# Centrifical force of earth

1. Oct 22, 2005

### Petrikovski

How long would a day be if the Earth were rotating so fast that objects at the equator had no weight?

I do Fcnet = m(v^2/r) --> mg = m(v^2/r) --> g = (v^2/r)

g = (v^2/r)
but g = 0 right? so then how do i do this problem? that would make v = 0 and ya...

2. Oct 22, 2005

### Janus

Staff Emeritus
g is the acceleration due to gravity. Its value is independent of any spin of the Earth.

3. Oct 22, 2005

### Petrikovski

ok thanks. unfortunately the answer isnt given so idk if im right until monday. i did:
g = (v^2)/r
9.8 = (v^2)/6.38e6
v =7907

40086722 (distance of Earth in m)/7907 = 5069 seconds
= 1.4 hours per day.

i have one more question and then im done.

A train traveling at a constant speed rounds a curve of radius 275m. A pendulum suspende dfrom the ceiling swings out to an anglr of 17.5 throughot the turn. What is the speed of the train?

Fcnet = m(v^2/r)
Fk = m (v^2/r)
coeff of friction * mg = m(v^2/r)

i dont know the coefficient of friction or v. im supposed to solve for v. am i not supposed to use friction as the force? and how do i factor the 17.5 degree swing fo the pendulum into this?

thanks a lot for the help

4. Oct 23, 2005

### Staff: Mentor

train problem

First figure out the acceleration of the train by analyzing the forces on the pendulum.

5. Oct 23, 2005

### Petrikovski

would this be correct then?
Fnet=ma
mg(cos17.5)=ma
9.8(cos17.5)=a
9.34=a

(v^2)/r = a
(v^2)/275 = 9.34
v^2 = 2568
v = 50.7 m/s

6. Oct 23, 2005

### Staff: Mentor

No. Start by identifying the forces acting on the pendulum mass. (There are two forces.) Then apply Newton's 2nd law to the vertical and horizontal components.