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I have researched angular acceleration and circular motion on google, but haven't found what I am looking for.

I hope you can help me find more information about the problem below, with particular emphasis upon the following:

(i) Closed formula for position. (ii) Derivation of closed formula and also velocity.

Problem: Suppose we take a point, and rather than have the point move with uniform speed (Always in the same direction), it's speed gets faster and faster: more precisely, the magnitude of its acceleration is a constant, and the point continues to rotate in the same direction.

What kind of mathematical model can depict this? I know that circular motion can be parametrised as (cost(t),sin(t)) on the plane though obviously that is with uniform speed and period two pi.

Thanks.

user227093
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1 Answers1

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Here's one approach: you get the equations for the evolution of $r$ and $\theta$:

$$r(t)=r_0 \\ \frac{d^2 \theta}{dt^2} = a(t) \\ \frac{d \theta}{dt}(0) = v_0 \\ \theta(0)=\theta_0$$

where $r_0$, $\theta_0$, and $v_0$ are numbers and $a(t)$ is a specified function describing the angular acceleration. Then you solve the $\theta$ equations by integrating twice:

\begin{align*} \theta(t) & = \theta_0 + \int_0^t v(s) ds \\ & = \theta_0 + \int_0^t \left ( v_0 + \int_0^s a(u) du \right ) ds \\ & = \theta_0 + v_0 t + \int_0^t \int_0^s a(u) du ds \end{align*}

Plugging this into $x=r\cos(\theta),y=r\sin(\theta)$ you get

$$x(t)= r_0 \cos \left ( \theta_0 + v_0 t + \int_0^t \int_0^s a(u) du ds \right )$$

and similar for $y$. Now you just have some integrations to do, which depend on the choice of the function $a$. You specifically asked about the case where $a$ is just equal to some constant $c$. In this case $\int_0^t \int_0^s c du ds = \frac{c t^2}{2}$, so you get

$$x(t) = r_0 \cos \left ( \theta_0 + v_0 t + \frac{c t^2}{2} \right )$$

and similar for $y$.

Ian
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