Why the body doesn't have radial component in uniform circular motion? If the body moves in an uniform circular motion when the object is at its lowest point it will be like this

there is a horizontal component of velocity which is $v_{x}$ and vertical force that creates $a_{c}$
but why when it's at the position the vertical force didn't create a vertical component of velocity and change the total velocity of the particle?
I've searched for this question and I ended up with the playlist with Ramamurti Shankar he was explaining a loop-the-loop example
and he said that this vertical acceleration  creates a vertical component of velocity he provided this image

but still if so it's creating a vertical component of velocity why the body moves with constant speed in case of uniform circular motion while its velocity must be equal to the total velocity of the vertical component and horizontal component which is $\sqrt {(v_{x})^2+(v_{y})^2}$  and also why it's direction also tangent while it must shift towards the center?
 A: The force on a body in circular motion is radially inwards so that force can do no work on the body and the speed of the body must stay constant.
Consider a body moving from position $A$ to position $B$ along an arc of a circle as shown in the left-side diagram below.

The speed does not change so $|\vec v_{\rm initial}|=|\vec v_{\rm final}|$.
The vector diagram on the right-hand side shows $\vec v_{\rm initial}+\Delta \vec v = \vec v_{\rm final}$.
What you must remember is that the changes are happening continuously or, put another way, in very, very small incremental steps which is equivalent to saying that one needs to look to see what happens as the angle $\theta$ gets smaller and smaller.
As $\theta$ tends towards zero so $\alpha$ tends towards $90^\circ$ and you will note that this means that the change in velocity produced by in force acting radially inwards is in the same direction as that force.
A: This is a really good question that I have personally struggled with.
The essential point over here is that the vertical velocity that would be produced will be $adt\hat j$, where $a$ is the centripetal acceleration.
Now total velocity at that point will then be $V^2=(adt)^2+(v_x)^2$, where $v_x$ is the initial velocity in the $x$ direction.
Now here comes the important point, at the next instant the centripetal acceleration is pointing in some other direction, thus the contribution of $adt$ this time will not point in the $\hat j$ direction, but in some other direction, call it $\hat b$.
Thus these contributions always add a differential value to the initial velocity vector in different directions, causing a change in velocity of negligible order.
Another way to think is that when the particle is at below most point then the contribution in velocity is $+adt\hat j$, while when the particle is at the topmost point the contribution is $-adt\hat j$, thus on average the net contribution is always zero for all points, and this average is an excellent approximation since the quantity in question always has a differential with it.
Another way is to argue via energy conservation. Since the force is always perpendicular to displacement we get
$$\int\vec F\cdot d\vec s = \Delta KE =0$$
Thus kinetic energy and by extension speed are conserved.
A: Consider the progression through a very small sequence of time intervals.
Starting at the point in your illustrations, the object's velocity is purely horizontal, directly to the right, and its acceleration is purely vertical, directly upwards.
A short time after that, the upwards acceleration has shifted the velocity so that it now has an upwards vertical component. At the same time, the object's position has moved towards the right. As a result of the change in position, the acceleration is now directed diagonally upwards and to the left.
After another short time interval, the upwards acceleration has further increased the upwards vertical component of velocity. However, the acceleration's new horizontal component pulling to the left is now acting to reduce the rightwards horizontal component of velocity. The object has also moved further to the right (and a tiny bit up), changing the direction of acceleration further, making the horizontal slowdown stronger and the vertical speedup weaker.
All of these changes in position, velocity, and acceleration all occur continuously and in sync with each other. The horizontal velocity causes a horizontal change in position, which causes an increasing horizontal component of acceleration, which causes a reduction in the horizontal component of velocity.
All of these interactions combine in a way that works out to always decreasing the horizontal component of velocity exactly enough to balance out the increase in the vertical component to keep the magnitude of velocity the same.
