Example 2-5: Path Known Problem

A particle moves along a curve described by the path equation [ y = 0.2x³ –2x –2 ], where x and y are in meters. If the particle’s y velocity is known to be –2 m/s, and its y acceleration is known to be  –1 m/s² at the location (x,y) = (0.514, -3) m, determine: 

(1)   The velocity of the particle, as a vector; 

(2)   The acceleration of the particle, as a vector; 

(3)  The magnitudes of a_n and a_t ; 

(4)    The radius of curvature of the curve at that point by two methods.

Apply the chain rule to the path equation to introduce time derivatives:

(a)
(b)
(c)

Remember we were given:

(x,y) = (0.514, -3)

Substitute the given information into:

The velocity and acceleration vectors of the particle may now be written, in cartesian and polar forms:

(3)  The angle, b , between the v  and a  vectors, as shown above, is   b = 61.5° – 46.83° = 14.67° degrees.

Knowing b , the normal and tangential components of acceleration are easily found:

(4)    Two methods for the radius of curvature, r :

(a)   Since

,  and knowing an and v (magnitude),

or

(b)  Since we know  y = f(x), we can take derivatives to evaluate r :

Important!  These are y' and y"--derivatives of y with respect to x.  These are not y dot or y dot dot, derivatives of y with respect to time.