the rate of change of acceleration is known as jerk.  I need the
direction and the magnitude of the jerk for a particle movign in a
circle of radius R at angular velocity w.  I need to draw a vector
diagram showing the instantaneous position velocity, acceration and
jerk.  escpecially how I can draw a jerk vector?

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Clarification of Question by diala-ga on 07 Oct 2004 20:25 PDT

I need an answere as soon as possible please

The acceleration of the particle you describe is constant, i.e. the
acceleration does not change and therefore has a rate of change (or
"jerk") of zero.

The magnitude of the acceleration is constant.  However, its direction is not.

As Helpfulperson has pointed out, medicineman was incorrect.  The
magnitude of the acceleration is constant, but it's direction is not.

The position of a particle in uniform circular motion (the situation
you are trying to describe) as a function of time is given by:

r(t) = R*cos(w*t)*i + R*sin(w*t)*j

where r(t) is a vector (x(t), y(t)), R is the radius of the circle, w
is the angular velocity, t is time, and i and j are the unit vectors
in the x and y directions, respectively (i.e., (1,0) and (0,1).

The velocity vector as a function of time is simply found by taking
the derivative of r(t) with respect to time:

v(t) = -w*R*sin(w*t)*i + w*R*cos(w*t)*j

Note that the velocity vector is always perpendicular to the position
vector, which can be proven by showing that the dot product of the two
vectors is equal to zero:

v(t).r(t) = -w*R*cos(w*t)*sin(w*t)*|i| + w*sin(w*t)*cos(w*t)*|j| = 0

The acceleration vector as a function of time is found by taking the
derivative of the velocity vector with respect to t:

a(t) = -w^2*R*cos(w*t)*i - w^2*R*sin(w*t)*j
     = -w^2*r(t)

Note that the acceleration vector is always parallel to r(t), but
pointed in the opposite direction.

Finally, the "jerk" vector as a function of time is found by taking
the derivative of a(t) with respect to time.  Because we already know
that the derivative of r(t) with respect to t is v(t), we can simply
write:

a'(t) = -w^2*v(t) = w^3*R*sin(w*t)*i - w^3*R*cos(w*t)*j

The jerk is parallel to the velocity vector, but oriented in the opposite direction.