Dgdrxrt

I have arrived to the following equations in polar coordinates

$$vecv=dotrhatr+rdotthetahattheta$$

$$veca=(ddotr-rdottheta^2 )hatr+(2dotrdottheta+rddottheta)hattheta$$

Then in the lecture note it is written that if the movement is in uniform angular velocity $omega = dottheta$ in radius $R$ then
$$veca=-Romega^2hatr$$

How did he arrived to this formula? maybe I written the formula wrong

First of all, let's sort out the notation: I take it that our OP newhere intends the "hat" symbol, as in "$hat x$", to indicate a unit vector in the direction of $vec x ne 0$:

$hat x = dfracvec xvert vec x vert, tag 1$

where the customary "arrow" is used for general vectors as in "$vec x$"

Second, the derivations of

$vec v = dot r hat r + rdot theta hat theta tag 2$

and

$vec a = (ddot r - rdot theta^2) hat r + (2dot r dot theta + r ddot theta)hat theta tag 3$

are in their own right worthy of presentation:

We may introduce cartesian coordinates $(x, y)$ on $Bbb R^2$ such that

$x = r cos theta, tag 4$

$y = r sin theta; tag 5$

if we write

$vec r = (x, y) = (rcos theta, r sin theta) = r(cos theta, sin theta), tag 6$

we see that

$vec n = (cos theta, sin theta) tag 7$

is the unit vector in the radial ($r$) direction, that is

$hat r = vec n; tag 8$

the unit vector $hat theta$ in the $theta$ direction is tangent to the curve (6) with constant $r$ at any point $(x, y) ne (0, 0)$; thus it must be collinear with

$dfracdvec n(theta)dtheta = (-sin theta, cos theta); tag 9$

since

$left vert dfracdvec n(theta)dtheta right vert = 1, tag10$

we infer that

$hat theta = (-sin theta, cos theta) = dfracdvec n(theta)dtheta tag11$

for $theta$ increasing in the counter-clockwise direction.

Now any curve

$phi(t): I to Bbb R^2, tag12$

where $I subset Bbb R$ is some open interval, may be written

$phi(t) = r(t)vec n(theta(t)) = r(t)(cos theta(t), sin theta(t)); tag13$

thus,

$vec v = dot phi(t) = dot r(t) vec n(theta) + r(t) dotvec n(theta); tag14$

by virtue of (7)-(11) this may be written

$vec v = dot phi(t) = dot r(t) hat r + r(t) dot theta hat theta, tag15$

thus establishing (2). We may then differentiate this equation and find

$vec a = dotvec v = ddot r hat r + dot r dothat r + dot r dot theta hat theta + r ddot theta hat theta + rdot theta dothat theta; tag16$

between (7)-(11) we see that

$dothat r = dot theta hat theta, tag17$

$dothat theta = -dot theta hat r; tag18$

assembling (16)-(18) together we reach

$vec a = ddot r hat r + dot r dottheta hat theta + dot r dot theta hat theta + r ddot theta hat theta - rdot theta^2 hat r$
$= (ddot r - rdot theta^2)hat r + (2dot r dot theta + rddot theta) hat theta, tag19$

which is (3).

Having derived (2) and (3), we take

$r = R = textconstant tag20$

and

$omega = dot theta = textconstant, tag21$

so that

$dot r = ddot r = dot omega = ddot theta = 0; tag22$

then (19) yields

$vec a = -Romega^2 hat r, tag23$

as was to be shown.