 
 
     
   
 
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