Centripetal Acceleration at 60 degrees

[wikidrox]

Can't quite figure out how to start this problem.

If earth's radius is 6378 km and has a period of 24 hrs. find:

a) centripetal acceleration of a particle at the equator.

I did this and got 0.03 m/s^2, but I can't figure out how to start the second part.

b) the centripetal acceleration of a particle at a 60 degree latitude.

and then

c) by what factor would the speed of earth's rotation have to increase for a body on the equator to require a centripetal acceleration of 9.8 m/s^2 to keep it on earth?

[turin]

Can't quite figure out how to start this problem.1) Assume the earth is a sphere. 2) Remember that centripetal acceleration is caused by axial rotation and don't confuse this with rotation about the center of the earth. 3) use trig. for part b. 4) Use the equation for centripetal acceleration that contains the angular velocity for part c. 5) Post any more questions like this in the HW Help forum, please. 6) Good luck.

[phaelax]

This might help explain it further.
http://www.cbv.ns.ca/rv/physics/Physics12/physics12notes.html#tth_chAp3

[bullroar_86]

damn... link doesnt work

anyone else have any ideas?

[russ_watters]

The distance from the point on earth at 60 degrees latitude to the axis of the earth is the radius for the second part. Draw a picture and apply some trig.

[bullroar_86]

ok great, thx for the help.

i got
a) .0337m/s^2
b) .0168m/s^2

this is an important question so I need to get it right..

is it the correct way to show my answers as:
a) 3.4 x 10^-2 m/s^2
b) 1.7 x 10^-2 m/s^2

?


now I'm on part c) by what factor would the speed of earth's rotation have to increase for a body on the equator to require a centripetal acceleration of 9.8 m/s^2 to keep it on earth?


Am i correct to assume that I'm just looking for the velocity required to give an answer of a = 9.8 m/s^2 ?

"to keep it on earth" is confusing me..

thanks

[Nylex]

ok great, thx for the help.

i got
a) .0337m/s^2
b) .0168m/s^2

this is an important question so I need to get it right..

is it the correct way to show my answers as:
a) 3.4 x 10^-2 m/s^2
b) 1.7 x 10^-2 m/s^2

?

Yes, or use m s^-2. Not sure about the last bit you had.