Question

Calculate the angular speed of the second hand, minute hand and hour hand of clock?

Answer 1

Step 1: Formula used:

Angular speed($\text{ω}$) = $\frac{Angulardisplacement\left(\theta \right)}{Totaltimetaken\left(t\right)}$

Step 2: Calculation of the angular speed of second hand of the clock:

The second hand of the clock completes one round in 60sec.

The angular displacement that is; the angle covered by the second hand of the clock in one complete round is $2\pi$ radian.

That is; $\theta =2\pi radian$

$t=60sec$

Thus,

The angular speed of second hand of the clock is;

${\text{ω}}_s=\frac{2\pi radian}{60sec}$

${\text{ω}}_s=\frac{\pi }{30}rad{s}^{-1}$

${\text{ω}}_s\approx 0.105rad{s}^{-1}$

Step 3: Calculation of angular speed of minute hand of the clock:

The second hand of the clock completes one round in 60minutes.

$60minutes=60\times 60sec$

$=3600sec$

The angular displacement that is; the angle covered by the minute hand of the clock in one complete round is $2\pi$ radian.

Thus,

The angular speed of the minute hand;

${\text{ω}}_m=\frac{2\pi radian}{3600sec}$

${\text{ω}}_m=\frac{\pi }{1800}rad{s}^{-1}$

${\text{ω}}_m\approx 1.745\times {10}^{-3}rad{s}^{-1}$

Step 4: calculation of angular speed of hour hand of the clock:

The second hand of the clock completes one round in 12 hours.

$12hours=12\times 60\times 60sec$

$=720\times 60sec$

$=43200sec$

The angular displacement that is; the angle covered by the hour hand of the clock in one complete round is $2\pi$ radian.

Then,

The angular speed of the hour hand of the clock will be;

${\text{ω}}_h=\frac{2\pi radian}{43200sec}$

${\text{ω}}_h=\frac{\pi }{21600}rad{s}^{-1}$

${\text{ω}}_h\approx 1.45\times {10}^{-4}rad{s}^{-1}$

Thus,

The angular speed of second hand of the clock = $\frac{\pi }{30}rad{s}^{-1}$ $\approx 0.105rad{s}^{-1}$ =

The angular speed of minute hand of the clock = $\frac{\pi }{1800}rad{s}^{-1}$ $\approx 1.745\times {10}^{-3}rad{s}^{-1}$

The angular speed of minute hand of the clock = $\frac{\pi }{21600}rad{s}^{-1}$ $\approx 1.45\times {10}^{-4}rad{s}^{-1}$