Deflection of Mass Homework Solution

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Homework Help Overview

The discussion revolves around the deflection of a particle launched along the z-axis in a rotating Earth frame. The original poster attempts to calculate the deflection based on the particle's fall and the Earth's rotation, raising questions about the assumptions made regarding the inertial frame and the direction of deflection.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the time of fall, gravitational acceleration, and the resulting deflection due to Earth's rotation. Questions arise regarding the validity of the assumptions, particularly whether the deflection direction aligns with the Earth's rotation.

Discussion Status

The discussion is active, with participants questioning the assumptions made about the inertial frame and the direction of deflection. Some guidance has been offered regarding the need to reconsider these assumptions, particularly in relation to the definition of points in the problem.

Contextual Notes

There is an ongoing debate about the reasonableness of the calculated deflection values and their implications, particularly concerning the physical reality of such deflections at the equator. The original poster acknowledges discrepancies but seeks clarification on specific points of confusion.

Pushoam
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Homework Statement


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Homework Equations

The Attempt at a Solution


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I assume that the particle is launched along the z- axis of the x-y-z frame which is fixed with the Earth and the Earth is rotating about the x – axis.

Wrt an inertial frame, the particle will fall on A.

The arc length between P and A is the required deflection.

Now, since the Earth is rotating with constant angular velocity, PA = ## \omega R T##, where T is the time of the fall.

Wrt an inertial frame, its acceleration is g ( let's take it constant).

Then, using ## s= \frac {at^2 } { 2} ## taking the initial speed to be 0.

We have, ## T = \sqrt{\frac {2h } { g } }##.

Is this correct?
 

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Last edited:
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There is no A in your figures ...

Do you get a reasonable result if you insert reasonable values for your parameters?
 
I can take h such that T = 1s.
Then the deflection is about 467 m.
As ##\omega = 7.3 * 10^{-5} rad/s ~ and ~ R = 6400 km##

Why should it not be reasonable?
 
That is a fall of ca 5 m. Do you find it reasonable that a fall of 5 m is deflected by almost 500 m? Do you not think people living at the equator would have serious problems with this?

Edit: On top of that, your deflection is in the wrong direction.
 
Orodruin said:
That is a fall of ca 5 m. Do you find it reasonable that a fall of 5 m is deflected by almost 500 m? Do you not think people living at the equator would have serious problems with this?
Then if I drop a ball from the 2nd floor of a building ( which is more than 5m above the ground) on the equator, it will deflect by about 500m.
Yes, this will be difficult. I realized it. Thank you for it.
Orodruin said:
your deflection is in the wrong direction.
The deflection is PA, in the anti - clockwise direction, opposite to the direction of rotation of the earth. Isn't this correct?
I have taken the clock - wise direction of the rotation to be in the +ve x- direction.
Sorry, I didn't take these direction things sincerely at start.

So, the above approach is wrong as it doesn't get veriffied by the experiment.
But, I don't know where is the mistake?
 
Pushoam said:
The deflection is PA, in the anti - clockwise direction, opposite to the direction of rotation of the earth. Isn't this correct?
No, it is not correct. It is even stated in the problem that the deflection is to the east, ie, in the same direction as the rotation. (Last time I checked, the Sun rises in the east.)

I suggest that you examine your assumptions. In particular your assumption that the object falls straight down in an inertial frame.

And you still have not defined A in this thread.
 
Orodruin said:
And you still have not defined A in this thread.
I have defined A in the picture.
 
Orodruin said:
In particular your assumption that the object falls straight down in an inertial frame.
The force acting on the object in the inertial frame is gravitational force (which is radial, in this case towards negative z- axis).

The Earth is rotating, but this doesn't affect the mass distribution.
I have assumed that the Earth is not orbiting. Is it this which is wrong?
Orodruin said:
No, it is not correct. It is even stated in the problem that the deflection is to the east, ie, in the same direction as the rotation. (Last time I checked, the Sun rises in the east.)

I have to measure the deflection wrt Earth frame. A person on the Earth will want the ball to fall on P, but the ball falls on A. So, for this person, PA is the deflection.

Isn't this correct?
 
Pushoam said:
The force acting on the object in the inertial frame is gravitational force (which is radial, in this case towards negative z- axis).

The Earth is rotating, but this doesn't affect the mass distribution.
I have assumed that the Earth is not orbiting. Is it this which is wrong?I have to measure the deflection wrt Earth frame. A person on the Earth will want the ball to fall on P, but the ball falls on A. So, for this person, PA is the deflection.

Isn't this correct?
No. The object is dropped from rest relative to the Earth frame so your assumption that it starts from rest in the inertial frame is obviously false.
 
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