# Uniform Circular Motion

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Uniform circular motion occurs when an object moves with a constant speed and is always at a fixed distance from a point. The fact that it has to remain at the same distance while maintaining a constant speed implies that it's velocity keeps changing. Velocity is defined by speed and direction, so although an object's speed is constant, its direction changes constantly as it moves around a circle.

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## Horizontal Circular Motion

Motion of a particle in a horizontal circle would be the most basic version of *Uniform Circular Motion*, since the force of gravitation doesn't affect the particle's trajectory. Before we get into the mathematical analysis, it's worth giving a shot at understanding what happens to the particle when it starts revolving in a circle, in this case a horizontal one. What we know so far is that the particle needs to maintain the *same speed*, but at the same time be able to *change it's direction*. But this appears to be a very unlikely scenario, because a change in velocity can only be brought about when an external force is applied, and applying a force should change the speed of the particle. Or does it really have to change(in every case)? Is there a possible case where both conditions satisfy?

Let's recall what we know about forces in general, in particular about how a particle's speed is affected under the influence of one. We know that when a force is applied along the particle's direction of motion, it's speed increases, what if the same force is applied at an angle to the particle? Well in that case, only a component of the force will be acting along the particle's direction of motion, thus a part of it doesn't affect it's speed at all(in that direction).

If we observe the general relationship between the angle \(\theta\) and the change in speed of the particle, we realize that when the force \(F\), acts perpendicularly, the speed never changes. Look more carefully and, alas! We find that when applied perpendicularly, it manages to keep the speed of the particle unchanged and yet manage to change the direction of motion of the particle. Well, this is exactly what we've been looking for! A force that fits perfectly into both definitions of uniform circular motion.

## Solving Using Newton's Laws of Motion

Consider this, a particle of mass \(M\) tied to an in-extensible string at one end, while it's opposite end remains fixed. The particle is now given some horizontal velocity of magnitude \(v\), and will tend to move in a horizontal circle of fixed radius \(R\). We will now try to define the force acting on the particle, using Newton's second law, which says,

\[\vec{\textrm{F}}=\frac{\Delta\vec p}{\Delta t}=M\frac{\Delta\vec v}{\Delta t}\]

Let's assume the particle is spinning anti-clockwise around the origin. If we compare the particle's motion now to what it is a very short time later (after it's swung through a small angle \(\Delta\theta,\) we can find an equation for \(\vec a,\) the acceleration we're looking for. If the initial and final velocities are \(\vec{v_\text{i}}\) and \(\vec{v_\text{f}},\) we can write:

\[\begin{align} \vec{v_i}&=v\sin\theta\mathbf{i}+v\cos\theta\mathbf{j}\\ \vec{v_f}&=v\sin(\theta -\Delta\theta)\mathbf{i}+v\cos(\theta -\Delta\theta)\mathbf{j} \end{align}\]

Since the distance between the points deferentially small we can safely say that \(\Delta\vec v=(v_f-v_i)\). We can evaluate the expression for the differential change in velocity by subtracting the expressions we have mentioned above as follows,

\[\begin{align}
\Delta\vec v &= v(\sin(\theta -\Delta\theta)-\sin\theta )\mathbf{i}+v(\cos(\theta -\Delta\theta)-\cos\theta)\mathbf{j}\\
&=-v\left[2\sin\left(\frac{\Delta\theta}2\right)\cos\left(\theta +\frac{\Delta\theta}2\right)\mathbf{i}-2\sin\left(\frac{\Delta\theta}2\right)\sin\left(\theta +\frac{\Delta\theta}2\right)\mathbf{j}\right]\\
&=-v\Delta\theta\left[\cos\left(\theta +\frac{\Delta\theta}2\right)\mathbf{i}-\sin\left(\theta +\frac{\Delta\theta}2\right)\mathbf{j}\right]\quad\quad(\because \sin\Delta\theta\approx\Delta\theta)

\end{align}\]

And the time interval in which the particle moves by an angle \(\Delta\theta\) will just be \(\Delta t=Rd\theta /v\). So the instantaneous acceleration can be calculated for any given point,

\[\vec a=\frac{\Delta\vec v}{\Delta t}=\dfrac{-v\Delta\theta\left(\cos\left(\theta +{\Delta\theta}/2\right)\mathbf{i}-\sin\left(\theta +{\Delta\theta}/2\right)\mathbf{j}\right)}{R\Delta\theta /v}=\frac{v^2}{R}(-\hat{\mathbf{r}})\]

Where, \(\hat{\mathbf{r}}=\cos(\theta +{\Delta\theta}/2)\mathbf{i}-\sin(\theta +{\Delta\theta}/2)\mathbf{j}\) is the direction vector which points away from the center. The negative sign indicates that the acceleration vector points towards the center.

?thought experiment?

?Qualitative analysis?

## Solving By Differentiating The Position Vector

Let us try to approach this problem in a different way, consider replacing the Cartesian coordinate system by a Polar coordinate system. So we have translated the motion of the particle around the circle of radius \(r\) in such a way that we can measure it's location, just with the help of the angle it makes with the x-axis. The position vector can be represented by

\[\mathbf{r}(\theta)=r(\cos\theta\mathbf{i} +\sin\theta\mathbf{j})\]

To get the velocity of the particle, all we have to do is differentiate the position vector with respect to time,

\[\mathbf{v}(\theta)=\frac{d\mathbf{r}}{dt}=r(-\sin\theta\mathbf{i}+\cos\theta\mathbf{j})\frac{d\theta}{dt}\]

But we end up with this new \(d\theta /dt\) term, which is the rate at which the angle made with x-axis is changing. Since the particle is moving with a constant speed, it would be covering equal angles in equal intervals of time so this new term must remain constant too, we refer to this term as angular velocity and it is denoted with \(\omega\). Also, we observe the magnitude of velocity to be,

\[|\mathbf{v}(\theta)|=|\mathbf{r}|\omega\]

One other thing that we might miss out is the fact that angular velocity is a vector quantity, a question which arises is, how to represent it? Vectors of this sort have a special name, curl vectors, and they are represented based on the sense of curling or rotation. Take out your right hand and curl your fingers along the sense of rotation, your thumb finger would be pointing along the vector, which would be along some normal \(\hat{\mathbf{n}}\) to the plane. This is true by definition; although it seems strange since the vector is perpendicular to the rotation, this definition turns out to be the only way to formulate a consistent vector theory of rotational forces. A simple analysis would lead us to this obvious yet beautiful result, \(\mathbf v=\mathbf{\omega}\times\mathbf r.\)

Differentiating the velocity vector using the chain rule, we get \[\begin{align} \mathbf{a}(\theta)=\frac{d\mathbf{v}}{dt}&=-r(\cos\theta\mathbf{i}+\sin\theta\mathbf{j})\left(\frac{d\theta}{dt}\right)^2+r(-\sin\theta\mathbf{i}+\cos\theta\mathbf{j})\frac{d^2\theta}{dt^2}\\ &=-\mathbf{r}\left(\frac{d\theta}{dt}\right)^2+\mathbf{v}\frac{d^2\theta}{dt^2}\\ &=-\mathbf{r}\omega^2\\ &=-\frac{v^2}{r}\hat{\mathbf{r}} \end{align}\]

We have ignored the term containing \({d^2\theta}/{dt^2}\), because \({d\theta}/{dt}\) is constant. The term is defined as the angular acceleration, or the rate of change of angular velocity, and it's significant during non-uniform circular motion where the speed of the particle is not constant(we don't have to worry about it now).

?vectors?

?problem?

## Dynamics Of Circular Motion

The motion of the particle changes when new forces are introduced, but then again we have already seen not all forces affect the speed of the system, in particular when the force is applied along the normal to the direction of the particle's trajectory. We will now be taking a look at what happens when one such force, friction, is present. Since friction would try to pull the particle from flying away it will act along the normal to the trajectory.

It is obvious now that we need to have a "track" on which the particle needs to tread on. So let's try to take the entire horizontal setup onto a rough circular track, but giving the particle and initial velocity would not be sufficient for it to continue on the track for long, friction would cause the particle to stop after a finite time. So let the particle have an engine of some sort, with which it can maintain the constant velocity required for a *uniform* circular motion. Drawing the FBD for the given setup, we get something like this

From the information we have gathered so far, we know that the normal force \(N\) from the ground would cancel out with the weight of the object hence there would be no net vertical force on the particle. This leaves us with the frictional force which is acting towards center all the time it would be acting as the centripetal force, which we know is \(=mv^2/r,\) so we can write

\[f_s=m\frac{v^2}r\]

But \(f_s\) has a maximum value, \(f_{s\text{max}}=\mu_sN=\mu_smg,\) so every circular track would have a speed limit beyond which we would no longer be able to remain on the track, we would start to slide and move away from the track. So to drive safely, the maximum velocity

\[\begin{align} f_{s\text{max}}=\mu_smg&=m\frac{v_{\text{max}}^2}r\\ v_{\text{max}}^2&=\mu_sgr\\ v_{\text{max}}&=\sqrt{\mu_sgr} \end{align}\]

Banking a circular track can increase the speed barrier so that traversing a the same curve would be possible at higher speeds, qualitatively we can see why this hapeens because the net normal

Free Body Diagram of a particle moving on an banked road

\[\begin{align}\ N\cos\theta &=f_s\sin\theta +mg\\ f_s\cos\theta+N\sin\theta &=m\frac{v^2}{R} \end{align}\]

## Projecting Uniform Circular Motion Into Simple Harmonics

Related Article: Periodic Motion.

From the previous discussions, we had concluded that the position vector of a particle undergoing uniform circular motion can be represented in polar coordinates as follows,

\[\mathbf{r}(\theta)=r(\cos\theta\mathbf{i}+\sin\theta\mathbf{j})\]

Projecting the position vector on either of the coordinate axes gives us a sinusoidal function, which is exactly how SHM's look like, this new discovery helps us in many ways for example, finding the resultant amplitude when super-posing two different SHM's which have different phases and amplitudes. Before we get into that, let's look at the projections first

\[\begin{align} \text{Projection along x-axis: } x(t)&=r\cos(\omega t)\\ \text{Projection along y-axis: } y(t)&=r\sin(\omega t)\quad\quad\quad (\because\theta = \omega t) \end{align}\]

These equations clearly represent SHM. An important thing to note here is that the radius of the circle is equal to the amplitude of the Simple Harmonic. Differentiating the above expressions would first give us the velocity and then the acceleration functions of the Simple Harmonic, we will just do it for the y-axis here

\[\begin{align} \frac{d(y(t))}{dt}=v_y(t)=r(\omega\cos(\omega t))&=\omega r\cos(\omega t)\\ \frac{d(v_y(t))}{dt}=a_y(t)=r\omega(-\omega\sin(\omega t))&=-\omega^2r\sin(\omega t) \end{align}\]

Let's now consider two different Simple Harmonics on the x-axis, each of them having a unique Amplitude and Initial phase, but the same frequency. What happens if we superpose them? How do we work out their resultant Amplitude? These questions can be answered simply by considering Simple Harmonics as a projection of uniform circular motion, as we've seen in the animation, we can easily project the position vector of a particle undergoing uniform circular motion to a Simple Harmonic and vice-versa.

\[\begin{align} x_1&=A_1\sin(\omega t+\phi_1)\\ x_2&=A_2\sin(\omega t+\phi_2) \end{align}\]

Note that the difference in the phases have separated the two position vectors by a certain angle and they remain separated by the same angle throughout, but if the particles have different frequencies as well then, the phase difference will not remain constant which results in the resultant amplitude to vary as the time passes.

\[A_{\text{res}}=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\Delta\phi}\]

## Non-Uniform Circular Motion

This must be a separate wiki I suppose. Non Uniform Circular Motion.

Up until now, we have tried to understand the simplest type of circular motion where the speed of the particle remains constant throughout it's motion, but things change when an external force field(e.g. gravity) acts on the system, the speed will vary. To understand the dynamics of vertical circular motion, we will have to begin from scratch, since we can't define the exact position function. What we can do is analyze the different forces acting on the particle at a particular point, a free body diagram would look like this

\[\textrm T-mg\cos\theta = \frac{mv^2}{R}\]

*TENSION comes into play*

*Conserving energy to eliminate variables*
\[\frac 12mu^2=\frac 12mv^2+mgR(1-\cos\theta)\]

*Velocity required to complete a full circle*

Case when the particle isn't tied to a string but is inside a loop

## References

[1] Derivation of the area of circle. *thecuriousastronomer.wordpress.com*. Retrieved 8:30, May 28, 2018 from https://thecuriousastronomer.wordpress.com/2014/07/15/derivation-of-the-area-of-a-circle/

**Cite as:**Uniform Circular Motion.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/circular-motion-dynamics/