Non inertial frame of reference(rotating)

In summary: It's because the centripetal force is constantly trying to balance the force of gravity, and as the car gets closer to the turn, the centrifugal force (which is due to the car's motion) becomes more significant.
  • #1
TimeRip496
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5

Homework Statement


$$g_o=\frac{GM}{R^2}e_R$$
where g0 is the gravitational acceleration, G = 6.67 * 10-11Nm2 /kg2 is the universal gravitational constant, M= 5.98 *1024 kg is the mass of the Earth, and R = 6.38 * 106 m its radius

$$g = g_0-ω*[ω*(r+R)]$$....(8.29)
where gravitational acceleration g that is felt near the surface of the Earth (i.e., on the surface of the Earth we cannot discerned between gravity g0 and the centrifugal acceleration -ω*[ω*(r+R)], we can only feel the resulting acceleration g )

The Earth’s angular velocity vector is given by ω= 7.3 * 10-5ez rad/s (i.e., it is directed along the z'-axis ), and we assume that it is a constant.

It is to be noted that because of the presence of the centrifugal acceleration
-ω*[ω*(r+R)] in this equation for the effective gravity, g and g0 will in general
not point exactly in the same direction. This effect is rather small, but measurable as
ωR^2/g_0= 0.0035 . It should also be clear from the equation (8.29) that the magnitude of the effect is a function of latitude.
upload_2016-7-31_15-45-16.png

Source:http://www.astro.uwo.ca/~houde/courses/PDF files/physics350/Noninertial_frames.pdf

where R is the radius of Earth, r is the position of the object measure from the rotating frame of reference which is moving at a constant velocity and constant angular velocity, r' is the position of the object measure from the center of the Earth

2. Homework Equations

ωR^2/g0= 0.0035 where R is the radius of Earth.
I have no idea how they get this equation. Besides I don't quite get what they mean about "that the magnitude of the effect is a function of latitude.".

The Attempt at a Solution


I have no idea how to get to this equation ωR^2/g0= 0.0035. I try to calculate the above equation but I get is 303234462.4 instead of 0.0035.
$$ωR^2/g_0=ωR^2*\frac{R^2}{GM}=(7.32*10^{-5})(6.38*10^6)^2*\frac{(6.38*10^6)^2}{6.67*10^{-11}*5.98*10^{24}}=303234462.4$$
 

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  • #2
I don't quite get what they mean about "that the magnitude of the effect is a function of latitude.".
They are working out a subjective feeling of the gravitational acceleration "g" in terms of the different effects that come into play.
One of these effects is centrifugal ... why would the centrifugal force vary with latitude and how?

As for the equation ...
Recall the work you have done on circular morion in an inertial frame: what is the equation for centripetal acceleration?
What is the relationship between centripetal acceleration and centrifugal force?
Now divide the centrifugal "acceleration" by the bare gravity to get the correct expression: compare with what the author did above.
 
  • #3
Simon Bridge said:
They are working out a subjective feeling of the gravitational acceleration "g" in terms of the different effects that come into play.
One of these effects is centrifugal ... why would the centrifugal force vary with latitude and how?

As for the equation ...
Recall the work you have done on circular morion in an inertial frame: what is the equation for centripetal acceleration?
What is the relationship between centripetal acceleration and centrifugal force?
Now divide the centrifugal "acceleration" by the bare gravity to get the correct expression: compare with what the author did above.
Ok thanks. I know the equatorial bulge at Earth can be explained by non-inertial frame of reference using centrifugal force. But can i explain it from the inertial frame of reference using centripetal force? Cause as we move from the equator to the poles, the centripetal force require to hold the mass that make up Earth decreases, therefore the gravitational force(remaining one not being use to provide the centripetal force) will increase from the equator to the pole, causing the bulge?
 
  • #4
TimeRip496 said:
I have no idea how to get to this equation ωR^2/g0= 0.0035.
This expression has a typographical error. The author of the notes misplaced the square.
 
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  • #5
... which was, of c
TimeRip496 said:
Ok thanks. I know the equatorial bulge at Earth can be explained by non-inertial frame of reference using centrifugal force. But can i explain it from the inertial frame of reference using centripetal force? Cause as we move from the equator to the poles, the centripetal force require to hold the mass that make up Earth decreases, therefore the gravitational force(remaining one not being use to provide the centripetal force) will increase from the equator to the pole, causing the bulge?
In the inertial frame, gravity is the same across the surface of the Earth. You need to think about why objects inside a car going around a corner tend to slide to one side.
 

1. What is a non-inertial frame of reference?

A non-inertial frame of reference is a coordinate system that is accelerating or rotating with respect to an inertial frame of reference. In other words, the laws of motion and the concept of inertia do not hold true in a non-inertial frame.

2. How does a rotating frame of reference affect motion?

In a rotating frame of reference, objects appear to experience fictitious forces, such as the centrifugal and Coriolis forces. These forces can cause objects to deviate from their expected path of motion as predicted by Newton's laws of motion.

3. Can we use Newton's laws in a non-inertial frame of reference?

No, Newton's laws of motion only hold true in an inertial frame of reference. In a non-inertial frame, we need to incorporate the fictitious forces into our equations of motion.

4. How do we account for the Coriolis force in a rotating frame of reference?

The Coriolis force is a fictitious force that acts on objects in a rotating frame and is perpendicular to the direction of motion. To account for this force, we need to use the Coriolis acceleration term in our equations of motion.

5. Is the Earth's surface considered an inertial frame of reference?

No, the Earth's surface is not an inertial frame of reference as it is rotating with respect to the fixed stars. However, it can be considered approximately inertial for small-scale motions and over short periods of time.

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