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The particle P moves along a space curve. At one instant it has velocity $v = (4i-2j-k)$ $m/s$. The magnitude of the acceleration is 8 $m/s^2$. The angle between the acceleration and the velocity vector is 20 degrees, so one can calculate that the acceleration in the direction of the velocity is 7.52.

How can I calculate the radius of curvature from this information? I am clueless... no formulas have been introduced, as I have seen there are throughout the Internet. The chapter that I'm working on is about three dimensional coordinate systems.

One of my attempts has been to try to imagine an infinitesimal change in velocity, v = r$\theta$. This implies $\frac{dv}{dt} = r\frac{d\theta}{dt}$. Could I perhaps know somehow what $\frac{d\theta}{dt}$ is?

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up vote 2 down vote accepted

Radius of curvature is governed by $a = v^2/r$. The radius of curvature thus calculated is good at that instant only, since 'v' will continue to increase; and, if 'a' remains constant, change 'r'.

The 'a' in the equation is the component of total acceleration which is normal to the velocity vector, or $sin(20^o)(8m/s^2)$

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The speed vector you gave is constant so there can't be any acceleration. However, for $|\textbf{v}|=u=const$, as Vintage said, $\textbf{a}=\frac{u^2}{R}\textbf{N}$, where $\textbf{N}$ is the unit vector normal to the curve. So $|\textbf{a}|=\frac{u^2}{R}|\textbf{N}|\Leftrightarrow R=\frac{u^2}{|\textbf{a}|}$.

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FYI - The formula for decomposing the acceleration vector is $ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{r} \vec{n} $ where $v$ is the speed, $\dot{v}$ the rate of change of speed (acceleration along the trajectory), the vectors $\vec{e}$ and $\vec{n}$ are along the trajectory and normal to it and $r$ is the curvature.

Look at this article for all the math the accompany this (Fenet Frames), or just look up Acceleration in wikipedia.

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