Suppose instead of looking at the rotation of an entire merrygoround, we examine only the change in the position of one rider on its surface, shown by the orange dot. For this initial discussion, we are going to assume that the merrygoround is rotating at a constant rate so that the rider moves through a circular path, or linear circumference, at a constant speed. Please be conscious of the fact that the rider's velocity is not constant since the direction of her motion is constantly changing as shown in the second diagram.
As long as the rider does not move along the merrygoround's surface, the magnitude of the rider's velocity (that is, her speed) equals where is the merrygoround's constant angular velocity and r is the rider's radial distance from the center of the merrygoround.
Although the merrygoround has no angular acceleration, the rider is experiencing a centripetal acceleration towards the center of the circle, or the axis of rotation. This type of acceleration acts parallel to the radius of the circle and is often referred to as radial acceleration.
As with linear acceleration, centripetal acceleration also points in the direction of the change in velocity. Vectorially, this is confirmed in the diagram to the left where you see the vector pointing to the center of the circle. Remember that and that subtracting a vector just means that you add the same magnitude vector pointing in 180º the opposite direction.
This type of acceleration is called uniform centripetal acceleration since the object's speed is not changing, just its direction is changing at a uniform rate based on the merrygoround's angular velocity.
If the force causing this centripetal (or radial  "along the radius") acceleration were to be removed, the object would "fly off at a tangent" and no longer move in a circular path.
As shown in the previous lesson, an object's centripetal acceleration is calculated with the formula . Since we can rewrite this equation as .
If asked to calculate the rider's total linear acceleration, you would also need to examine any tangential acceleration present, . In our case, the angular acceleration equals zero since we are demanding that the merrygoround rotate at a constant angular velocity. Since and the rider's net acceleration just equals the centerseeking acceleration along the radius, or .
