We're going to see how to derive the equation for the acceleration of an object moving in a circular path with a constant speed. Otherwise, no one has a uniform circular motion. Let's have a look at a small part of the trajectory or the motion of the object. Let's say it starts here. This is the initial position. And it ends up over here in a very small amount of time. The time that passes, let's call that delta T. Now, let's draw a bisector. Okay? So at the middle of its travel, it's at this point. And what I want to draw in here is that this represents the center of the circle along which this object is travelling or the center of the curvature of the path of the object. This distance is R as is this. And let's call this angle theta and then angle theta because this line is the bisector. Now, one of the first things we want to look at is we want to see how the velocity of the ball or the object has changed as it moves from its initial position. Let's represent the velocities of the object at the two moments in time by these blue arrows. In order for us to analyse how the velocity has changed, we have to have a coordinate system. The coordinate system that we'll use will be the radial and tangential coordinate system. The radial and tangential coordinate system moves with the particle. The radial direction always points outwards from the radius. Now, in this case, in order for us to analyse the change, we need to have a fixed coordinate system. So, what we'll do is say that our fixed coordinate system consists of what the radial and the tangential directions would be if the particle were frozen at this point. So, the frozen are the radial and tangential directions at this point here. We can now analyse the velocity vector in terms of these coordinates of these coordinates directions. Okay. This, the velocity at this point can be broken up into components along the radial and tangential directions. Now, with a bit of geometry, we can realise that this angle here must be theta because if I draw a line parallel to this one, these two are alternate. If that's theta, then this is the right angle. This must be 90 minus theta and that's theta. We can apply a similar argument over here to determine that this angle over here must be theta. We can now break up the velocity of the particle into the radial and tangential vectors as shown by the red and green arrows. Now, let's have a look at the final velocity first. The final velocity in the theta direction will be V cosine theta. In the tangential direction, that's what this sine here means. Okay. Please don't confuse this theta with that theta. This theta is just simply means it's the accepted convention for saying that this is the tangential direction. Over here, we have a positive, actually sorry, it's a negative velocity because the red arrow here is points in the opposite direction to the radial direction. So, the radial component of the velocity over here will be minus V sine theta in the radial direction. Let's do the same for the initial velocity. The initial velocity is as a radial component that is pointing outwards from the center and so that component here must be positive. And the component in this direction, in the tangential direction, is going to be V cosine theta. When we subtract these two vectors to determine what delta V is, we find that the terms in the tangential directions cancel and the radial direction we get minus to the sine theta in the radial direction. Okay. If we wish to determine the acceleration, we need to know the time between the object travel from here to there. Well, we know from math that theta in radians is equal to the arc length over the radius. If we say that the length of the black line is S, the arc length, we know that two theta, two theta, because we're going from here to there. That's an angle of two theta. It's going to equal S over R. So therefore, two theta R equals S. Now, we know the speed is constant. So, and we know speed is going to equal distance, the S over delta T. So therefore, delta T must equal S over V, which is two theta R over V. If we now divide delta V by delta T, we get the acceleration. The acceleration is going to be minus two V sine theta all over two theta R over V in the radial direction. And what we end up with is V squared sine theta over R over theta. Let's have to get the minus. Now, if theta is small, we can use the small angle approximation to simplify this equation. So, for small values of theta, sine theta approximately equals theta. So, this equation here goes to A is equal to minus V squared over R in the radial direction.