# Thread: motion of a particle

1. ## motion of a particle

Suppose that the equations of motion of a particle moving along an elliptic path are x= a cos omega t , y=sin omega t.

a) show that the acceleration is directed toward the origin.

b) show that the magnitude of the acceleration is proportional to the distance from the particle to the origin.

2. Originally Posted by Jenny20
Suppose that the equations of motion of a particle moving along an elliptic path are x= a cos omega t , y=sin omega t.

a) show that the acceleration is directed toward the origin.

b) show that the magnitude of the acceleration is proportional to the distance from the particle to the origin.
Let's call omega w until LaTeX is back up...
a) Suppose the particle is at the point (x, y). Then
x = a cos(wt)
y = sin(wt)

The acceleration is s'' where s is the displacement vector and each derivative is taken with respect to time. The displacement is a vector from the origin to the point in question, so in terms of the unit vectors i and j:
s = x*i + y*j = a cos(wt)*i + sin(wt)*j

Then taking the derivatives:
s' = -aw sin(wt)*i + w cos(wt)*j

s'' = -aw^2 cos(wt)*i - w^2 sin(wt)*j

Let's play with this a bit and factor a -w^2:
s'' = -w^2 (a cos(wt) + sin(wt)) = -w^2*s

So s'' is a vector that is w^2 longer than s and, most importantly, points in exactly in the opposite direction to s. Since s points from the origin to the point, s'' points from the point to the origin.

b) The magnitude of the acceleration is |s''|, so
|s''| = w^4 * |s|
Thus the two are proportional.

-Dan

Note carefully that the centripetal acceleration does NOT point toward the origin in general. The centripetal acceleration is the acceleration component in a direction perpendicular to the velocity.

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# a partical movesnso that its position vector is given by r= (sinwt) i (coswt)j, sow that the velocity of oartical is perpendicular andmd the acceleration a is directed tiwards the orgin and has magnitude proportional ti the distance fron the orgin

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