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I'm interested in a purely mathematical argument as to why when I have a constant velocity perpendicular to a constant force, that it produces uniform circular motion?

If I was given this information, how would I derive that it was uniform circular motion mathematically?

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  • $\begingroup$ I don't know about mathematically, but intuitively it makes sense. The perpendicular force changes direction; the tangential velocity keeps it moving. As long as the velocity and force are constant, every forward movement has the same direction change associated with it at every point; so the radius/curvature cannot change. $\endgroup$ – JMac May 10 '17 at 11:12
  • $\begingroup$ I don't think your question is very physical, at least I cannot think of a physical system where your conditions would apply a priori. Typically what you study in such problems is: you have an object which is subject to forces and you know its position and velocity at some point in time (e.g. at $t=0$ and you want to know its position and velocity at any future point in time ($t>0$). Here you say that the velocity (you really mean "speed") is constant, but this should rather be a result of calculation. $\endgroup$ – user1583209 May 11 '17 at 10:09
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Since speed is constant, we can write the velocity as $$\vec{v} = \begin{bmatrix}v_x(t) \\ v_y(t)\end{bmatrix} = v\begin{bmatrix}\cos(\theta_v(t)) \\ \sin(\theta_v(t))\end{bmatrix}$$ for some unknown function of time $\theta_v(t)$. This is because $$|\vec{v}| = v = \sqrt{v_x^2 + v_y^2} = const.$$

Similarly, for a constant magnitude force $$\vec{a} = \vec{F}/m = a\begin{bmatrix}\cos(\theta_a(t)) \\ \sin(\theta_a(t))\end{bmatrix}$$ for another unknown function $\theta_a(t).$

With the force perpendicular to velocity and with $\vec{v}$ and $\vec{a}$ non-zero, we have $$\vec{a}\cdot\vec{v} = av\left(\cos(\theta_a(t))\cos(\theta_v(t)) + \sin(\theta_a(t))\sin(\theta_v(t))\right) = 0$$ $$\cos(\theta_a(t))\cos(\theta_v(t)) + \sin(\theta_a(t))\sin(\theta_v(t)) = 0.$$ This expression simplifies to $$\cos(\theta_a(t) - \theta_v(t)) = 0,$$ which implies that $$\theta_a(t) - \theta_v(t) = \left(n + \frac{1}{2}\right)\pi$$ for some integer $n$. So, $$\cos(\theta_a(t)) = \cos\left(\left(n + \frac{1}{2}\right)\pi + \theta_v(t)\right) = -\sin(\theta_v(t))$$ $$\sin(\theta_a(t)) = \sin\left(\left(n + \frac{1}{2}\right)\pi + \theta_v(t)\right) = \phantom{-}\cos(\theta_v(t))$$

From the definition of acceleration, we have \begin{align} \vec{a} &= \frac{d\vec{v}}{dt} \\ a\begin{bmatrix}\cos(\theta_a(t)) \\ \sin(\theta_a(t))\end{bmatrix} &= v\frac{d\theta_v}{dt}\begin{bmatrix}-\sin(\theta_v(t)) \\ \phantom{-}\cos(\theta_v(t))\end{bmatrix}. \end{align} Using the $\cos$ and $\sin$ relations above, \begin{align} a\begin{bmatrix}-\sin(\theta_v(t)) \\ \phantom{-}\cos(\theta_v(t))\end{bmatrix} &= v\frac{d\theta_v}{dt}\begin{bmatrix}-\sin(\theta_v(t)) \\ \phantom{-}\cos(\theta_v(t))\end{bmatrix}. \end{align} Thus, $$a = v\frac{d\theta_v}{dt}$$ Since, $a$ and $v$ are both constant, $d\theta_v/dt$ must also be constant. $$\frac{d\theta_v}{dt} = \omega$$ $$\theta_v(t) = \omega t + \theta_0$$ for constant $\theta_0$ and $\omega = a/v$.

Substituting back into the original $\vec{v}$ equation results in $$\vec{v} = v\begin{bmatrix}\cos(\omega t + \theta_0) \\ \sin(\omega t + \theta_0)\end{bmatrix}.$$ Integrating yields $$\vec{x} = \frac{v}{\omega}\begin{bmatrix}\phantom{-}\sin(\omega t + \theta_0) \\ -\cos(\omega t + \theta_0)\end{bmatrix} + \begin{bmatrix}x_0 \\y_0\end{bmatrix}$$ which is circular motion about the point $(x_0, y_0)$ with a radius of $$r = \frac{v}{\omega} = \frac{v^2}{a} = \frac{mv^2}{F}.$$ This gives us the formula for centripetal force $$F = \frac{mv^2}{r}$$ and acceleration $$a = \frac{v^2}{r}.$$

As an addendum, the following assumptions are redundant:

  • Speed is constant.
  • Force is perpendicular to velocity.

Since $$\frac{d(|\vec{v}|^2)}{dt} = \frac{d(\vec{v}\cdot\vec{v})}{dt} = 2\vec{v}\cdot\frac{d\vec{v}}{dt} = 2\vec{v}\cdot\vec{a}$$ So, if speed is constant, then the acceleration vector will be zero or perpendicular to the velocity vector.

Addendum the second

This derivation is only valid in 2D. In 3D, you can add a velocity component that is perpendicular to the original velocity and the force (a constant $v_z$, for example). This results in helical motion. Charged particles in a uniform magnetic field do this. The magnetic force is $\vec{F} = q\vec{v} \times \vec{B}$, which is always perpendicular to the velocity of the particle. In fact, as long as the magnetic field is of constant magnitude, the force will be of constant magnitude, and the particle will follow a helical path around a curved line that follows the curved magnetic field lines.

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  • $\begingroup$ Comment: in 3D you have way more freedom than even that, as helical motion results from the two assumptions that $\|\vec a(t)\|=\text{const}$ and $\vec a(t)\cdot \hat z = 0$ and we can relax the latter. $\endgroup$ – CR Drost May 11 '17 at 15:49
  • $\begingroup$ @CRDrost Good point. I've added a bit to the end. $\endgroup$ – Mark H May 11 '17 at 20:11
  • $\begingroup$ Cool, but note that the link you put in does not display due to markdown formatting rules. $\endgroup$ – CR Drost May 12 '17 at 12:56
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I am not giving a general rigorous proof (so basically, I'm not answering your question), but rather, a simple, semi-intuitive illustration (so basically, I am shedding some alternative mathematical light onto this). The logic would be as follows:

  • In the case of a constant velocity perpendicular to a constant force, the force can not do any work. This can be seen as the infinitesimal amount of work done in producing a small displacement $d{\vec r}$ would read $$dW = {\vec F} \cdot d{\vec r} = {\vec F} \cdot {\vec v} \ dt = 0,$$ as ${\vec F} \cdot {\vec v} = 0$ owing to the two being perpendicular.

  • Work-Energy Theorem tells us that no work done $\Rightarrow$ no change in energy. Thus, the point particle subjected to these conditions (mutually perpendicular force and velocity, both of constant strength), would move in a manner such that its energy stays constant.

  • Since the energy expression depends on the magnitude of ${\vec v}$, and not explicitly on its direction, if only the direction of ${\vec v}$ changes, and not its magnitude, energy stays the same. Thus, the above conditions would stand respected in this case, even with the continuous motion of the point particle.

  • The problem is thus mathematically down to ** finding the locus of this moving point particle, such that its velocity remains the same in magnitude, but keeps changing direction?**

Clearly uniform circular motion is a special case where this is true. The whole of central force motion, i.e. elliptical orbits (cf. Toby Peterken's answer ) is another situation where this holds.

So clearly, uniform circular motion is not a unique solution of this problem.

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I will explain briefly but I will add a link to a more detailed explanation of how to do the vector calculus.

enter image description here

The location of a particle $\mathbf{r}$ can be described as $\mathbf{r}=r\mathbf{e_r}$. The vector $\mathbf{e_r}$ depends on $\theta$ and so this vector includes both radial and angular position.

From this we can get expressions for velocity and accelleration. I will just state them however will provide a link that explains their derivation.

Velocity $$ \mathbf{v}= \dot{r}\mathbf{e_r} + r\dot{θ}\mathbf{e_θ} $$

Acceleration: $$ \mathbf{a}= (\ddot{r} − r\dot{θ}^2)\mathbf{e_r} + (r\ddot{θ}+ 2\dot{r}\dot{θ}) \mathbf{e_θ}$$

Now lets look at circular motion, $\dot{r}=0$ and $\ddot{r}=0$ as the radius is not varying. You also said uniform circular motion so $\ddot{\theta}=0$. Let us put this in for velocity and acceleration:

Velocity $$ \mathbf{v}= r\dot{θ}\mathbf{e_θ} $$

Acceleration: $$ \mathbf{a}= − r\dot{θ}^2\mathbf{e_r} $$

It should be clear that $\mathbf{e_θ}$ is perpendicular to $\mathbf{e_r}$ and so the velocity and acceleration vectors are perpendicular.

The first few pages of https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf are much more detailed.

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    $\begingroup$ I don't think this answers the question. You assume circular motion, while that is what you should prove. $\endgroup$ – user1583209 May 10 '17 at 12:46
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Suppose the orbit of mass m is an arbitrary closed trajectory around some point $C$, in wich a mass $M$ is placed. Let us call the point at which the mass is the farthest away from $C$ call $l$ and the point at which the mass is closest to $C$, $d$

Energy conservation requires:

$$E_{{kin}_d}+\frac{MG}{d}=E_{{kin}_l}+\frac{MG}{l}$$

from which follows:

$$MG(\frac 1 d - \frac 1 l)=\frac 1 2 m{v_l}^2-\frac 1 2 m {v_d}^2$$

Subtracting the RHS from the LHS allows us to write:

$$\frac 1 d -\frac 1 l=0$$

$${v_l}^2={v_d}^2$$

So $d=l$, and $v_d=v_l$, which are the conditions for a circular motion. Off course we can take also another force than gravitational force to provide the inward directed force, but the result will stay the same (for conservative forces).

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