What is the length of the hour hand divided by the length of the second hand? 
A clockmaker wants to design a clock such that the area swept by each hand (second, minute, hour) in one minute is the same (all hands move continuously). What is the length of the hour hand divided by the length of the second hand?

Attempt
We have that $\pi r_s^2 = \dfrac{1}{60}\pi r_m^2 = \dfrac{1}{3600} \pi r_h^2$. Thus, $\dfrac{r_h}{r_s} = \sqrt{3600} = 60$. The answer to this question was actually $\sqrt{720}$, so I am confused where I went wrong.
 A: Why $3600$, that should be $12\cdot 60$.
A: The hour hand takes $12$ hours to sweep a complete circle. During this time the second hand makes $12\times 60=720$ circuits. Therefore the area swept by the second hand must be $\frac{1}{720}$ of the area swept by the hour hand.
In other words, the relevant numbers are "how many turns of the minute hand per turn of the hour hand", and "how many turns of the second hand per turn of the minute hand". But it is immaterial how many seconds there are in a minute -- no matter how many small units we divide the minute into, the second hand needs to move by the same angular velocity, namely one revolution per minute.
A: Denote by $\omega_h$, $\omega_s$ the angular velocity of the hour hand and the second hand, respectively, and let $r_h$, $r_s$ be their radii. If they must sweep out the same area in equal times we have to require that
$${1\over2} r_s^2\>\omega_s={1\over2} r_h^2\>\omega_h\ .$$
It follows that
$${r_h\over r_s}=\sqrt{\omega_s\over\omega_h}=\sqrt{\mathstrut720}\ ,$$
since the second hand performs $720$ cycles in the $12$ hours the hour hand takes to perform one cycle.
