# 2.2.4.2: Circular Motion


Consider an ''object'' -e.g., a stone or a ball -tied to a string, which is being swung in circles in horizontal plane (e.g., by holding the other end of the string above one's head -probably, everyone once did such an ''experiment''). The object moves along a circular path. The string is tensioned, so, apparently, the object pulls it ''outwards''. Now, recall the Third Newton's Law of Dynamics -it says that if Object A exerts a force $$\vec{F}_{A \rightarrow B}$$ on Object B (called ''action force''), then Object B exerts a ''reaction force'' $$\vec{F}_{B \rightarrow A }$$ on Object A, which is of the same magnitude and of opposite direction as the action force -i.e.,

$\vec{F}_{B \rightarrow A} =-\vec{F}_{A \rightarrow B}$

So, if the stone is tensioning the string "outwards'', the string is pulling the stone ''inwards'', always towards the center of the circular path.

The First Newton's Law states that if there is no net force acting on an object, then it either rests or moves along a straight line with constant velocity -meaning that its acceleration is zero, $$\vec{a} = 0$$. The Second Law states that if there is a non-zero force, the object experiences acceleration that is proportional to the force, and inversely proportional to the object's mass:

$\vec{a} = \dfrac{\vec{F}}{m}.$

The ''stone on the string'', as we concluded above, is always pulled towards its circulat motion center by the force of the tensioned string. So, as follows from the above equation, it must experience an acceleration always directed towards the center. Is there indeed such an acceleration in circular motion? Let's examine such motion in closer detail in mathematical terms. Suppose that a small object -or, even better, a ''point-like object'' -of mass $$m$$ is attached to a string of length $$R$$, and the other end of the string is fixed at the origin of a Cartesian system with axes $$X$$ and $$Y$$. Let the object rotate in the $$XY$$ plane. If $$T$$ is the time it takes the object to make one full revolution, its speed along the circular path is $$v = 2\pi R/T$$.

Let's choose the instant at which the object intersects the $$X$$ axis as $$t=0$$. Then, at any time $$t$$ the angle $$\varphi$$ (expressed in radians) the string string makes with the $$X$$ axis is

$\varphi = \dfrac{2\pi}{T} t$

The stretched string in the experiment can be thought of as a vector $$\vec{R}$$ connecting the origin of the $$XY$$ coordinate frame with the circulating object (usually, it's referred to as the ''radius vector''). The projections of $$\vec{R}$$ on the $$X$$ and $$Y$$ axes -call them $$R_x$$ and $$R_y$$, respectively -are:

$R_x =R\cos\varphi = R\cos \left (\dfrac{2\pi}{T}t\right )\;\;\;{\rm and}\;\;\;R_y = R\sin\varphi = R\sin \left (\dfrac{2\pi}{T}t\right )$

The expression $$(2\pi/T)$$ is commonly referred to as the ''angular velocity'' and is denoted as $$\omega$$:

$\omega = \dfrac{2\pi}{T}$

Using the $$\omega$$ symbol, the two projections can be written in a more compact form, $$R_x =R\cos \omega t$$ and $$R_y = R\sin\omega t$$. These two expressions are the Cartesian components of the the vector $$\vec{R}$$. In vector algebra -in which one can perform various mathematical operations on vectors -a vector can be written as a list of its components, separated by commas, in parentheses\footnote{If $$\vec{R}$$ were a vector in a three-dimensional space, it would have a three components corresponding to projections on three Cartesian axes, $$X$$, $$Y$$, and $$Z$$, and it would be written then as a three-element list: $$\vec{R} = (R_x,\; R_y,\; R_z)$$.}

$\vec{R}(t) = (R_x,\;R_y) = (R\cos\omega t, R\sin \omega t)=R(\cos\omega t, \sin \omega t)$

(constant factors can be factored out). Note that the left-side term is written as $$\vec{R}(t)$$, because it should be kept in mind that this vector is a function of time. Now, we can calculate the velocity of the object, taking advantage of the fact that the velocity vector is a time derivative of the time-dependent position vector:

$\vec{v} = \dfrac{d}{dt}\vec{R}(t)$

For differentiating a vector, one uses a simple receipt -each component is individually differentiated, so we get:

$\dfrac{d}{dt}\vec{R}(t) = \left(\dfrac{d}{dt}R_x(t),\; \dfrac{d}{dt}R_y(t)\right ),$
$\dfrac{d}{dt}R_x(t)= \dfrac{d}{dt}R\cos\omega t = R\omega(-\sin \omega t) =-R\omega\sin\omega t,$
$\dfrac{d}{dt}R_y(t)= \dfrac{d}{dt}R\sin\omega t = R\omega(\cos \omega t) =R\omega\cos\omega t,$

so, finally:

$\vec{v} = (-R\omega\sin\omega t,\;R\omega\cos\omega t) = R\omega(-\sin\omega t,\;\cos\omega t).$

Now, in a similar fashion, we can calculate the acceleration experienced by the circulating object -by definition, acceleration is a time derivative of the velocity:

$\vec{a} = \dfrac{d}{dt}\vec{v}(t).$

Using an analog procedure as the one done step-by-step when differentiating the $$\vec{R}(t)$$ vector, we obtain:

$\vec{a}=\dfrac{d}{dt}\vec{v}(t)=R\omega\dfrac{d}{dt}(-\sin\omega t,\;\cos\omega t)= \omega^2R(-\cos\omega t,\; -\sin\omega t),$
which, taking into account the Eq. 2.24, can be rewritten in a final compact form:

$\vec{a}(t) = \omega^2[-\vec{R}(t)].$

Note the minus sign -the result means that the object's acceleration vector is exactly parallel to the radius vector $$\vec{R}$$, but pointing in the opposite direction -i.e., towards the center of the object's circular trajectory. It's the famous centripetal acceleration.2

We have obtained the velocity and the acceleration vectors, but we haven't yet mention their magnitudes. In general, the magnitude of a ''generic'' vector, call it $$\vec{G}$$, is a square root of the sum of the squares of its Cartesian components, and the symbols commonly used for it are $$|\vec{G}|$$ (also called the vector's ''modulus''), or simply the letter symbol alone, without the tiny arrow at the top. So, for a two-dimensional vector:

$G \equiv |\vec{G}| = \sqrt{G_x^2 + G_y^2}$

where the $$\equiv$$ symbol means ''equivalent to'' -and for a three-dimensional one:

$G \equiv |\vec{G}| = \sqrt{G_x^2 + G_y^2 + G_z^2}$

The radius vector $$\vec{R}$$ used in our considerations is simply a sector a connecting two points on the $$XY$$ plane. Its magnitude is, of course, the same as its length:

$\vec{R} = \sqrt{R_x^2 + R_y^2} =\sqrt{(R\cos \varphi)^2+(R\sin \varphi)^2} = R\sqrt{\cos^2 \varphi+\sin^2 \varphi} = R$

By the same token, one can find the magnitude of the velocity vector given by the Eq. 2.25:

$v = |\vec{v}| = \omega R \sqrt{(-\sin \omega t)^2+ (\cos\omega t)^2} = \omega R \sqrt{\sin^2\omega t + \cos^2\omega t} = \omega R .$

In combination with the Eq. 2.23 it yields:

$v = \omega R = \dfrac{2\pi R}{T}$

which is the expected result, considering that over a period of $$T$$ the object travels the length of the circle, $$2\pi R$$. Usually, $$v$$
is called the ''linear velocity'', to distinguish it from the angular velocity $$\omega$$, or ''tangent velocity", because the $$\vec{v}$$ vector is tangent to the circular trajectory.

Finally, in a similar way one can find that the magnitude of the centripetal acceleration be equal to $$a = \omega^2 R$$. Then, by multiplying this expression by another expression, which is usually called "a well-chosen unity" -namely, by $$1=R/R$$, and by combining such product with the Eq. 2.27, we obtain an equivalent expression for $$a$$ in terms of the linear velocity:

$a = \omega^2 R = \omega^2 R\dfrac{R}{R} = \dfrac{(\omega R)^2}{R} = \dfrac{v^2}{R}.$

The next step is to consider the \textbf{dynamics} of circular motion, i.e., the forces involved. For this purpose, we need to recall the Second Newton's Law of Dynamics, the mathematical form of which was given earlier as the Eq. 2.17 -let's look at it again:

$\vec{F} = m\vec{a}$

What this equation states, can be expressed by words as follows: ''If an object of mass $$m$$ is experiencing an acceleration $$\vec{a}$$, there must be a force $$\vec{F}$$ responsible for it, equal to $$\vec{a}$$ multiplied by $$m$$.'' Hence, in the circular motion considered, there must be a force acting on the object -considering the Eq. 2.26, we obtain:

$\vec{F} = m \vec{a} = m \omega^{2}\left ( -R \right ) \text{ and }\vec{F} = m \omega^{2} R$

The first of the above equations tells us that at any moment there is a force acting on the object, parallel to the radius-vector $$\vec{r}$$, and pointing towards the motion center. It can be called a \textbf{\textbf{centripetal force}}. Its origin can be easily identified -it's the force with which the stretched string pulls the object towards the motion center.

From the Newton's Third Law of Dynamics it follows that if the string pulls the object inwards, there must be a reaction force the same magnitude and opposite direction, with which the object pulls the string outwards. It is called the \textbf{centrifugal force}.

Students (and not only them) often confuse the centripetal force with the centrifugal force. There is much material related to that confusion in the Web, with many illustrative examples, and it is worth reading it -one can access many such sites simply by typing ''centrifugal + centripetal + confusion" in one of the many available search engines. For instance, one can find instructive definitions in the ''American Heritage Dictionary of the English Language''-namely:
Centripetal force: The component of force acting on a body in curvilinear motion that is directed toward the center of curvature or axis of rotation, and: Centrifugal force: The apparent force, equal and opposite to the centripetal force, drawing a rotating body away from the center of rotation, caused by the inertia of the body. These two definitions make it clear that only the centripetal force is a ''real force'' -i.e., the only force an observer placed in a non-rotating frame will see; whereas the centrifugal force is an apparent force which only an observer placed in the rotating frame would register.

In other words, the only force needed to make a body traveling along a circular route is the centripetal force. Telling the truth, the stone on a string is not the best example in support of that one can think of -because the string not only ''transmits'' the centripetal force, but it also partially limits the stone's freedom of motion, i.e., it acts as a \textbf{ constraint}.
The centrifugal force emerges from the interaction between the stone and the string.

Much more convincing is to consider a fully unconstrained motion -a good example is the motion of a body in a gravitational field, e.g., of a small artificial satellite of Earth. The only force acting on the satellite is the Earth's gravity, always pointing towards the Earth center. It's the centripetal force. So, does it mean that there is no centrifugal force? Not exactly: there is no such force anywhere nearby the satellite. But the Newton's Third Law is merciless, there always must be a reaction to any force. So, there is a reaction to the said centripetal force -it's the force the satellite exerts on Earth.

1. To get convinced that the equation is true, please check: if, for instance, the object makes one quarter of a full turn, then $$\varphi = 90^{\circ} = \pi/2$$, and $$t = \dfrac{1}{4} T$$. Now, put the latter into the above equation and you get, indeed, that $$\varphi = (2\pi/T)\times(T/4) = \pi/2$$; you may make a similar check for any other fraction of a full turn.