Why's radial acceleration zero at north pole

AI Thread Summary
Radial acceleration at the equator is calculated to be 9.9337 m/s² due to the Earth's rotation. At the North Pole, the radial acceleration is zero because the radius of rotation effectively becomes zero, resulting in no centripetal force. This means that while the Earth is spinning, a point at the North Pole does not experience any radial acceleration. The discussion highlights the difference in motion between the equator and the poles, emphasizing the unique conditions at the poles. Understanding these concepts is crucial for grasping the physics of rotational motion on Earth.
SS2006
Messages
80
Reaction score
0
a)Compute hte radial accleration of a point at the equator of the earth. b)repeat for the north pole of hte earth. Take the radius of the eartk to be 6370 km.

so for a) i got 9.9337 m/s square which is right
but for b) the answer is just zero
why?
 
Physics news on Phys.org
You are at the "center" of rotation. You really are not moving anywhere are you? Just spinning...
 
Whoa!

What Force could cause an acceleration as quick as 9.934 m/s^2 ?

I don't recall having to be tied down in Equador, to stay on the ground. (;->
 
lightgrav said:
Whoa!

What Force could cause an acceleration as quick as 9.934 m/s^2 ?

I don't recall having to be tied down in Equador, to stay on the ground. (;->

Yeah, but have you actually ever BEEN to the equator? ;)
 
SS2006 said:
a)Compute hte radial accleration of a point at the equator of the earth. b)repeat for the north pole of hte earth. Take the radius of the eartk to be 6370 km.

so for a) i got 9.9337 m/s square which is right
but for b) the answer is just zero
why?

Your part "a" isn't correct it says you need to find the acceleration not velocity, the velocity you find also incorrect.
 
hey.there's zero acceleration at poles because the R in the mrw^2 is approaching zero.or rather zero in this case.as it reaches the pole, the centripetal radius becomes zero, Fc=0 therefore, no acceleration.hope this helps.heh I'm new here.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top