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[tex]\pi[/tex]r 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 [tex]\theta[/tex] 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 [tex]\theta[/tex]?

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:

[tex]v=r\omega[/tex]

Every point on the Earth has the same angular velocity [tex]\omega[/tex], 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 ([tex]\inline{\frac{1}{4} v=\frac{1}{4}r\omega}[/tex])

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 [tex]\theta[/tex]=arccos(1/4)
because 2[tex]\pi[/tex]r (cos[tex]\theta[/tex])/t =1/4 2[tex]\pi[/tex]r/t

it came out to be 75.5 degrees

 

[clementc]

yep that's right =)

 

[SammyS]

Yes !

 

[Lfrizz]

yayyy