A dot started to move around circle with constant angular acceleration $\alpha = 0.25\frac{\text{rad}}{s^2}$. At what time tangential and perpendicular acceleration of the dot will become equivalent?

Very easy problem, but one thing is not clear to me. This is the first way how I solved it.

We know that $\omega_0=0$ because the dot has no velocity at point $t=0$. Let $t = t_1$ be the point in time at which $a_\tau=a_n$ (equivalent tangential and perpendicular acceleration). From well-known equation we know that $$a_n=\frac{v^2}R$$ where $R$ is the radius of the circle. Also $$a_\tau=\alpha R$$ and $$v=\omega R$$ Combining these equations, we get $$\begin{align}\alpha&=\omega^2\\&=\left(\omega_0+\alpha t_1\right)^2\\&=\alpha^2t_1^2\end{align}$$ which gives us $$\begin{align}t_1&=\sqrt{\frac{\alpha}{\alpha^2}}\\&=\sqrt{\frac1\alpha}\\&=\sqrt{\frac1{0.25}}s\\&=2s\end{align}$$ This is also the correct solution provided in my book. But, after solving this I tried to solve the same problem using different approach. This is my second way.

By definition, we know that $$a_\tau=\frac{dv}{dt}$$ From equation $a_\tau=a_n$ we get $$\frac{dv}{dt}=\frac{v^2}R\\\frac{dv}{v^2}=\frac{dt}R\\\int_{v_0}^v\frac{dv}{v^2}=\int_{t_0}^t\frac{dt}R\\\frac1{v_0}-\frac1v=\frac tR$$ Surprisingly, for $v_0\to0$ this equation gives $t\to\infty$. Possibly, I missed something abvious, but I still cannot figure out where is my mistake in the second solution. What I actually did wrong?

  • $\begingroup$ Hello @sbisit and welcome to Physics.SE! You mention in a comment elsewhere that you've solved your problem. Let me encourage you to write an answer to your own question rather than trying to get the question deleted. $\endgroup$
    – rob
    Oct 22, 2016 at 2:40

1 Answer 1


You can't just randomly set things equal to each other. The accelerations $a_\tau(t)$ and $a_n(t)$ are independent quantities, and the question is asking for the time $t_0$ where they happen to be equal. Instead, you set $a_\tau(t) = a_n(t)$ for all times, which doesn't make any sense.


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