Another Angular Velocity question

[Lfrizz]

The Earth has a radius of 6.38 x 106 m and turns on its axis once every 23.9 h. (a) What is the tangential speed (in m/s) of a person living in Ecuador, a country that lies on the equator? (b) At what latitude (i.e., the angle in the figure, in degrees) is the tangential speed 1/4 that of a person living in Ecuador?

From the question I got help with earlier, I think I know how to go about this, but I want to make sure.

use the equation v=d/t where d= 2\pir and r=6.38x10⁶

I then convert the 23.9 hours into seconds and divide the d found above by the seconds

To find the latitude I have to find \theta in this picture that I am not sure how to post here...
Question is, do I need to convert the radians to degrees to find \theta?

Thank you!
L

 

[SammyS]

The method looks good.

You can do it in either radians or degrees. Make sure to check what mode your calculator is in. Usually Latitude is given in degrees.

 

[Lfrizz]

Now I went and confused myself b/c I wasn't using the tangential variables... I THINK I need to divide v/4.

and I am confused how to get back to theta.

 

[Lfrizz]

http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c08/ch08p_33.gif

 

[SammyS]

It will be at a Latitude where the distance from the Earth's axis is 1/4 of what it is at the equator.

 

[clementc]

remember the equations linking the rotational counterparts with their linear equivalents:

v=r\omega

Every point on the Earth has the same angular velocity \omega, but depending on the radius of the circular path which it traces, its tangential velocity will vary.

To have a tangential velocity 1/4 that of the point on the equator, r must be 1/4 of the radius at the equator since angular velocity is constant (\inline{\frac{1}{4} v=\frac{1}{4}r\omega})

To find the angle at the point where the radius of the circular path is 1/4 of the Earth's radius...
DRAW A DIAGRAM! Assume the Earth is a circle and use simple right angle trig.

 

[Lfrizz]

:) thank you, that helped alot-

I found \theta=arccos(1/4)
because 2\pir (cos\theta)/t =1/4 2\pir/t

it came out to be 75.5 degrees

 

[clementc]

yep that's right =)

 

[SammyS]

Yes !

 

[Lfrizz]

yayyy