If it is given that component of acceleration perpendicular to the velocity of a body has a constant, non-zero magnitude, how can we mathematically prove that the trajectory of the body will be circular?

I have basic knowledge of calculus, vectors and co-ordinate geometry. I can prove that the trajectory of a body in projectile motion is parabolic. I tried to follow a similar method to get an equation of circle but couldn't succeed. I tried forming parametric equations of x and y in terms of time t (the way we proceed when proving the parabolic path of a projectile) but could not reach the conclusion.

Edit: I have edited the question in order to improve it. However I still can not share my work on this question as I had asked the question almost 6 months ago and now I can not find where I tried solving it. Sincerely sorry for that.

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    $\begingroup$ Hi Aumkaar Pranav Shukla. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Feb 9, 2018 at 18:27
  • $\begingroup$ You need a mathematical proof of the following velocity/acceleration decomposition $$ \begin{aligned} \mathbf{v} & = v \mathbf{e} \\ \mathbf{a} & = \dot{v} \mathbf{e} + \frac{v^2}{r} \mathbf{n} \end{aligned} $$ where $\mathbf{e}$ is the tangential direction and $\mathbf{n}$ the normal direction vectors. Here $r$ is the radius of curvature of the path, and $v$ and $\dot{v}$ the speed and speed rate. $\endgroup$ Feb 9, 2018 at 19:44
  • $\begingroup$ @ja72 All I can read in your comment is a mess of symbols. I think the app of SE doesn't convert those symbols into the actual mathematical terms when use them in comments. $\endgroup$ Feb 11, 2018 at 3:42
  • $\begingroup$ @AumkaarPranavShukla - I am sorry to hear that MathJax doesn't render properly for you. Physics supports math objects in posts and comments, so my guess is that there is something with the browser or the firewall that prevents proper rendering. $\endgroup$ Feb 11, 2018 at 17:48
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    $\begingroup$ Possible duplicate of Circular motion - vectors $\endgroup$ Feb 13, 2018 at 16:59

1 Answer 1


Ok let's try. I think that there are a lot of way to do so and I will try one (maybe not the faster way, but should be clean enough). To understand the motion we need just two dimensions, so we work on a plane. We take a point and define its velocity $\vec{v}=(v_x,v_y)$ and say that it is subject to a constant (in module) acceleration that is orthogonal to $\vec{v}$, say $\vec{a}=(a_x,a_y)$. Let's fix an origin so that we can write $$ v_x = v\cos\theta \qquad v_y=v\sin\theta $$ where $\theta$ is an angle with respect to some origin point. Since $\vec{a}$ must be orthogonal, than it is in the form $$ a_x = -a\sin\theta \qquad a_y=a\cos\theta $$ so that $\vec{v}\cdot\vec{a}=0$ (you can also exchange the signs, is the same). To find the trajectory we write the equations of motion of the components which are $$ \frac{d v_x}{dt} = a_x \qquad \frac{d v_y}{dt} = a_y $$ Before starting doing derivatives, we can simplify the calculation proving that under our assumptions the module of the velocity $v$ is constant. Indeed $$ \frac{dv^2}{dt}=\frac{d}{dt}\left(v_x^2+v_y^2\right) =2\left(v_x\frac{dv_x}{dt}+v_y\frac{dv_y}{dt}\right) =2\left(v_x a_x+v_y a_y\right) = 2\vec{v}\cdot\vec{a}=0 $$ so we proved that a perpendicular acceleration cannot vary the module of the velocity but only rotate it. Now we substitute the expression of the components in the equations of motion, obtaining $$ \frac{d v_x}{dt}=-v\sin\theta\frac{d\theta}{dt}=-a\sin\theta \qquad \frac{d v_y}{dt}=v\cos\theta\frac{d\theta}{dt}=a\cos\theta $$ so $$ \frac{d\theta}{dt}=\frac{a}{v} $$ where the parameter on the right is a constant. Integrating $$ \theta(t)=\frac{at}{v}+\phi $$ where it is nice to define $\omega=a/v$ and $\phi=\theta(0)$ which is just the initial angle at $t=0$. Then, putting it into the definitions of the velocities we have $$ v_x = v\cos(\omega t+\phi) \qquad v_y = v\sin(\omega t+\phi) $$ Last step: integrate the positions. The coordinates are $$ \frac{dx}{dt}=v_x \qquad \frac{dy}{dt}=v_y $$ and it is easy to integrate them obtaining $$ x=x_0+\frac{v}{\omega}\sin(\omega t+\phi) \qquad y=y_0-\frac{v}{\omega}\cos(\omega t+\phi) $$ where $x_0$ and $y_0$ can be imposed by setting initial conditions. Notice that here you can recover the "standard" cosine-sine assignments just by adding a phase $\pi/2$ to $\phi$. Now, finally, we can recognize that this is a circular motion. Indeed, summing the square of the previous equations one finds that $$ (x-x_0)^2+(y-y_0)^2=\left(\frac{v}{\omega}\right)^2 $$ which is the circle equation with center in $(x_0,y_0)$ and radius $R=|v/\omega|$.


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