Why do pendulums swing more slowly at the equator?

  • Thread starter Thread starter Blue_Jaunte
  • Start date Start date
  • Tags Tags
    Equator Swing
AI Thread Summary
Pendulums swing more slowly at the equator due to the Earth's oblate shape and rotation, which affects gravitational acceleration. The period of a pendulum is influenced by gravity, which is slightly lower at the equator—approximately 9.780 m/s² compared to the poles. This reduction in gravity results from both the centrifugal force caused by Earth's rotation and the increased distance from the Earth's center due to its bulging shape. The centrifugal force contributes about 65% to the difference in gravitational acceleration. Additionally, altitude and local density variations play minor roles in gravitational differences.
Blue_Jaunte
Messages
21
Reaction score
0
I was asked this recently and the only explanation I could come up with was that the Earth is oblate and the difference in R would account for the difference in the period. Is this wrong? Is this even a real phenomenon (the longer period at the equator than at other latitudes)?
 
Physics news on Phys.org
The period of a pendulum is 2\pi\sqrt{l/g} so if g is less the period (time for a swing) becomes longer. g varies by a few percent around the world dependign on the density of local rocks.
Tt also depends on latititude because, as you said, the Earth bulges at the equator and so you are futrther from the centre of the Earth and the force of gravity is slightly lower.
 
A pendulum swings slower at the equator because the Earth is rotating. The rotation acts to make the pendulum swing slower directly and indirectly. The direct effect is easiest to envision from the perspective of the rotating Earth-fixed frame. In this frame, a centrifugal force of r\omega^2 arises directly from the Earth's rotation. At the equator, this direct effect alone amounts to 0.034 m/s2. There is of course no centrifugal force at the poles.

For those who insist there is of course no such thing as centrifugal force anywhere, you will be forced to look at things from the perspective of an inertial frame. I leave this as an exercise to you curmudgeons. Keep in mind that the answer you get will be the same.

I also cited an indirect effect that results from rotation. The Earth's rotation makes the Earth bulge at the equator. As mgb_phs noted, this makes things at the equator further from the center of the Earth. Together, the direct and indirect effect act to make Earth's gravitational acceleration 9.780 m/s2 at the equator. This is about 0.052 m/s2 less than it is at the poles, or a 0.53% reduction in the gravitational acceleration. The direct effect accounts for about 65% of the difference.

The next leading factor in variations in gravitational acceleration after Earth rotation and the J2 non-spherical harmonic term is altitude above the spheroid. Things weigh 0.28% less at the top of Everest than they do at sea level at the same latitude. Local variation in density is a distant fourth.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...

Similar threads

Back
Top