You have seen polar coordinates in math class in which a radius and an angle are given rather than an x and y coordinate.  Polar coordinates are really the same thing as a trigonometric unit circle that is not a unit radius, but rather has a radius r.  Given an angle [math]\theta[/math]measured counter clockwise from the x-axis, and a radius [math]r[/math], we can write the Cartesian coordinates as [br][br][center][math]\vec{r}=r\cos\theta\hat{i}+r\sin\theta\hat{j}.[/math][/center] [br]If that position vector describes the position of an object traveling in a circular path of radius r, it must change in time.  As it is, that vector doesn't describe motion because it doesn't have time-dependence.  We can easily fix that.  If the path is along a circle, then the radius r should be constant in time, but [math]\theta[/math] should change in time.  Therefore if we consider what the corresponding velocity vector should be, it is just the time derivative of position, as it always is, while understanding that in taking the derivative of the angle [math]\theta[/math] that we must employ the chain rule since [math]\theta[/math] is a function of time, or [math]\theta=\theta(t).[/math]  This leads to:[br][br][center][math]\vec{v}=-r\sin\theta\frac{d\theta}{dt}\hat{i}+r\cos\theta\frac{d\theta}{dt}\hat{j}.[/math][/center][br]Looking at this equation, take note that transcendental functions like sine and cosine do not have units associated with them, and neither do unit vectors. The units of velocity must come from the radius term and the derivative expression. The magnitude of the velocity (or speed) associated with this motion is obtained the same way it always is - as the square root of the sum of the squares of the components. If you write out that expression it looks like this:[br][br][center][math]|\vec{v}|=\sqrt{(-r\sin\theta\frac{d\theta}{dt})^2+(r\cos\theta\frac{d\theta}{dt})^2}\\[br]|\vec{v}|=\sqrt{r^2\sin^2\theta(\frac{d\theta}{dt})^2+r^2\cos^2\theta(\frac{d\theta}{dt})^2}\\[br]|\vec{v}|=\sqrt{r^2(\frac{d\theta}{dt})^2(\sin^2\theta + \cos^2\theta})\\[br]|\vec{v}|=r\frac{d\theta}{dt}.[/math][/center]

The derivative term [math]\frac{d\theta}{dt}[/math] is referred to by two different names depending on context - angular frequency and angular velocity. Mathematically angular frequency makes sense since its how quickly or frequently the angle in radians is changing. Physically if we are describing a system that is actually moving around in a circular path obviously angular velocity makes sense too. Just realize that the term does not have units of velocity on its own. It only will be a velocity after multiplying by the radius of the path.  This quantity is denoted by the Greek lower case omega (which in Greek sounds similar to our 'o'), or [math]\omega\equiv\frac{d\theta}{dt}.[/math]   [b]So taking the result from the derivation above, the speed of an object undergoing circular motion can be written [math]|\vec{v}|=r\omega.[/math] [/b] Since this velocity is constantly changing direction there is no easy way to refer to it using rectangular coordinates.  Its direction is always tangent to the path and thus it is often referred to as [math]v_{tan}[/math] for tangential. Another option is to use polar unit vectors and to write it [math]\vec{v}=r\omega \hat{\theta}.[/math] This means the velocity is directed in the direction that increases the angle theta. That fact should be clear from the animation above.

Since forces are related to acceleration via Newton's 2[sup]nd[/sup] law, we need to take another time derivative to arrive at an expression for acceleration if we wish to look at the relation between circular motion and the forces responsible for it.  Taking the derivative of the velocity while assuming the radius is constant, since we are on a circular path, gives us: [br][br][center][math][br]\vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}(-r\sin\theta\frac{d\theta}{dt}\hat{i}+r\cos\theta\frac{d\theta}{dt}\hat{j}) \\ [br]\\[br]\vec{a}=-r\cos\theta(\frac{d\theta}{dt})^2\hat{i}-r\sin\theta\frac{d^2\theta}{dt^2}\hat{i}-r\sin\theta(\frac{d\theta}{dt})^2\hat{j}+r\cos\theta\frac{d^2\theta}{dt^2}\hat{j} \\[br]\text{By regrouping terms this becomes} \\[br]\vec{a}=-r\cos\theta(\frac{d\theta}{dt})^2\hat{i}-r\sin\theta(\frac{d\theta}{dt})^2\hat{j}-r\sin\theta\frac{d^2\theta}{dt^2}\hat{i}+r\cos\theta\frac{d^2\theta}{dt^2}\hat{j}. \\[br]\text{If we inspect these pairs, we see that the first one points opposite the original position vector, or}\\[br]\vec{a}=-\vec{r}\omega^2-r\sin\theta\frac{d^2\theta}{dt^2}\hat{i}+r\cos\theta\frac{d^2\theta}{dt^2}\hat{j}. \\[br]\text{The second pair of terms looks just like } v_{tan} \text{ but contains the second time derivative of angle.} \\[br]\text{Thus we write it as } a_{tan} \text{ and define \frac{d^2\theta}{dt^2}\equiv\alpha, or angular acceleration.} \\[br]\vec{a}=-\vec{r}\omega^2+r\alpha\hat{v} \\[br]\text{Using \vec{r}=r\hat{r}, we can also write this:}\\[br]\vec{a}= -r\omega^2\hat{r}+r\alpha\hat{v}. \\[br][/math][/center]

Let's take a look at that acceleration expression and make sense of it.  Keep it mind that it arose while we considered the simple situation of an object traveling along a circular path.  What's clear from the expression is that it has two terms which are orthogonal to one another.  One points radially inward and the other points along the tangential direction parallel to the velocity.[br][br]The first term depends on a constant radius and a constant angular velocity [math]\omega[/math]. This tells us that even when the speed of an object traveling around a circular path is constant, this term will be present. This is the acceleration that is due to the rate of change of the direction of the velocity vector while the speed remains constant. Clearly the direction of this acceleration is toward the center of the path, or in the inward radial direction. This is the centripetal acceleration that we discussed in the opening section to this chapter. Any time a car rounds a turn there must be a force to push inward toward the center of the curve that gives rise to this acceleration. It is also worth noting that we can write it either in terms of [math]\omega[/math] or [math]v[/math], since we have [math]v=r\omega[/math] from above. Using a 'c' subscript for centripetal we define:[br][br][center][math]\vec{a}_{c}=-r\omega^2\hat{r}=-\frac{v^2}{r}\hat{r}.[/math][/center][br][br]The second term in the acceleration expression is directed tangent to the path, but is only present when the angular velocity is changing. The tangential acceleration is written[br][br][center][math]a_{tan}=r\alpha\hat{v}.[/math][/center][br][br]Tangential acceleration is entirely due to the rate of change of the magnitude of the velocity vector, or speed. For a car rounding a turn, it will only be non-zero if the driver is speeding up or slowing down while in a turn. Going at constant speed around a turn corresponds to zero tangential acceleration, but there will still be a non-zero centripetal acceleration.[br][br]While riding on a Ferris wheel, the tangential acceleration term would only be non-zero if the wheel is speeding up at the beginning of the ride or slowing down at the end. In such cases we have both a centripetal acceleration due to turning, and a tangential acceleration due to speeding up or slowing down. Since the two acceleration terms are orthogonal, the total acceleration vector may be expressed as[br][br][center][math]\vec{a}=\vec{a}_c+\vec{a}_{tan}, \\[br]\text{and the magnitude is just} \\[br]|\vec{a}| = \sqrt{a_c^2+a_{tan}^2}.[/math][/center] [br]

If you want an easy way to see how the acceleration must contain these two terms, another way is to expand the velocity as a product of its magnitude and direction, and do the derivative as shown below - while not forgetting the product rule for derivatives:[br][br][center][math][br]\vec{a} = \frac{d\vec{v}}{dt}=\frac{d}{dt}(v\hat{v})=\frac{dv}{dt}\hat{v}+v\frac{d\hat{v}}{dt}.[br][/math][/center][br]Just looking at this expression, it is clear that the first term is related to the rate of change of speed and is directed tangent to the velocity, and that the second term depends only on the rate of change of direction of travel. Therefore, in this view, in the context of circular motion the first term must be the tangential acceleration and the second term is the centripetal acceleration. [br][br]In order to make more sense of this expression, we need to figure out what the rate of change of the velocity unit vector in the second expression means. Let me note here that Cartesian unit vectors for a fixed coordinate system, like [math]\hat{j}[/math] , do not change in time, but obviously a velocity unit vector can change in time. Note also that [b]the rate of change of a unit vector must only depend on its change of direction since it will always have a constant magnitude of one unit[/b].[br][br]To determine the rate of change of the velocity unit vector in the second term of the acceleration expression above we can use the expression of the velocity from the first part of this section which is [math]\vec{v}=-r\sin\theta\frac{d\theta}{dt}\hat{i}+r\cos\theta\frac{d\theta}{dt}\hat{j}.[/math] Dividing out the radius and the derivative term (which as you recall is angular velocity) gives us the velocity unit vector. You can be sure that it's a unit vector since the square root of the sum of the squares of the terms is equal to one, due to the familiar trigonometry identity [math]\sin^2(x)+\cos^2(x)=1.[/math] Using that velocity unit vector we can find an expression for its rate of change as follows: [br][br][center][math]\hat{v}=-\sin\theta\hat{i}+\cos\theta\hat{j} \\[br]\text{and} \\[br]\frac{d\hat{v}}{dt}=-\cos\theta\frac{d\theta}{dt}\hat{i}-\sin\theta\frac{d\theta}{dt}\hat{j} \\[br]\frac{d\hat{v}}{dt}=-\omega\cos\theta\hat{i}-\omega\sin\theta\hat{j} \\[br]\frac{d\hat{v}}{dt}=-\omega(\cos\theta\hat{i}+\sin\theta\hat{j}) \\[br]\text{Note that the unit position vector $\hat{r}=\cos\theta\hat{i}+\sin\theta\hat{j}$ may be factored out, giving:} \\[br]\frac{d\hat{v}}{dt}=-\omega\hat{r}=-\frac{v}{r}\hat{r}[br].[/math][/center][br][br]If we plug this back in as the second term in the acceleration expression, we get the following:[br][br][center][math][br]\vec{a} =\frac{dv}{dt}\hat{v}-\frac{v^2}{r}\hat{r}.[br][/math][/center][br]This form makes understanding acceleration in circular motion very clear if you understand the expression. It tells us that we can have tangential acceleration directed parallel to velocity due to an object's changing speed, and we can separately have centripetal acceleration directed radially inward due to an object's changing direction. [br][br]Recall that the centripetal acceleration may also be written in terms of [math]\omega[/math] and that the tangential acceleration may be written in terms of angular acceleration [math]\alpha[/math]. This must be true since with constant radius we had [math]v_{tan}=r\omega[/math] and taking a time derivative of that expression gives [math]a_{tan}=r\alpha.[/math]