Question: ![enter image description here](https://i.stack.imgur.com/6EXpa.png)

Why does the object not go inward, into the circle if the acceleration is inward? I think its because the velocity to outward? So they sort of cancel each other out? But if the speed is kept constant and the acceleration is inward, won't the object eventially move inward? 

Also, what keeps the direction of acceleration always point inward?

Answer: 
No, a uniform circular motion has both constant speed and constant acceleration and it will continue indefinitely, never going inwards.

If there were no inward acceleration, the object would move along the straight line in the direction of $\vec v$, and therefore away from the circle. To keep the object on the circle, there has to be a radial attractive force – the centripetal force – whose magnitude must be $F=ma=mv^2/r$.

We may also describe the situation from the viewpoint of the frame rotating together with the object. In that frame, there is a centrifugal, outward force $mv^2/r$, which tries to "repel" the object from the origin. This centrifugal (fictitious) force must be compensated by a real, centripetal force if the object is supposed to stay at a fixed distance from the origin.

Those statements may be easily derived from a particular coordinate parameterization of the motion. The coordinates are
$$ x(t) = r\cdot \sin(\omega t), \quad y(t) = r\cdot \cos(\omega t)$$
The velocities are derivatives with respect to $t$
$$ v\_x(t) = r\omega \cdot \cos(\omega t), \quad v\_y(t) = -r\omega\cdot \sin(\omega t)$$
and the accelerations are derivatives of the velocities
$$ a\_x(t) = -r\omega^2\cdot \sin(\omega t), \quad a\_y(t) = -r\omega^2\cdot \cos(\omega t)$$
and they are proportional to $(x,y)$ themselves, so the force is inward and radial. Note that $r\omega=v$ so $r\omega^2=v^2/r$.
> 
> Why does the object not go inward, into the circle if the acceleration is inward? I think its because the velocity to outward? So they sort of cancel each other out?
> 
> 
> 

No, the velocity on the object is not outwards, which you can see if you recall that the velocity of an object is $\vec {v} = \frac d {dt} \vec r$. Over a time $dt$, the change in the position vector $d\vec r$ points in the same direction as the tangent to the trajectory of the moving object. Dividing this by the scalar $dt$ gives the velocity vector with the direction still along the tangent.

> 
> But if the speed is kept constant and the acceleration is inward, won't the object eventially move inward? 
> 
> 
> 

Over a time $dt$ the displacement from the velocity is $Vdt$, that from the acceleration is $\frac 1 2 a {dt}^2$. The displacement is therefore dominated by the velocity because ${dt}^2$ becomes far smaller than ${dt}$ as $dt\rightarrow 0$

> 
> Also, what keeps the direction of acceleration always point inward?
> 
> 
> 

The force causing the acceleration.