Deflection of Mass Homework Solution

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
8 replies · 2K views
Pushoam
Messages
961
Reaction score
53

Homework Statement


upload_2017-12-9_14-36-6.png


Homework Equations

The Attempt at a Solution


upload_2017-12-9_15-58-25.png


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?
 

Attachments

  • upload_2017-12-9_14-36-6.png
    upload_2017-12-9_14-36-6.png
    11.5 KB · Views: 747
  • upload_2017-12-9_14-44-22.png
    upload_2017-12-9_14-44-22.png
    7.4 KB · Views: 483
  • upload_2017-12-9_15-58-25.png
    upload_2017-12-9_15-58-25.png
    7.2 KB · Views: 756
Last edited:
Physics news on Phys.org
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?
 
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.
 
  • Like
Likes   Reactions: Pushoam