Posted by Nikita on Monday, October 26, 2009 at 11:58am.
How long would it take for the Earth to complete a full turn if a person at 49.2° northern geographical latitude floats apparently weightlessly across the room? Use REarth = 6,385 km for the radius of Earth.
Not sure if I understood your question.
I interpret it as saying "if the Earth is rotating at a yet unknown angular velocity ω such that a person would float weightlessly at latitude 49.2°N, find &omega."
It is not as simple as it sounds, because the acceleration due to gravity acts towards the centre of the Earth. On the other hand, the rotation of the Earth is around a N-S axis, causing the centripetal force to be at an angle θ with the vertical, where θ is the latitude.
Assuming that the vertical (towards the centre of the earth) components balance, and the person floating is restrained from flying south by a horizontal rope, then we can do the following calculations:
Acceleration due to gravity, g = 9.8 m/s²
Radius of the Earth, R = 6385 km = 6385000 m
Latitude = 49.2°
We will find r, the distance of the surface of the earth to the axis of rotation, i.e. measured along the equatorial plane.
r = Rcos(θ)
Centripetal acceleration, a
= rω² (perpendicular to axis of rotation)
Vertical component of centripetal acceleration, av
= a cos(θ)
= rω² cos(&theta)
= Rω&up2; cos^{2}(&theta)
Equate av and g, solve for &omega.
I get 0.0019 radians/sec. which translats to a full rotation in 55 minutes and 14 seconds.
Editorial correction:
Vertical component of centripetal acceleration, av
= a cos(θ)
= rω² cos(θ)
= R ω² cos²(θ)
Here's an article complete with figures for supplementary reading:
http://galitzin.mines.edu/INTROGP/notes_template.jsp?url=GRAV%2FNOTES%2Flatitude.html&page=Gravity%3A%20Notes%3A%20Latitude%20Variations
how did you translate 0.0019 radians/sec into minutes?