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In my physics book "University Physics", there is a chapter on relating linear and angular kinematics.

I understand the parts where it shows $v = r\omega$ and $a_{\text{tan}} = r\alpha$.

However in the part where they show $a_{\text{rad}} = r\omega^2$, they use the formula $a_{\text{rad}} = \frac{v^2}{r}$ which had previously been shown for when $v$ is constant. It mentions that it is also true when $v$ is not constant, but does not provide proof.

Is this hard to prove, or is it something I should easily be able to see? Can anyone help make it clear?


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up vote 1 down vote accepted

You can prove it with a little trick. For a particle on a circle of radius $r$ we have $\vec{x}(t)\cdot\vec{x}(t)=r^2$ at each point in time. Differentiating with respect to time we get $\vec{x}\cdot\vec{v}=0$. Differentiating again we get $v^2+\vec{x}\cdot\vec{a}=0$. But $\vec{x}\cdot\vec{a} = - r a_{\mathrm{rad}}$, and the relation $a_{\mathrm{rad}} = v^2/r$ follows.

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Thanks! That was very helpful. – zrbecker Aug 7 '12 at 22:07

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