What is the derivation of the centripetal acceleration in terms of position?

[unknown_2]

This problem has been discussed before, but one of the parts were not discussed.

Homework Statement

Your calculation is actually a derivation of the centripetal acceleration. To see this, express the acceleration of the particle in terms of its position r(t).

r(t) = $$Rcos(\omega t)\textbf{i} + Rsin(\omega t)\textbf{j}$$

Homework Equations

$$v = \omega R$$
$$a = \frac{v^2}{R}$$
$$a = \omega v$$

The Attempt at a Solution

$$v = \frac{d}{dt}r(t) \times \omega$$

btw, this is a Mastering engineering question, so i submitted and it says that:
The correct answer does not depend on the variables and functions: d, d, r.

is there another way to display this or am i missing something?

thx,

 

[rl.bhat]

Acceleration = V^2/R = w^2*R. In a uniform circular motion w and R are constant.

 

[unknown_2]

Acceleration = V^2/R = w^2*R. In a uniform circular motion w and R are constant.

omg...too stressed out about finals...i think i got it =_=

please correct me if I'm wrong (again)...

a(t) = $$-\omega^{2}r(t)$$

 

[rl.bhat]

omg...too stressed out about finals...i think i got it =_=

please correct me if I'm wrong (again)...

a(t) = [tex] -\omega^{2}r(t)[/tex]

If you express in the vector form, it is correct.