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Here is a picture illustrating my question:

https://upload.wikimedia.org/wikipedia/commons/7/71/Uniform_circular_motion.svg

I must not understand what acceleration is, because I don't understand how the acceleration of the rotating body can be towards the center of rotation when the velocity is not.

Do we simply say the acceleration is towards the center of rotation because that is the direction of centripetal force? When in reality there is no acceleration towards the center of rotation?

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    $\begingroup$ If the acceleration pointed in the same direction as the velocity then the direction of the velocity wouldn't change. $\endgroup$ Jul 9 at 21:02
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As you surmise in your comment upon the answer given by levitopher, you are confused because you associate acceleration with a change of speed. You should instead consider it to be the cause of a change in velocity, which can be a change in the direction of motion at a constant speed. Circular motion arises when a force is applied that is always normal to the velocity of the moving body.

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The definition of (average) acceleration is

$$\vec{a}=\frac{\Delta \vec{v}}{dt}$$

Where $\Delta \vec{v}=\vec{v}_f-\vec{v}_i$ (final minus initial). The important thing you're missing is that $\vec{v}$ has a vector sign over it - the direction matters, and if the direction changes then the acceleration is non-zero. Let's try that calculation in your specific picture.

enter image description here

I've added a coordinate system to your figure, and I'm calculating the acceleration between the velocity at 3 o'clock and 12 o'clock positions (that's why I placed the acceleration vector at about 1:30). The components of the acceleration vector are

$$a_x=-v_f/\Delta t, \qquad a_y=-v_i/\Delta t$$

(of course, as magnitudes $|v_f|=|v_i|$ so the acceleration is at a 45 degree angle, down to the third quadrant.)

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    $\begingroup$ Since the speed of the rotating body is constant, the change in velocity is due exclusively to the body constantly changing direction as it moves along a curved path, right? And we only say there is acceleration since acceleration is a measure of the change of velocity over time and a change in direction is a change in velocity. Typically, I think acceleration is thought of as a change in speed not a change in the direction of a moving object, and that's why this is confusing to some degree. $\endgroup$
    – 228
    Jul 9 at 22:05
  • $\begingroup$ @228 Yes, you've got it. The acceleration tells us about the change in the velocity vector, including changes in both magnitude and direction. In uniform circular motion, the magnitude of the velocity (the speed) stays constant, but the direction of the velocity is constantly changing, and the acceleration vector describes this change in direction. This is just something you have to get used to when working with motion in more than one dimension. $\endgroup$
    – d_b
    Jul 9 at 22:43
  • $\begingroup$ @228: Just to emphasize, we are not just "saying there is acceleration", there is acceleration in every sense. For example, since there is an acceleration, there is a net force, which is directed towards the center of the circle. If this was a ball on a rope, the net force would be the tension, and we would call the net force "the centripetal force". $\endgroup$
    – levitopher
    Jul 11 at 17:33
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Think about how the position changes despite the fact that the distance remains constant because position depends on direction to the origin and not just the distance to the origin. Similarly velocity can change even when speed is constant as velocity depends on the direction of motion, not just how fast something is moving.

An equation for a unit circle is

$$x=\cos t$$ $$y=\sin t$$

Velocity is the time derivative of position and acceleration is the time derivative of velocity and the derivative of $\cos t$ is $-\sin t$ while the derivative of $\sin t$ is $\cos t$

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  • $\begingroup$ Each of these equations should include an angular frequency. If you use these derivatives to find the components of the velocity and acceleration vectors, you find that the velocity is perpendicular to the radius, and the acceleration is parallel to the radius (but in the opposite direction). $\endgroup$
    – R.W. Bird
    Jul 11 at 17:45
  • $\begingroup$ @R.W.Bird Well velocity is perpendicular to the position, and acceleration is in the opposite direction in the case of circular motion with a constant speed, so I'm not sure what the problem is here. $\endgroup$ Jul 11 at 21:38
  • $\begingroup$ No problem. Including the angular velocity expresses your arguments as angles. Then I was pointing out that your functions do give an answer to the question. $\endgroup$
    – R.W. Bird
    Jul 12 at 12:51
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Velocity is a vector, i.e. it has both magnitude (speed) and direction. Changing magnitude (speed) is a result of the component of the acceleration in the direction of velocity, and changing the direction of velocity is a result of the component of the accelaration perpendicular to the velocity. In a uniform circular motion, the speed is constant, which means the component of acceleration in the direction of the velocity (tangent to the circle) is zero, but the direction of the velocity is changing, so the accelaration is perpendicular to the circle (=radial).

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  • $\begingroup$ What if both the speed and direction were changing for the rotating object? Then what direction do we say the acceleration is in? $\endgroup$
    – 228
    Jul 10 at 18:05
  • $\begingroup$ @228 Just treat it like any other vector. If it has components in both the radial and tangential directions then the resultant vector will have a direction between radial and tangential. $\endgroup$ Jul 10 at 18:27

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