The relative angular speed of the hour hand a | vedclass

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Question

(B) The angular speed of the hour hand $(\omega_h)$ is given by the angle covered in one full rotation ($2 \pi$ radians) divided by the time taken $(1

Vedclass · Accepted Answer

$\frac{11 \pi}{21600}$

Vedclass · Answer

(B) The angular speed of the hour hand $(\omega_h)$ is given by the angle covered in one full rotation ($2 \pi$ radians) divided by the time taken $(12 	ext{ hours} = 12 	imes 3600 	ext{ seconds})$: $\omega_h = \frac{2 \pi}{12 	imes 3600} 	ext{ rad/s}$.The angular speed of the minute hand $(\omega_m)$ is given by the angle covered in one full rotation ($2 \pi$ radians) divided by the time taken $(1 	ext{ hour} = 3600 	ext{ seconds})$: $\omega_m = \frac{2 \pi}{3600} 	ext{ rad/s}$.The relative angular speed is the difference between the angular speeds of the two hands: $\Delta \omega = \omega_m - \omega_h$.Substituting the values: $\Delta \omega = \frac{2 \pi}{3600} - \frac{2 \pi}{12 	imes 3600} = \frac{2 \pi}{3600} \left(1 - \frac{1}{12}ight) = \frac{2 \pi}{3600} \left(\frac{11}{12}ight) = \frac{22 \pi}{43200} = \frac{11 \pi}{21600} 	ext{ rad/s}$.

The relative angular speed of the hour hand and minute hand of a clock is (in $rad/s$):

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