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[SOLVED] Tangential and normal acceleration of a particle moving in a plane curve in the cartesian coordinates

Dhamnekar Winod

Active member
Nov 17, 2018
163
How did the author compute the highlighted term 2 from the highlighted term 1 in the following answer to the given question?

If $\rho =\frac{d\psi}{ds}$, then the term 2 should be $\upsilon^2 \frac{d\hat{T}}{d\psi}\rho$, but instead, it was written $\frac{\upsilon^2}{\rho}\frac{d\hat{T}}{d\psi}$

How is that computed? How to compute radius of curvature($\kappa$) if $\frac{d\hat{T}}{ds}= \kappa\hat{N}$

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topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,274
It's a typo. Look at the units real quick. s has units of length and \(\displaystyle \psi\) in is radians (or so I suppose, to make the derivatives right), which means \(\displaystyle \psi\) is essentially unitless. So \(\displaystyle \dfrac{d \psi }{ds}\) has units of 1 / length. Thus it's reasonable to suppose that \(\displaystyle \dfrac{d \psi }{ds} = \dfrac{1}{ \rho }\). You can look up the formula online... I checked it.

-Dan
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
To add to topsquark 's answer, the radius of curvature is the radius of the circle that "fits" the curve.
If we traverse an angle $d\psi$ on a circle with radius $\rho$, we traverse an arc length of $ds=\rho\,d\psi$.
In other words, by definition we have $\rho = \frac{ds}{d\psi}$, which we can also write as $\frac{d\psi}{ds}=\frac 1\rho$ by the inverse function theorem.

This is also what the actual equation (4) and subsequent equations show. They just misquoted the definition of radius of curvature, but they did apply the real definition correctly.
 
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