Find Gravitational & Centripetal Forces on Mass in Rotating Sphere

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Homework Help Overview

The discussion revolves around a physics problem involving a sphere with the mass and radius of the Earth, rotating at a constant speed. A mass is placed on the surface of this sphere, and participants are tasked with finding the gravitational force, centripetal force, the difference in their magnitudes, and suggesting a value for the acceleration due to gravity for the mass.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between gravitational and centripetal forces, questioning how these forces interact and the implications of latitude on the problem. There are attempts to clarify definitions, such as "free fall," and discussions about the significance of the mass's position relative to the axis of rotation.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the relationship between forces, but there is no explicit consensus on the interpretations of the forces involved or the implications of the mass's position.

Contextual Notes

Participants note the absence of latitude information and its potential impact on the calculations. There is also a recognition that the centripetal force is not an additional force but rather a component of the resultant forces acting on the mass.

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Homework Statement


There's a sphere with the mass of the Earth and same radius.It rotates at constant speed. It hhas the period of the earth. a mass of a kg sits on the surface.
1) find grav force on mass
2) find centripetal force on mass
3) find the difference in magnitude between them
4) suggest a value of acceleration of free fall for the mass.

Homework Equations

The Attempt at a Solution


Cant do last part; the way I see it is that the centripetal force does not exist separately; it is criteria for circular motion at constant speed. So clearly the gravitational force over-provides, but I don't understand how this excess is related to the acceleration.
 
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What is the definition of "free fall"?

Note for (2): latitude is not given.
 
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Motion where the only force acting is weight
 
So what is the weight of the object close to the surface?
 
mg, where g is the GFS at that point, so is it the grav force?
 
When you say latitude isn't given, what is the significance of that?
 
mg, where g is the GFS at that point, so is it the grav force?
No - you are asked to find the acceleration.
Acceleration and force are different physical concepts - important not to mix them up.
What the question is basically asking is what the acceleration would be if there was no ground to stop it falling.
(Note: the centripetal force is the difference between the object's weight and the normal force to the surface.)

When you say latitude isn't given, what is the significance of that?
Compare the centripetal force sitting on the equator to that when sitting on a pole.
What about latitudes in between?
 
Jamessamuel said:
When you say latitude isn't given, what is the significance of that?
How far is the mass from the axis of rotation.
By the way, I agree with you that it is not right to ask what is the centripetal force on the mass. It would be better to ask what centripetal force is required by the mass in order to stay on the surface.
 
Haruspex - its a whole radius away isn't it?
Mr.Bridge- so if we removed the surface the reaction force would disappear. The only acting force would be the grav force?
 
  • #10
1433761790376.jpg
 
  • #11
Jamessamuel said:
Haruspex - its a whole radius away isn't it?
How far are you away from the Earth's axis of rotation now?
 
  • #12
Pretty far , I am english so certainly on a smaller radius than the mass in the question, which is on the equator.
 
  • #13
The acceleration due to gravity and centripetal force are in the same direction if you are on the equator. I would use the equator as your teacher probably expected. Then it's easy to add the two acceleration vectors together. And g is easy if we use the standard r at the equator. And then Ac is easy if we use the equator. Note you will have to find v by looking up r. I'm thinking this is what your teacher wants you to do.

And the gravitational force idea just states that any two masses attract each other with the same force. But one can also see that an object on the surface is also changing direction( if not exactly on a rotational pole) so there must be centripetal force as well. And Fc Is supplied by the force of gravity.
 
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  • #14
The actual question:
1433777769338.jpg
 
  • #15
Jamessamuel said:
Pretty far , I am english so certainly on a smaller radius than the mass in the question, which is on the equator.
This is the first time you have stated it is on the equator. This is why Simon asked for the latitude.
 
  • #16
pgardn said:
The acceleration due to gravity and centripetal force are in the same direction if you are on the equator. I would use the equator as your teacher probably expected. Then it's easy to add the two acceleration vectors together. And g is easy if we use the standard r at the equator. And then Ac is easy if we use the equator. Note you will have to find v by looking up r. I'm thinking this is what your teacher wants you to do.

And the gravitational force idea just states that any two masses attract each other with the same force. But one can also see that an object on the surface is also changing direction( if not exactly on a rotational pole) so there must be centripetal force as well. And Fc Is supplied by the force of gravity.
This is backwards.
Centripetal force is not some extra force that you can add on to gravity. It is a required component of the resultant force, namely, that necessary to keep the object traveling on the arc. See https://www.physicsforums.com/insights/frequently-made-errors-pseudo-resultant-forces/
The two forces acting on the mass are gravity and the normal force from the surface. The resultant of the two is the centripetal force. The apparent weight of the mass is equal and opposite to the normal force.
 
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  • #17
Ah

Because the normal is a 3rd law reaction to what is the weight pushing down
 
  • #18
Jamessamuel said:
Ah

Because the normal is a 3rd law reaction to what is the weight pushing down
To the apparent weight, yes.
 
  • #19
Technically the 3rd law reaction to the weight of the object due to the Earth is the weight of the Earth due to the object. This force acts at the center of mass of the Earth and points towards the object.

The normal force to the ground is the reaction to the force of the object pressing into the ground. It can be a tad confusing because "how hard we press against the ground" is what we commonly think of as "weight". In physics, "weight" is a technical term for the force of gravity on an object, so it should be used carefully.
 

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