Solved Problem on Circular Motion
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In an old clock of hands, find the period and frequency of rotation of the hours hand, minutes hand, and seconds hand (answers in seconds and hertz).


Problem diagram:

Figure 1

Solution

The hour's hand gives a turn in 12 h. The period will be
\[ \begin{gather} T=\left(12\;\mathrm{\cancel h}\right)\times\left(\frac{3600\;\mathrm s}{1\;\mathrm{\cancel h}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {T=43200\;\mathrm s} \end{gather} \]
the frequency as a function of period is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {f=\frac{1}{T}}\tag{I} \end{gather} \]
\[ \begin{gather} f=\frac{1}{43200\;\mathrm s} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {f\approx 0.000023=2.3\times 10^{-5}\;\mathrm{Hz}} \end{gather} \]
The minute's hand gives a turn in 60 min. The period will be
\[ \begin{gather} T=\left(60\;\mathrm{\cancel{min}}\right)\times\left(\frac{60\;\mathrm s}{1\;\mathrm{\cancel{min}}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {T=3600\;\mathrm s} \end{gather} \]
for the frequency, applying equation ()
\[ \begin{gather} f=\frac{1}{3600\;\mathrm s} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {f\approx 0.00028=2.8\times 10^{-4}\;\mathrm{Hz}} \end{gather} \]
The second's hand takes a turn in 60 s. The period will be
\[ \begin{gather} \bbox[#FFCCCC,10px] {T=60\;\mathrm s} \end{gather} \]
for the frequency, applying equation (I)
\[ \begin{gather} f=\frac{1}{60\;\mathrm s} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {f\approx 0.017=1.7\times 10^{-2}\;\mathrm{Hz}} \end{gather} \]
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