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{\bf Question}

Estimate the centripetal acceleration  of
\begin{description}
\item[(a)] an item of clothing in a tumble dryer;
\item[(b)] the rim of a car tyre going at 100kmh$^{-1}$
\item[(c)] the earth going around the sun (assume it has a circular
orbit of radius $1.5 \times 10^{11}$m);
\item[(d)] a child on a roundabout.
\end{description}

Make clear your assumptions

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{\bf Answer}

\begin{description}
\item[(a)]
Tumble dryer rotates about once per second, i.e. $\dot \theta =
2\pi\ {\rm rad \ s}^{-1}$

The radius of the drum $\approx 0.3$m

Therefore the centripetal acceleration $ = 0.3 \times(2\pi)^2
\approx 9.2{\rm ms}^{-1}$



\item[(b)]
$ { }$

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\begin{eqnarray*} 100{\rm kmh}^{-1} & = & \frac{10^5}{3.6 \times
10^3} = \frac{10^2}{3.6}{\rm ms}^{-1} \\ {\rm as\ } v & = & r \dot
\theta \hspace{.1in} r \approx 0.2{\rm m} \\ \dot \theta & = &
\frac{10^2}{3.5 \times 0.2} \approx 138.9 {\rm rad\ s}^{-1}
\end{eqnarray*}

Centripetal acceleration $= r \dot \theta ^2 = 0.2 \times 138.9^2
= 3658{\rm ms}^{-1}$

\item[(c)]
\begin{eqnarray*} {\rm Angular\ velocity\ } & = & \frac{2\pi}{365
\times 24 \times 3600}  \\ & = & 1.9 \times 10^{-7}\, {\rm rad\
s}^{-1}
\end{eqnarray*}

Centripetal acceleration = $ 1.5 \times 10^{11} \times 1.9^2
\times 10 ^{-14}  =  5.415 \times 10^{-3}{\rm ms}^{-1}$


\item[(d)]
Roundabout rotates $\approx$ once per second.  Therefore angular
velocity $\approx \, 2\pi \approx 6 {\rm rad\ s}^{-1}$

Radius $\approx$ 1m.  Therefore centripetal acceleration is
36ms$^{-1}$

\end{description}



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