Angular speed and rotation problem

[suspenc3]

what is the angular speed $$\omega$$ about the polar axis of a point on Earth's surface at a latitude of $$40^o$$

I know that the radius of the Earth is $$6.37 x 10^6m$$

I also know that the Earth rotates about this axis ($$40^o$$)

I don't really understand what their asking...obviously angular speed, but I can't picture it
What do I do?

 

[Tide]

The Earth rotates about the polar axis - not the "40 deg" axis.

Do you know the definition of angular velocity? Do you think latitude would matter?

 

[Homer Simpson]

Are you sure you got the total question read properly?? I don't know what you consider angular speed to be. Look up Angular Velocity. It is a 'rotations per unit time' sort of thing. You can relate this to Linear Velocity, maybe that is what you are after. That would be how fast you would be traveling in unit dist per unit time as observed from space.

 

[andrevdh]

Angular speed is defined as
$$\omega=\frac{\Delta \theta}{\Delta t}$$
where $\Delta \theta$ is the angle, in radians that a point (on the surface of the Earth at latitude 40 degrees in this case) rotates through during the time interval $\Delta t$. The Earth rotates through $2\pi$ radians in 24 hours, irrespective where on Earth you are.

 

[-Christastic-]

The way I would do this problem is to first assume a spherical Earth for simplicity. Next draw a circle on a piece of paper, this will be your cross sectional area of the earth. Draw a line from the center of the Earth to equator, this is your Earth radius(this is only for reference). Now make a line 40 degrees from your first line connecting the Earth center to the surface(this is your Earth radius as well). Now draw a line from the surface of your 40 degree line(vertex) directly "down" to your first line. You should now have a right triangle. Solve for the bottom leg of this right triangle and that gives you the radial distance the point 40 degrees latitude is away from the axis of the earth. Since you want to know the angular acceleration of that point(the speed of that point revolving) you need to find the circumference of that revolution simply 2(pi)r. Now since omega is just the rate of change in revolution vs rate of change of time, you simply have 2(pi)r/24hrs where r is your radial distance from the Earth's axis of rotation. This should give you an angular acceleration in m/s if you convert the 24hrs. Hope this helps, sorry if it's confusing