Physics - Rotation Question with Solution | TestHub

Angular speed of the minute hand is ${\omega }_m=\frac{2\pi }{T_m}$

Since the minute hand completes one revolution in $60$ minutes, $T_m=60\times 60=3600$ s.

${\omega }_m=\frac{2\pi }{3600}=\frac{\pi }{1800}$ rad/s

Angular speed of the second hand is ${\omega }_s=\frac{2\pi }{T_s}$

Since the second hand completes one revolution in $60$ seconds, $T_s=60$ s.

${\omega }_s=\frac{2\pi }{60}=\frac{\pi }{30}$ rad/s

The difference between their angular speeds is $\Delta \omega ={\omega }_s-{\omega }_m$

$\Delta \omega =\frac{\pi }{30}-\frac{\pi }{1800}=\frac{60\pi -\pi }{1800}=\frac{59\pi }{1800}$ rad/s

Answer: $\frac{59\pi }{1800}$