 Previously we looked at uniform circular motion and the velocity of objects undergoing uniform circular motion. In this video we'll derive the acceleration of objects undergoing uniform circular motion, which is also known as the centripetal acceleration. This is a pretty involved derivation. Don't worry if it seems a bit complicated as you wouldn't be expected to invent it on your own without having seen it. So before we get started, how can an object undergoing uniform circular motion be accelerating if it's traveling at a constant speed? What we need to remember is that velocity is a vector. While it may be traveling at a constant speed, this vector is constantly changing direction for the object in question. Since the velocity is therefore changing, the object is accelerating. So let's find the acceleration of the object. Well, firstly we know that the acceleration a is equal to the change in velocity v divided by the change in time t. So let's try to work out what each one of these is separately. Let's start by working out the change in velocity delta v. Now you may be able to see that the change in velocity is dependent on where along the circle the object is. For example, if we waste for the object to undergo a complete revolution and looked at it in the exact same spot, then the change in velocity will be zero because it will be the exact same velocity vector. So what we really want to do is we want to find the instantaneous change in the velocity to find the instantaneous acceleration. So we're going to take a very, very tiny slice of the object's motion, infinitesimally small, to find the change in velocity at this point over the time it's taken to undergo this change. But to make our lives easier, we'll draw it far away, and we'll just understand this change is actually meant to be very small. Now, the velocity at the first point along the object's path will call v1, and the velocity at the second point along the object's path will call v2. So we're off to delta v, the change in velocity. This is the final velocity, v2, minus the initial velocity, v1. So let's draw it out. We now have a triangle three sides, v1, v2, and delta v. To get an expression with delta v, we'll need to work out some information about the angles and the sides of the triangle v1, v2, and delta v. We'll start by calling this angle theta. And remember that theta is really small as we work this out. Now, we're going to use some geometry to get the angle between v2 and minus v1. If we draw dashed lines out from our original v1 and v2 positions, we know that each of these dashed lines is perpendicular to the radial lines of the circle. We can find the angle in between our dashed lines by remembering that the sum of all angles in a quadrangle is 360 degrees. We know that two of these angles are 90 degrees, and the other one is theta, so the upper angle must be 180 degrees minus theta. Now, we also know that minus v1 in our triangle is parallel to the dashed line extension of v1, our original vector. Since angles on a line must sum to 180 degrees, we know that this angle here must be theta. And since we have two parallel lines, the angle between v2 and minus v1 will also be theta. This is known as the Alternate Angles Rule. So now we're going to zoom in on this triangle to work out an expression for theta. What else do we know about this triangle? Remember, we're considering uniform circular motion, so the speed is constant. This means that the length of v2 is equal to the length of v1, as these are both the magnitude of our velocity vector, and these are both equal to the speed v. As both of these sides are equal, we also know that our triangle is an isosceles triangle. Isosceles triangles can really help us out in problems, because if we cut it in half, we know that we'll get two right angle triangles. This means we can use Sokotoa, our trigonometry rule. So, if we consider our diagram, v is the hypotenuse, half of delta v is the opposite side, and the angle is half theta. So we can write sin of theta on 2 is equal to delta v on 2 divided by v. We can actually simplify this expression down even further by using the small angle approximation. Remember that theta is really, really tiny and infinitesimally small. The small angle approximation allows us to approximate sin of an angle as the angle itself. So sin of theta on 2 will be approximately equal to theta on 2. And for the limit case, which is the case we're considering, as we approach an infinitesimally small angle, sin theta on 2 will actually become theta on 2 exactly. So plugging this back in, we find theta on 2 is equal to delta v on 2 divided by v. Rearranging, we get that the change in velocity delta v is equal to v times theta. Therefore, we have the first ingredient for our acceleration. The change in velocity is equal to v theta. Now, let's move on to the second ingredient on our acceleration, which is the change in time. Since we know that the speed of our uniform circular motion is v, we can work out the change in time delta t from the distance it travels. So we'll redraw a diagram with the object at two points along the circle. Now, as we know from circle geometry, the length that the object travels is equal to the radius of the circle times theta, the angle in radians over which the object is traveled. Therefore, we can find that the distance the object has traveled is r times theta. We also know that the speed the object is traveling at is equal to v, the speed. We can now plug this back into our change in time to give us the change in time is equal to r theta on v. We now have expressions for both delta v and delta t, so we're almost there. So the acceleration is equal to the change in velocity over the change in time, and we're looking at an instantaneous moment in time. If we plug in our values, we find that our acceleration is equal to v theta divided by r theta divided by v. And if you cancel out the theaters and move the v up to the top, we get that the acceleration a is equal to the velocity v squared divided by r, the radius. We have now found the magnitude of the acceleration of an object undergoing uniform circular motion, also known as the centripetal acceleration. But we know that acceleration is a vector, so what about the direction? Well, the key factor to remember here is that the speed is constant. Now this means that there can't be any component of acceleration that is in the same direction as the velocity, otherwise the speed would change. We know from our previous velocity derivation that the velocity always points tangentially to the circle. This must mean that all of the acceleration must be perpendicular to the tangent, which is in the radial direction. The final bit we need to do is to figure out whether this acceleration is pointing radially inwards or radially outwards. Now we know that the acceleration must be in the same direction as delta v, our change in velocity, so if we draw out a triangle with v1, v2, and delta v, it's clear that delta v is pointing in towards the center of the circle and not away from the center of the circle. This means that the acceleration must point radially inwards. And there we have it. The acceleration of an object undergoing uniform circular motion is v squared on r and it points radially inwards. This is called the centripetal acceleration.