Centripetal force on a person on the Earth

[barryj]

Homework Statement

Would a person’s weight change if the Earth stopped rotating?

Relevant Equations

F = GM1M2/R^2
A = w^2R

I am confused. See my diagram below. With the Earth rotating, I think the scale would read the force of gravity or 448.4 N. If the Earth were not rotating, would the scale read less or more due to the effect of centripetal force? I tend to think more by an amount of 1.78 N. Is this correct?

If there were no Earth, and the 50 kg person was rotating around a point the centripetal force should be 1.78N. So how does the centripetal force add or subtract from gravity.

 

[Orodruin]

The weight shown by the scale is the force from the person on the scale. There are two forces acting on the person, the gravitational force and the force from the scale on the person. These need to add up (vectorially) to the centripetal force needed to maintain the person in circular motion. Draw a free body diagram with the forces and consider that the force resultant needs to be 1.78 N towards the center of the Earth.

 

[barryj]

Would the centripetal force affect the scale reading?

 

[Delta2]

Would the centripetal force affect the scale reading?

The answer is yes and it would make the scale reading 1.78N smaller than the real weight. Just do as @Orodruin says to understand why. All I have to add to post #2 is that

(force from scale to person)=-(force from person to scale)

due to Newtons 3rd law of mechanics.

The force from the scale to person is part of the centripetal force. The force from the person to the scale is what the scale reading shows (in absolute value).

 

[barryj]

It is still confusing. So, if the Earth were not turning, then the scale would read the weight of the person and in this case 448.4 N. If the Earthis turning, then the scale would read 448.4 - 1.78 N = 446.2 N Is this correct?

 

[barryj]

Here is my confusion. Centripetal force acts toward the center of rotation, yes? Gravity acts toward the center of mass. So force due to gravity is 448.4N Centripetal force is 1.78 N so one would think that the total force on the person is the sum of these two forces but this is not the case. It seems that the centripetal force acts against gravity? If the Earth would begin to spin more rapidly, then the Centripetal force would increase and intuition would say the scale would gradually show less weight. At some point, the person would levitate off the surface.

 

[vela]

Your confusion is a common one, which is why some teachers don't talk of a centripetal force but a centripetal acceleration. When an object moves along a circular path, it experiences a centripetal acceleration of magnitude $a_c = \omega^2 R$. When you apply Newton's 2nd law in this problem, there are the two forces, gravity and the force of the scale on the person, that appear on one side of the equation, and there's the centripetal acceleration which appears on the other side. There's no separate centripetal force which contributes to the net force.

 

[Delta2]

The total force on the person is the centripetal force.

It is equal to (centripetal force)=(gravitational force) - (force from scale to the person).

yes it appears to act against gravity because to find the force from the person to the scale (that is the scale reading) we subtract the centripetal force from gravity. But in fact the centripetal force is what is left from gravitational force if we subtract the force from the scale to the person as that equation above says. For the rest you are right, if the Earth would spin too fast the centripetal force would become equal to the gravitational force which means that (again from the above equation) that the force from the scale to the person will become zero, equivalently that the person levitates.

 

[barryj]

Thanks for the explanation

 

[haruspex]

Would the centripetal force affect the scale reading?

No. Centripetal force is not an applied force. Adding up all the applied forces (in this case, gravity and the reaction from the scale) produces the net force. We can choose to resolve the net force into a component collinear with the velocity and a component normal to it. Centripetal force is just the name we give to the normal component.

Confusion arises because of the way we think about cause and effect here. As described above, the reduced scale reaction results in the centripetal acceleration. Yet it seems more natural to say that the spin of the Earth results in the acceleration and hence the reduced scale reaction.

The reaction force between two solid bodies in contact is the minimum magnitude force that prevents interpenetration. The spin of the Earth allows you to accelerate towards its axis to some extent without breaking the scale. Hence the reaction force reduces exactly so as to produce the appropriate centripetal acceleration.

The line of thought works when the scale is in an accelerating elevator, ignoring Earth's spin.

 

[Delta2]

No. Centripetal force is not an applied force. Adding up all the applied forces (in this case, gravity and the reaction from the scale) produces the net force. We can choose to resolve the net force into a component collinear with the velocity and a component normal to it. Centripetal force is just the name we give to the normal component.

Your comment is very correct here (about centripetal force not being an applied (or I would prefer the term interaction) force), however why you say "No." since to find the scale reading we subtract the centripetal force from the gravitational force.

 

[haruspex]

Your comment is very correct here (about centripetal force not being an applied (or I would prefer the term interaction) force), however why you say "No." since to find the scale reading we subtract the centripetal force from the gravitational force.

Being able to calculate the scale reading from the centripetal force is not the same as saying the centripetal force affects the scale reading. I can determine how much rain has fallen by looking at the water in my rain gauge, but I would not say my rain gauge affects the weather.

 

[Delta2]

Being able to calculate the scale reading from the centripetal force is not the same as saying the centripetal force affects the scale reading. I can determine how much rain has fallen by looking at the water in my rain gauge, but I would not say my rain gauge affects the weather.

Let me ask you something ,for you it is correct to say that the rotation of Earth affects the scale reading?
For me it is equivalent to say that the rotation of Earth affects it or the centripetal force affects it. Because we can't define centripetal force if there isn't some sort of rotation or some sort of movement and centripetal force is the component of the net force perpendicular to the movement.

 

[haruspex]

Let me ask you something ,for you it is correct to say that the rotation of Earth affects the scale reading?
For me it is equivalent to say that the rotation of Earth affects it or the centripetal force affects it. Because we can't define centripetal force if there isn't some sort of rotation or some sort of movement and centripetal force is the component of the net force perpendicular to the movement.

The situation is strongly analogous to an elevator accelerating downwards. The acceleration permits a reduction in the reaction force from the floor/scale. Since the reaction force is always the minimum necessary, it reduces.
The downward acceleration of the elevator does not cause a downward net force which in turn somehow affects the scale reading.

 

[Delta2]

Sorry @haruspex I am afraid I don't agree. For me the logic goes like this:

rotation of Earth with angular velocity $\omega$--->definition of term $m\omega^2r$--->math and physics tell us that this term is equal to the net perpendicular force--->we subtract this term from the gravitational force to find the scale reading..

hence it is obvious that this term affects the scale reading. But ok deep down the rotation of Earth is the root cause.

 

[Steve4Physics]

Hi. It might be worth adding these points...

1. Just in case there is any possibility of a misunderstanding, note that in the Post #1 diagram, the 50kg person is shown on ‘top’ of the Earth. This might imply that the person is at the North Pole (rather than at the equator say).

If at the North (or South) Pole, the person is on the axis of rotation, so their distance from the centre of rotation is r=0. The person’s centre of mass is not then rotating. This would mean ω²r = 0 and there is zero centripetal force associated with the earth’s rotation.

It would be clearer to show the person on the ‘side’ (at the equator), if that is what is intended.

2. The wrong value is used for the earth’s radius. Post #1 states R = 6.67E6 (with no units!). In fact the approximate equatorial radius is 6.38x10⁶ m (and the polar radius is 6.36x10⁶m). Looks like the value for R has been muddled with the value for the gravitational constant (G).

The centripetal force for a 50kg mass at the equator is then:
$m \omega ^2 r = 50 \times (\frac {2\pi}{86400})^2 \times 6.38 \times 10⁶ = 1.69$N (not 1.78N)

Therefore, depending on the person’s latitude, the centripetal force can be anywhere between 0 and 1.69N.

3. I’d like to add that the term ‘centripetal force’ is simply an alternative name for whatever force is perpendicular to an object’s velocity. This means that for constant speed circular motion, ‘centripetal force’ is basically just another name for the resultant force.

4. Centrifugal force is a useful concept if used properly. Centrifugal force is a fictitious force that appears to exist in a rotating frame of reference. Think of a car turning a tight bend. Inside the car you experience being forced sideways. The apparent sideways force is called centrifugal force. It is the result of you tending to go in a straight line (Newton’s 1st law) while the car changes direction. The loss of weight due to the earth’s rotation is centrifugal force.

5. Also, here are some general points:

- never forget units when giving values; for example a planet’s radius is sometime given in metres and sometimes given in kilometres. If you don’t bother including the unit you risk a serious mistake!

- you calculated ω and rounded to 7.3E-5 rad/s; but you then used this rounded value to calculate another value to a higher precision ($T_{string}$ = 1.78N). Avoid introducing rounding errors by not rounding intermediate values in a calculation.

[Edit: minor change.]

 

[PeroK]

If we consider, for the sake of simplicity, someone standing on the equator, then gravity is the centripetal force. That force, however, is greater than that required to maintain the circular motion. Some of the gravitational force, therefore, provides the required centripetal acceleration and the remainder must be balanced by an upward force from the ground. The faster the Earth is spinning, the more of the gravitational force is needed for the centripetal acceleration and the less needs to be counterbalanced.

 

[haruspex]

gravity is the centripetal force

That's not how I would put it. Gravity provides a component of the centripetal force. By definition, the centripetal force is that component of the net force which is normal to the velocity.

 

[Steve4Physics]

That's not how I would put it. Gravity provides a component of the centripetal force. By definition, the centripetal force is that component of the net force which is normal to the velocity.

I agree.

This means, for circular motion at constant speed, the centripetal force is one-and-the-same as the resultant force.

For a person at the equator, the centripetal force is the resultant force, $\vec F_{gravity} + \vec F_{normal-reaction}$.

For a satellite in a circular orbit, the centripetal force is the resultant force, $\vec F_{gravity}$.

 

[jbriggs444]

I agree.

This means, for circular motion at constant speed, the centripetal force is one-and-the-same as the resultant force.

For a person at the equator, the centripetal force is the resultant force, $\vec F_{gravity} + \vec F_{normal-reaction}$.

For a satellite in a circular orbit, the centripetal force is the resultant force, $\vec F_{gravity}$.

For a person not at the equator, the centripetal force is still the resultant force, $\vec F_{gravity} + \vec F_{normal-reaction}$. However, $\vec F_{gravity}$ and $\vec F_{normal-reaction}$ will not be anti-parallel in such a case. The Earth is not perfectly spherical.

 

[Steve4Physics]

For a person not at the equator, the centripetal force is still the resultant force, $\vec F_{gravity} + \vec F_{normal-reaction}$. However, $\vec F_{gravity}$ and $\vec F_{normal-reaction}$ will not be anti-parallel in such a case. The Earth is not perfectly spherical.

If the Earth were a perfect Maclaurin spheroid I would agree with you. But it isn’t. E.g. see https://en.wikipedia.org/wiki/Maclaurin_spheroid

That means (apart from at the equator and poles) the resultant (centripetal) force on a person (P) standing on ‘level’ ground is not $\vec F_{gravity} + \vec F_{normal-reaction}$, though for a perfect Maclaurin spheroid it would be.

I believe that there must be a third force acting on P so that the resultant of all three forces acts towards P’s centre of rotation. The third force on P is friction in this case.

Apart from at the equator and the poles, the resultant force on P is therefore $\vec F_{gravity} + \vec F_{normal -reaction}+ \vec F_{friction}$.

Edit - typo' corrected.

 

[jbriggs444]

If the Earth were a perfect Maclaurin spheroid I would agree with you. But it isn’t. E.g. see https://en.wikipedia.org/wiki/Maclaurin_spheroid

That means (apart from at the equator and poles) the resultant (centripetal) force on a person (P) standing on ‘level’ ground is not $\vec F_{gravity} + \vec F_{normal-reaction}$, though for a perfect Maclaurin spheroid it would be.

I believe that there must be a third force acting on P so that the resultant of all three forces acts towards P’s centre of rotation. The third force on P is friction in this case.

Apart from at the equator and the poles, the resultant force on P is therefore $\vec F_{gravity} + \vec F_{normal -reaction}+ \vec F_{friction}$.

Edit - typo' corrected.

The ground is not necessarily level at the equator and poles. There is no need to exclude those locations.

 

[kuruman]

The original question is

Homework Statement:: Would a person’s weight change if the Earth stopped rotating?

The answer to that question is simply, "No". The weight of a person is the force with which the Earth attracts the person at a given location. As long as the person is at the same location, that force will not change, whether the Earth is rotating or stands still, or the person is standing on a scale or whatever.

 

[jbriggs444]

The original question is

The answer to that question is simply, "No". The weight of a person is the force with which the Earth attracts the person at a given location. As long as the person is at the same location, that force will not change, whether the Earth is rotating or stands still, or the person is standing on a scale or whatever.

There are two competing definitions of "weight". One is based on the true force of gravity as judged from an inertial frame: $\frac{Gm_1m_2}{r^2}$. The other is based on the apparent force of gravity as judged from a frame of reference rotating along with the Earth.

I prefer the latter definition, but understand that preferences vary.

 

[kuruman]

There are two competing definitions of "weight". One is based on the true force of gravity as judged from an inertial frame: $\frac{Gm_1m_2}{r^2}$. The other is based on the apparent force of gravity as judged from a frame of reference rotating along with the Earth.

I prefer the latter definition, but understand that preferences vary.

I prefer the former definition for consistency reasons. The gravitational force is mass times gravitational field in direct analogy to electric force being charge times electric field. If something else is going on at the same time, e.g. a rotating Earth for the mass or a magnetic field for the moving charge, it can be taken care of separately.

 

[haruspex]

I prefer the former definition for consistency reasons. The gravitational force is mass times gravitational field in direct analogy to electric force being charge times electric field. If something else is going on at the same time, e.g. a rotating Earth for the mass or a magnetic field for the moving charge, it can be taken care of separately.

According to https://en.m.wikipedia.org/wiki/Weight, the ISO definition is "operational", i.e. is measured in whatever frame of reference. But I suspect the question setter intends the "gravitational" definition you prefer.
The short answer to the question appears to be: it depends how you are defining weight.

 

[kuruman]

According to https://en.m.wikipedia.org/wiki/Weight, the ISO definition is "operational", i.e. is measured in whatever frame of reference. But I suspect the question setter intends the "gravitational" definition you prefer.
The short answer to the question appears to be: it depends how you are defining weight.

If the question setter is OP's instructor, then the definition of weight is whatever this instructor said it is. It seems that OP thinks that it is whatever a scale measures at one's location. If that's the case, then the answer to the original question is that "the person's weight will change when the Earth stops rotating unless the person is at one of the geographic poles."

I am uneasy about this definition, however. As the Wikipedia article notes, "$\dots~$the famous apple falling from the tree, on its way to meet the ground near Isaac Newton, would be weightless." How does one draw a free-body diagram of a weightless apple? More to the point, what are the rules for drawing FBDs when the system's weight depends on its state of motion?

 

[pbuk]

More to the point, what are the rules for drawing FBDs when the system's weight depends on its state of motion?

I have never seen weight on a FBD. Instead for a body near the surface of the Earth you may have the force $mg$, and $g$ is defined as the resultant of the gravitational and centripetal force (either locally or according to a standard).

 

[kuruman]

I have never seen weight on a FBD. Instead for a body near the surface of the Earth you may have the force $mg$, and $g$ is defined as the resultant of the gravitational and centripetal force (either locally or according to a standard).

I was referring specifically to drawing a FBD of an apple in free fall using the Wikipedia definition according to which the apple is weightless. I guess that kind of FBD would not be very helpful.

Weight on a FBD is easy to find on the web. This one is indicative of how the person who drew it has defined weight.

 

[pbuk]

Weight on a FBD is easy to find on the web. This one is indicative of how the person who drew it has defined weight.
View attachment 284267

Do you think that diagram would be improved by ignoring the definition of g, making $F_g$ larger and adding another arrow in the same direction as $F_{normal}$? How would you label that arrow: $F_{centrifgugal}$? $F_{reaction\ force\ to\ the\ centripetal\ force}$?

 

[kuruman]

Do you think that diagram would be improved by ignoring the definition of g, making $F_g$ larger and adding another arrow in the same direction as $F_{normal}$? How would you label that arrow: $F_{centrifgugal}$? $F_{reaction\ force\ to\ the\ centripetal\ force}$?

The simple answer to your question is "No". If the diagram needs improvement it's not along the lines you suggest in my opinion. FBDs of this sort are used for solving problems almost exclusively in introductory physics courses. The underlying approximation in these is that the Earth is an inertial frame. Some textbooks mention the approximation explicitly but most don't bother. In this approximation, a plumb bob points in the direction of $\vec g$ which is towards the center of a perfectly spherical Earth.

It is a good approximation meant to bring to the forefront the application of Newton's 2nd law to solutions of dynamics problems. Ignoring air resistance serves a similar purpose in projectile motion problems. So the label F_g = W = mg in the diagram expresses the approximation and the definition of weight: the force with which the Earth attracts the object is the same as the weight which is the same as the mass multiplied by the magnitude of the local acceleration of gravity.

If we are going to refine the approximation because we are thinking of the non-inertial frame of a rotating Earth, we are moving into the realm of an intermediate mechanics course. However, why consider only the Earth's rotation and not add other sources that affect $\vec g$, e.g. the gravitation effects of the Moon and the Sun which cause the observable effects of ocean tides?

Also, in my opinion, the egregious error in this FBD is the label F_normal = - W which confuses the magnitude of a vector with its component. We have $\vec W=|\vec W|(-\hat y)$ which implies that $W_y=-|W|$. The convention is that labels in FBDs are magnitudes of vectors whilst the direction of the vector is indicated by the direction of the arrow. Thus, the label on the arrow pointing straight up sets a magnitude equal to a negative number. It's this sort of thing that students see and ask "is $g$ positive or negative?"

 

[jbriggs444]

How does one draw a free-body diagram of a weightless apple?

One includes all forces on the apple other than gravity -- which you have apparently decided to transform away based on calling the apple "weightless".

If for instance, the apple is sitting on a table, it is subject to a real force equal to mg and is experiencing an upward proper acceleration of g in the freely falling frame that you have selected.

 

[vela]

As the Wikipedia article notes, "$\dots~$the famous apple falling from the tree, on its way to meet the ground near Isaac Newton, would be weightless." How does one draw a free-body diagram of a weightless apple? More to the point, what are the rules for drawing FBDs when the system's weight depends on its state of motion?

The FBD would be the same as before: a single force directed downward with magnitude $mg$. You just don't refer to that force as the weight. In other words, you still represent all of the forces acting on a body in the FBD. Whether you call any of those forces the weight of the object is irrelevant.

The more common example is the person standing on a scale in an elevator. As the elevator accelerates upward or downward, the scale's reading changes. Some would say the person's weight is changing; the rest of us say the person's apparent weight changes. In either case, the FBD would generally have two forces acting on the person (as long as $a<g$): $mg$ downward and the normal force upward.