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Calculate the angular speed of the minute hand of a clock and second hand of a clock​

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The second-hand goes through  $2 \pi$ radians in 1 min,

so $\omega=\pi / 30 \mathrm{rad} . \mathrm{s}^{-1}=0.105 \mathrm{rad} . \mathrm{s}^{-1}$
The minute hand goes through  $2 \pi$ radians in 60 min
so

\omega=2 \pi / 3600$ rad.s $^{-1}=1.7 \times 10^{-3} \mathrm{rad} . \mathrm{s}^{-1}$

Posted by

avinash.dongre

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