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Contents Index Dark Mode Prev Up Next \(\newcommand{\vf}[1]{\mathbf{\boldsymbol{\vec{#1}}}}
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\)
Section 23.8 Velocity & Acceleration
Newton’s Laws require a knowledge of velocity and acceleration. In
Section 23.7 , we chose plane polar coordinates, so now we must deal with the problem of how to compute velocity and acceleration as time derivatives of the position vector
\(\rr=r\rhat\) in terms of the coordinates
\(r\) and
\(\phi\) and the basis vectors
\(\rhat\) and
\(\phat\text{.}\) A difficulty arises because
\(\rhat\) (and
\(\phat\) ) are not independent of position and therefore are not independent of time. This problem does not present itself in Cartesian coordinates because
\(\xhat\text{,}\) \(\yhat\text{,}\) and
\(\zhat\) are independent of position. Using the chain rule, the general velocity vector is given by:
\begin{equation}
\vv
= \frac{d\rr}{dt}
= \frac{d}{dt}(r\,\rhat)
= \frac{dr}{dt}\,\rhat + r\,\frac{d\rhat}{dt}\tag{23.8.1}
\end{equation}
To evaluate
(23.8.1) , we need the derivatives of
\(\rhat\) (and
\(\phat\) ) with respect to time.
One method for finding these time derivatives is to exploit the time independence of the Cartesian basis. From
Figure 23.11 , we see that
\(\rhat\) and
\(\phat\) are given, in terms of
\(\xhat\) and
\(\yhat\text{,}\) by
\begin{align}
\rhat\amp= \;\;\;\cos\phi\, \xhat + \sin\phi\, \yhat\notag\\
\phat\amp= -\sin\phi\, \xhat + \cos\phi\, \yhat\tag{23.8.2}
\end{align}
Figure 23.11. The polar basis vectors at the point \(P\) can be found in terms of the Cartesian basis vectors.
You should recognize this basis change as a rotation performed on the \(\xhat\text{,}\) \(\yhat\) basis.
\begin{equation}
\begin{pmatrix}
\rhat \\ \phat
\end{pmatrix}
= \begin{pmatrix}
\;\;\;\cos\phi \amp \sin\phi \\
-\sin\phi \amp \cos\phi
\end{pmatrix}
\begin{pmatrix}
\xhat \\ \yhat
\end{pmatrix}
= R(\phi) \begin{pmatrix} \xhat \\ \yhat \end{pmatrix}\tag{23.8.3}
\end{equation}
Using the chain rule, the general velocity vector is given by:
\begin{equation}
\vv
= \frac{d\rr}{dt}
= \frac{d}{dt}(r\,\rhat)
= \frac{dr}{dt}\,\rhat + r\,\frac{d\rhat}{dt}\tag{23.8.4}
\end{equation}
To evaluate
(23.8.1) , we need the derivatives of
\(\rhat\) (and
\(\phat\) ) with respect to time. Using the definitions in
(23.8.3) above, we obtain:
\begin{align}
\frac{d\rhat}{dt}
\amp= \frac{d}{dt} (\cos\phi\,\xhat + \sin\phi\,\yhat)\tag{23.8.5}\\
\amp = -\sin\phi\frac{d\phi}{dt}\,\xhat + \cos\phi\frac{d\phi}{dt}\,\yhat \notag\\
\amp = \frac{d\phi}{dt}\,\phat\notag\\
\frac{d\phat}{dt}
\amp= \frac{d}{dt} (-\sin\phi\,\xhat + \cos\phi\,\yhat)\notag\\
\amp = -\cos\phi\frac{d\phi}{dt}\,\xhat - \sin\phi\frac{d\phi}{dt}\,\yhat\notag\\
\amp = -\frac{d\phi}{dt}\,\rhat\notag
\end{align}
where we have used the Cartesian expressions
(23.8.2) for the polar basis vectors in the last equalities of each calculation.
Notice that we have used the convenient notation of putting a dot over a symbol to denote time derivative.
Combining these expressions with equation
(23.8.1) gives:
\begin{equation}
\vv =
\dot{r}\,\rhat+ r\dot\phi\,\phat\tag{23.8.6}
\end{equation}
Alternatively, in
Section 23.9 , we use only reasoning about the orthonormality of
\(\rhat\) and
\(\phat\) to find the same result.
Activity 23.3 . Acceleration in Polar Coordinates.
Taking another derivative of
(23.8.6) with respect to time, show that the acceleration is given by
\begin{align}
\aa
\amp = \dot{\vv}\notag\\
\amp = \left( \ddot{r} - r\dot\phi^2 \right) \rhat
+ \left( r\ddot\phi + 2\dot{r}\dot\phi \right) \phat\tag{23.8.7}
\end{align}
