Question

You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While there, you notice that the elf Sneezy, when hanging from a rope, produces a tension of 505 N in the ropeA. If Sneezy hangs from a similar rope while delivering presents at the earth's equator, what will the tension in it be? (Recall that the earth is rotating about an axis through its north and south poles.)

Answer 1

To solve this problem it is necessary to apply the concepts related to the Rotational Force described from the equilibrium and Newton's second law.

When there is equilibrium, the Force generated by the tension is equivalent to the Force of the Weight. However in rotation, the Weight must be equivalent to the Centrifugal Force and the tension, in other words:

$W = F_T + m\omega^2r_E$

Where

$\omega = \frac{2\pi}{T} \rightarrow$ Angular velocity is equal to the Period, at this case Earth's period

$r_E = 6.371*10^6m \rightarrow$ Radius of the Earth

m = mass

$F_T$= Force of Tension

$W = mg \rightarrow$ Newton's second law

Replacing and re-arrange to find the Tension we have,

$F_T = W- \frac{W}{g} (\frac{2\pi}{T})^2r_E$

$F_T = W(1-(\frac{2\pi}{T})^2\frac{r_E}{g})$

$F_T = (505)(1-(\frac{2\pi}{24hours})^2\frac{6.371*10^6}{9.8})$

$F_T = (505)(1-(\frac{2\pi}{24hours(\frac{3600s}{1hour})})^2\frac{6.371*10^6}{9.8})$

$F_T = (505)(1-(\frac{2\pi}{86400})^2\frac{6.371*10^6}{9.8})$

$F_T = 503.26N$

Therefore when Sneezy is on the equator he is in a circular orbit with a Force of tension of 503.26N