Acceleration associated with the Earth’s spin

In summary, the acceleration associated with the Earth's rotation about its axis at an equatorial point is zero, as there is no torque acting on the Earth. However, the component of acceleration at the equator due to Earth's spin is Rω^2 towards the center, where ω is the angular speed of the Earth's rotation and R is the distance between the center of the Earth and its equator. This acceleration is directed radially inward, as determined by Newton's 2nd Law of motion.
  • #1
Pushoam
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Homework Statement



Calculate the acceleration associated with
the Earth’s rotation about its axis (assume an equatorial point).

Homework Equations



About center of mass motion
Γ=Iα

The Attempt at a Solution


Since the external forces acting on the Earth are gravitational forces and hence there is no torque acting on the earth.
Hence the acceleration associated with
the Earth’s rotation about its axis (assume an equatorial point) is zero.
 
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  • #2
You've found the angular acceleration, but I believe the question is asking for the acceleration of a point of the Earth's surface on the equator due to Earth's spin. In other words, if you were standing at the equator, what would the component of your acceleration associated with the Earth's rotation be?
 
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  • #3
o.k. I got it.
Thank you.
 
  • #4
Then, the acceleration would be Rω2 towards the center where ω is the angular speed of the Earth rotation and R is the distance between the center of the Earth and its equator,wouldn't it?
 
  • #5
That looks about right. And what direction is the acceleration in?
 
  • #6
The acceleration is towards the center of the earth, radially inward.
 
  • #7
Is it? Remember we're talking about the acceleration associated with the Earth's spin, not its gravity.
 
  • #8
Let's model the Earth as a spinning ball for this problem.
I am sitting on the equator.
Now from an inertial frame,I will be in a uniform circular motion with angular speed ω and radius R.
according to Newton's 2nd Law of motion,
F = ma = 2 R(-## \hat r ##)
So ,the direction is radially inward.
 
  • #9
Pushoam said:
So ,the direction is radially inward.
I apologize, I was mistaken. Of course the direction is radially inward. Otherwise the point wouldn't be kept in circular motion. It's been awhile since I took my last physics class. :rolleyes:
 
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