Angular speed and displacement of clock hands
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10. Let $\omega_1$ and $\omega_2$ be the angular speed of the second hand and minute hand of a smoothly running analog clock respectively. If $x_1$ and $x_2$ are the respective angular displacement in 1 minute then (1) $\frac{\omega_1}{x_1} = \frac{\omega_2}{x_2}$ (2) $\omega_1 x_1 = \omega_2 x_2$ (3) $\omega_1 x_1^2 = \omega_2 x_2^2$ (4) $\omega_1^2 x_1 = \omega_2^2 x_2$
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Hi Mohammad, let's solve this problem about an analog clock together.
Analyzing Clock Hand Dynamics
First, let's recall the definition of angular speed. Angular speed is defined as the change in angular displacement divided by the time interval taken.
In this problem, the angular displacement is denoted as x, and the time interval is one minute. So we can write the formula as omega equals x over t.
We can rearrange this formula to isolate the ratio of angular speed to angular displacement. Notice that omega divided by x equals one over t.
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