Understanding centripetal force and free body diagram

[jsmith613]

Homework Statement

a person is standing stationairy on Earth (whilst it is rotating) there is a net force towards the centre of the earth.

What would the free body diagram look like.


Please could someone explain why

thanks

Homework Equations

I know the formula for acceleration (v^2/r) and Force = m(v^2/r).

The Attempt at a Solution

I would suggest that the free body digram shows:
R + mv^2/r = mg
so mv^2/r = mg-R
But I get really confused by R and mg as they should cancel out.

Please could someone explain it

 

[PeterO]

Homework Statement

a person is standing stationairy on Earth (whilst it is rotating) there is a net force towards the centre of the earth.

What would the free body diagram look like.


Please could someone explain why

thanks

Homework Equations

I know the formula for acceleration (v^2/r) and Force = m(v^2/r).

The Attempt at a Solution

I would suggest that the free body digram shows:
R + mv^2/r = mg
so mv^2/r = mg-R
But I get really confused by R and mg as they should cancel out.

Please could someone explain it

First let's ignore the centripetal force.

If you stand in a stationary lift, the reaction force from the floor will be equal in magnitude and opposite in direction to mg.
If the lift was accelerating up, the Reaction force from the floor will be greater than mg [you feel heavier when the lift starts going up], Net force is up - you accelerate up.
If the lift was accelerating down, the Reaction force from the floor will be less than mg [you feel lighter when the lift starts going down] Net force is down - you accelerate down.

Now let's get centripetal Force involved.
When you are on the spinning Earth you have a centripetal acceleration down [like in the lift accelerating down] but the size of the acceleration will be MUCH smaller than what you achieve in the average lift.
As such the reaction force will be slightly smaller than mg. The difference possibly only shows up in the decimal places of the Force magnitude.

So R and mg only cancel out if the acceleration is zero.
In a lift, the acceleration can be significant. The centripetal acceleration is very small.

 

[jsmith613]

so we have a free body diagram looking like:

Up: reaction force
down: mg, mv^2/r

so there is no reaction force to mv^2/r?

therefore
mv^2/r = mg - R

what confuses now is what is the Newton pair of the centripetal force?

 

[jsmith613]

also, if we have a body orbiting a planet what is the reaction force? or does it not exist?

 

[jsmith613]

http://www.youtube.com/watch?v=1CD8Q1WGiVA&feature=related

could someone please explain what is the forces are here (especially Fn)

 

[jsmith613]

Finally,
Q:If the Earth rotated faster on its axis would would happen to your weight (as measured by a scale)
A: decrease

Q:As the rotational speed of a space habitat increases what happens to the apparent weight of the people inside
A: increases

could someone please explain the answers (i have tried 2 as below but can't do 1)


Example: Q2:

The force, mg, of the space station acts towards the earth

F = ma
F = mv²/r
BUT v = omega (w) * r
SO

F = m*w²*r

F = (GMm)/r²

(GMm)/r² = m*w²*r

If w (omega) increases then Force increases
this seems wrong because GMm/r² should always be constant and therefore not affected by speed :S

 

[Doc Al]

so we have a free body diagram looking like:

Up: reaction force
down: mg, mv^2/r

Never put 'centripetal force' on a free body diagram. There are only 2 forces acting on the body: The weight, mg, and the force of the Earth pushing up, the 'reaction force'.

so there is no reaction force to mv^2/r?

Don't think of mv^2/r as a separate force; think of it as applying Newton's 2nd law in the case of centripetal acceleration.

What we call 'centripetal force' is just the net force that produces the centripetal acceleration.

therefore
mv^2/r = mg - R

OK. (For the simple case where the mass is on the equator.) But think of it as:
ΣF = ma
mg - R = m(v^2/r)

what confuses now is what is the Newton pair of the centripetal force?

Again, the term 'centripetal force' is just another name for the net force. The only forces involved in this are mg and R, and they both have Newton's 3rd law pairs. (What are they?)

 

[jsmith613]

I would presume there has to be something wrong with my working because it is not possible for it to be correct but it follows the advice I've be given :S

 

[Doc Al]

I would presume there has to be something wrong with my working because it is not possible for it to be correct but it follows the advice I've be given :S

What working are you referring to? Which post?

 

[jsmith613]

post 6

 

[Doc Al]

Finally,
Q:If the Earth rotated faster on its axis would would happen to your weight (as measured by a scale)
A: decrease

A scale measures 'apparent weight', which is what you called R (the 'reaction' force) above.

Q:As the rotational speed of a space habitat increases what happens to the apparent weight of the people inside
A: increases

could someone please explain the answers (i have tried 2 as below but can't do 1) Example: Q2:

The force, mg, of the space station acts towards the earth

F = ma
F = mv²/r
BUT v = omega (w) * r
SO

F = m*w²*r

F = (GMm)/r²

(GMm)/r² = m*w²*r

If w (omega) increases then Force increases
this seems wrong because GMm/r² should always be constant and therefore not affected by speed :S

You'll have to redo this one. They are talking about a space station 'habitat' that creates its own artificial 'gravity' by rotation. No real gravity involved. (Hopefully they have some sort of diagram of this thing in action.)

 

[jsmith613]

Q:If the Earth rotated faster on its axis would would happen to your weight (as measured by a scale)


Answer
Here, the centripetal force increases due to an increase in speed.
As we said before, mv²/r = mg - R

Is this correct


For the next question I will break it into two questions


Q1:As the rotational speed of a space habitat increases what happens to the apparent weight of the people inside

I don't understant how this would be done as I have never come across the idea of simulated gravity

Q2:As the rotational speed of a satellite increases what happens to the apparent weight of the people inside
for this question, which is what the working was meant to be for (post 6), I presume the above working is still flawed

 

[jsmith613]

oh an also, just quickly back to this question:
http://www.youtube.com/watch?v=1CD8Q1WGiVA&feature=related

could someone please explain what is the forces are here (especially Fn)

 

[jsmith613]

post #12

I just to check also, is your weight given by the reaction force or mg?

 

[Doc Al]

Q:If the Earth rotated faster on its axis would would happen to your weight (as measured by a scale)


Answer
Here, the centripetal force increases due to an increase in speed.
As we said before, mv²/r = mg - R

Is this correct

But what about your 'weight'?

 

[Doc Al]

post #12

I just to check also, is your weight given by the reaction force or mg?

The scale measures the force pressing on it, which is the reaction force.

 

[jsmith613]

Q:If the Earth rotated faster on its axis would would happen to your weight (as measured by a scale)


Answer
Here, the centripetal force increases due to an increase in speed.
As we said before, mv2/r = mg - R

My weight = the reaction force so R = mv^2/r + mg therefore weight increases
I have said weight = reaction force as you said this in the last post

 

[Doc Al]

Q:If the Earth rotated faster on its axis would would happen to your weight (as measured by a scale)


Answer
Here, the centripetal force increases due to an increase in speed.
As we said before, mv2/r = mg - R

This is correct.

My weight = the reaction force so R = mv^2/r + mg therefore weight increases

This is not correct.

Take the correct equation and solve for R.

 

[jsmith613]

This is correct.


This is not correct.

Take the correct equation and solve for R.

sorry
R = mg - mv^2/r

 

[Doc Al]

sorry
R = mg - mv^2/r

Good. Now how would you answer the question?

 

[jsmith613]

as speed increases, Reaction force decreases so apparent weight decreases

 

[jsmith613]

This matches the answer
could we now go through Q2 (with the change of "space habitat" to "satellite")

 

[Doc Al]

as speed increases, Reaction force decreases so apparent weight decreases

Good.

could we now go through Q2 (with the change of "space habitat" to "satellite")

Please state the problem you are asked to solve.

 

[jsmith613]

Q:As the rotational speed of a satellite increases what happens to the apparent weight of the people inside

The force, mg, of the space station acts towards the earth

F = ma
F = mv2/r
BUT v = omega (w) * r
SO

F = m*w2*r

F = (GMm)/r2

(GMm)/r2 = m*w2*r

If w (omega) increases then Force increases
this seems wrong because GMm/r2 should always be constant and therefore not affected by speed :S

this is where i get stuck

 

[Doc Al]

Q:As the rotational speed of a satellite increases what happens to the apparent weight of the people inside

I assume they mean orbital speed, not that the satellite is rotating about it's center of mass.

Note that the satellite is in free fall as it orbits. So what's the apparent weight of the people inside? No calculation needed.

 

[PeterO]

http://www.youtube.com/watch?v=1CD8Q1WGiVA&feature=related

could someone please explain what is the forces are here (especially Fn)

I don't like the presenters explanation very much. I look at the problem this way:

When the bucket is overhead, the water will tend to accelerate down with acceleration g, just as it would if there was no bucket there at all.

In order to not have the water coming out of the bucket, you must accelerate the bucket down at g or higher.

You could pull the bucket down with great acceleration, but it will soon hit the ground.

If you rotate the bucket fast enough, the bucket will have acceleration [centripetal acceleration towards the centre] given by V² / R. Provided that is at least equal to g, we have the bucket accelerating down with a sufficiently large acceleration for the water not to fall out of the bucket.

 

[jsmith613]

I assume they mean orbital speed, not that the satellite is rotating about it's center of mass.

Note that the satellite is in free fall as it orbits. So what's the apparent weight of the people inside? No calculation needed.

I copied the question exactly from a website. The website has the answer as an increase.
The only force acting on the satellite is mg (towards the earth) and we would use the F = GmM/r^2 to find this force (I presume)

I don't see why no calculation is needed for your interpretation of the Q

 

[jsmith613]

I don't like the presenters explanation very much. I look at the problem this way:

When the bucket is overhead, the water will tend to accelerate down with acceleration g, just as it would if there was no bucket there at all.

In order to not have the water coming out of the bucket, you must accelerate the bucket down at g or higher.

You could pull the bucket down with great acceleration, but it will soon hit the ground.

If you rotate the bucket fast enough, the bucket will have acceleration [centripetal acceleration towards the centre] given by V² / R. Provided that is at least equal to g, we have the bucket accelerating down with a sufficiently large acceleration for the water not to fall out of the bucket.

i would presume the logic is
if the bucket accelerates fast than the water downwards then the bucket will move before the water has a chance to get out.
My next question is why won't the bucket and the water accelerate at the same speed downward as they are effectively one object

 

[Doc Al]

I copied the question exactly from a website. The website has the answer as an increase.

So they are talking about the rotation of the satellite about its center of mass, not its orbiting around the earth.

The only force acting on the satellite is mg (towards the earth) and we would use the F = GmM/r^2 to find this force (I presume)

What you need to consider is the force exerted by the rotating satellite (a space 'habitat') on the passenger. That's his apparent weight. The force of the Earth on the satellite is not relevant to this question.

Imagine the 'satellite' as a rotating cylinder and a person is standing on the inner surface of that cylinder. As the rotational speed increases, what happens to the force with which he is pressed against the 'floor'?

FYI: 'mg' is only valid near the Earth's surface.

I don't see why no calculation is needed for your interpretation of the Q

Think of astronauts in the space shuttle once they are in orbit. What's their apparent weight?

 

[jsmith613]

What you need to consider is the force exerted by the rotating satellite (a space 'habitat') on the passenger. That's his apparent weight. The force of the Earth on the satellite is not relevant to this question.


Imagine the 'satellite' as a rotating cylinder and a person is standing on the inner surface of that cylinder. As the rotational speed increases, what happens to the force with which he is pressed against the 'floor'?

FYI: 'mg' is only valid near the Earth's surface.

ok so with a free body diagram,
mg = down
Reaction force = towards centre
he is accelerating towards the centre of the satellite (i presume this is up)
F = ma
ma = R - mg
mv^2/r = R - mg
mg = R + mv^2/r

so as the satellite rotates faster, his weight increases :)

I am confused though as to why on Earth the effect of a faster rotation is completley different than on a satellite

Think of astronauts in the space shuttle once they are in orbit. What's their apparent weight?

well weight = mass * acceleration
I would presume their weight is dependant on how fast the rocket is accelerating
Once in orbit, they must have a net acceleration towards the Earth (v^2/r) so the guys weight = mass * (v^2/r)

 

[Doc Al]

ok so with a free body diagram,
mg = down
Reaction force = towards centre
he is accelerating towards the centre of the satellite (i presume this is up)
F = ma
ma = R - mg
mv^2/r = R - mg
mg = R + mv^2/r

so as the satellite rotates faster, his weight increases :)

The only force on the person that you need to consider is the 'reaction force' from the 'floor' of the satellite.

I am confused though as to why on Earth the effect of a faster rotation is completley different than on a satellite

On the Earth you are standing on the outside of the rotating body; in the satellite, you're standing on the inside. (Look up 'artificial gravity'.)

well weight = mass * acceleration
I would presume their weight is dependant on how fast the rocket is accelerating
Once in orbit, they must have a net acceleration towards the Earth (v^2/r) so the guys weight = mass * (v^2/r)

So I guess you've never see pictures of the astronauts in orbit floating around in their ship?

 

[jsmith613]

of course, but they must weight something
Although, if your weight is given by the reaction force then floating astronouts have no weight

 

[jsmith613]

The only force on the person that you need to consider is the 'reaction force' from the 'floor' of the satellite.

so was my answer wrong?

(please also see last post for the previous question)

 

[PeterO]

My next question is why won't the bucket and the water accelerate at the same speed downward as they are effectively one object

I don't like the way you used speed [I underlined it]!
They do accelerate at the same rate [speed is not a term that relates well to acceleration]. The fact is that acceleration is greater than g, so the only way the water can accelerate at a rate greater than g, is if something applies an extra force to it [gravity is not enough] The thing that applies the extra force is the bucket - but that can only happen if the water remains in contact with the bucket - it doesn't come out.

 

[PeterO]

of course, but they must weight something
Although, if your weight is given by the reaction force then floating astronouts have no weight

Common usage refers to people's weight, and their apparent weight.

Weight is mg - and for an orbiting satellite g might be only 8.7 if the satellite is high enough.

Apparent weight refers to how heavy you feel.

If you ride a roller coaster, you feel light as you go over the top of a hump, you feel heavy as you go through a dip, and who knows how you feel as you go through the inside of an inverted loop, or a spiral.

At all times you mass is the same, and g is 9.8, so your weight is constant, but your apparent weight varies wildly.

When people say an astronaut is weightless, they mean her apparent weight is zero.

 

[jsmith613]

I don't like the way you used speed [I underlined it]!
They do accelerate at the same rate [speed is not a term that relates well to acceleration]. The fact is that acceleration is greater than g, so the only way the water can accelerate at a rate greater than g, is if something applies an extra force to it [gravity is not enough] The thing that applies the extra force is the bucket - but that can only happen if the water remains in contact with the bucket - it doesn't come out.

What I still don't get is that the water is accelerating downwards at the top of the flight, no?

 

[PeterO]

What I still don't get is that the water is accelerating downwards at the top of the flight, no?

Anything traveling in circular motion is accelerating towards the centre of the circle. If the circle is a vertical circle - as in this case - the centre is down, when the object is at the top of the circle.
Just because an object accelerates in a particular direction doesn't mean it is going to travel in that direction.

If you are cruising in a car, then apply the brakes - you will be accelerating towards the direction you have come from, but merely reduce the rate at which you continue the way you were going. [If you take on that [negative?] acceleration for long enough you will stop and start going back where you came from.

A pendulum does that - in fact it is the strangest thing for when it is traveling at its maximum speed it has minimum acceleration, and when it is traveling at minimum speed [at the extremes], it has maximum acceleration.

 

[jsmith613]

Anything traveling in circular motion is accelerating towards the centre of the circle. If the circle is a vertical circle - as in this case - the centre is down, when the object is at the top of the circle.
Just because an object accelerates in a particular direction doesn't mean it is going to travel in that direction.

If you are cruising in a car, then apply the brakes - you will be accelerating towards the direction you have come from, but merely reduce the rate at which you continue the way you were going. [If you take on that [negative?] acceleration for long enough you will stop and start going back where you came from.

A pendulum does that - in fact it is the strangest thing for when it is traveling at its maximum speed it has minimum acceleration, and when it is traveling at minimum speed [at the extremes, it has maximum acceleration.

thanks for this :)

 

[jsmith613]

ok so based on this:

If I am on a roller coaster and I react a point such that my vertical reaction force becomes zero and I feel weighless, what expression can I write for mg in terms of mv^2/r and why?

http://www.edexcel.com/migrationdoc...E January 2010 - QP/6PH04_01_que_20100128.pdf

Question 12e)

So my free body diagram would just have mg down.
BUT I am also accelerating downwards with mv^2/r
so mg = mv^2/r

is this correct?

JUST TO SAY: the mv^2/r force will actually be at a slight angle to mg from the diagram, no?

 

[PeterO]

ok so based on this:

If I am on a roller coaster and I react a point such that my vertical reaction force becomes zero and I feel weighless, what expression can I write for mg in terms of mv^2/r and why?

http://www.edexcel.com/migrationdoc...E January 2010 - QP/6PH04_01_que_20100128.pdf

Question 12e)

So my free body diagram would just have mg down.
BUT I am also accelerating downwards with mv^2/r
so mg = mv^2/r

is this correct?

JUST TO SAY: the mv^2/r force will actually be at a slight angle to mg from the diagram, no?

Yes you are correct, the only force acting is mg, and the only force needed is mv²/R They will be equal in both magnitude AND direction.

The forces acting on you on a roller coaster are your weight [mg] and the Reaction force from the seat.
If your motion is such that the reaction force has magnitude zero, then the only force acting on you is mg.

As you go over the top of a hill on a roller coaster, you will be traveling in part of a circle - even if the hump is actually parabolic, the actual top of the hill will follow the path of a some circle.
While traveling in that circle, you have need of a centripetal Force of size mv²/R

Lets say F_c is 300 N [down], and your weight is 650 N. The reaction force from the seat must be 350 N [UP] to achieve that. You would feel light since before the ride started the reaction force from the seat was 650 N.
If you were traveling faster, F_c might be 500 N, so the reaction force from the seat would be only 150 N. You would feel even lighter.
If you were traveling faster, F_c might be 650 N. There is now no need for a reaction force at all. As a result you would feel weightless.
If you were traveling faster, F_c might need to be 700 N. Now you have a problem!
If this is a modern Roller coaster, which has 2 sets of wheels - one above and one below the "rail" - it is capable of generating a downward force. If the "safety Bar" has been correctly fitted - it will transfer sufficient downward force [50 N] to supplement gravity so you stay on the track.
If this was one of the old-fashioned Roller Coaster - you are in big trouble as you and the cart have just become a projectile.

 

[jsmith613]

Lets take a look at the 700N example.

I have 650N down (weight) and a reaction force of -50N
How can my reaction force be in the same direction as my weight?
this is odd :S

(forces have always baffled me, quite literally)

 

[PeterO]

Lets take a look at the 700N example.

I have 650N down (weight) and a reaction force of -50N
How can my reaction force be in the same direction as my weight?
this is odd :S

(forces have always baffled me, quite literally)

As I said - it has to be a new design Roller Coaster

http://www.madehow.com/Volume-6/Roller-Coaster.html

have a look at part 6 and you will see that modern roller-coasters have double wheel set-ups [above and below a tubular rail] so that as well as the standard upward force, they can supply a downward force if necessary. That is why modern roller coasters have a safety bar - to enable a downward force to be transferred to the occupants if necessary. It is also why they don't allow "little" people to ride, since the "safety bar" leaves room for small people to slip out from under.

 

[jsmith613]

no, what I am asking is how can the reaction force be -50 (i.e: in the same direction as weight).
As you said the Fc = 700N and mg = 650N so reaction force must be -50N
how does this make sense?

 

[jsmith613]

do you see my problem?

 

[PeterO]

no, what I am asking is how can the reaction force be -50 (i.e: in the same direction as weight).
As you said the Fc = 700N and mg = 650N so reaction force must be -50N
how does this make sense?

A reaction force is exactly that: a reaction to the situation.

When the cart goes slowly, or is stationary, the weight of the cart tries to make it sink through the structure, towards the Earth [down]. The wheels of the cart contact the rails, and the rails react by supplying sufficient upward force to prevent that happening - works fine if the rails/wheels are strong enough.
In the scenario where Centripetal Force needs to be 700 N, the cart will be trying to fly off the track. In a modern roller coaster, this will cause the lower set of wheels to push up on bottom of the track - with the reaction that the track pushes DOWN on the cart.

Did you look at the picture of a modern roller coaster - where there is a set of wheels above the rail as well as a set of wheels below the rail?
That is also why modern roller coasters can go upside down!
If the cart is moving as expected, the cars will remain in contact with the rails anyway. However, if someone's jacket fell out and became entangled with the wheels the cart may slow down too much and the cart would fall off the track [down] from the inside of an inverted loop. That possibility MUST be catered for, so the modern roller coaster has a complex wheel arrangement to prevent that. This arrangement also enables the coaster to pass over a hill at a speed which was previously too fast to remain on the track.

 

[PeterO]

do you see my problem?

I think so. The reaction force is not a reaction to mg, it is a reaction to what ever is trying to happen.
When you stand on the floor, the reaction force from the floor is equal in magnitude and opposite in direction to your weight, but does not constitute an action-reaction pair in the Newton's 3rd law sense.

The 3rd law pairs, or couples as they are often called, are.

The Earth pulls you down [with a force of mg down] - you pull the Earth up [with a force of mg up]

Because the Earth is pulling on you we also get;

You push down on the floor [with a force of size mg] - the floor pushes up on you [with a force of size mg]

When we do a free body diagram, we look at those 4 forces, and note that two of them are acting on YOU [the other two act on the Earth and the floor] and do a little picture representing you and those two forces.

We call one of those forces weight - and often give the tag mg to it.
We call the other force the Reaction Force - and often give it the tag F_R.
We sometimes use the tag F_N to emphasise that the force is perpendicular to the surface and is called the Normal Reaction Force.

**** that term Normal refers to "perpendicular", it is not some reference to the situation being common or the usual situation.

 

[jsmith613]

this partly cleared upon the netwon pair

What is confusing me is that I know Force = mv^2/r BUT this force is produced by other forces.
In our example (exam paper) the roller coaster is above the track so if we assume this is an old type (just for the sake of the question) what is providing the extra 50N to lead to a 700N force given the weight is constant?

is that clearer??

(by the way thanks a lot for all the help you've been giving me! it means a lot :)

 

[Doc Al]

What is confusing me is that I know Force = mv^2/r BUT this force is produced by other forces.

Best to think of mv^2/r as being the required net force to produce the given circular motion.

In our example (exam paper) the roller coaster is above the track so if we assume this is an old type (just for the sake of the question) what is providing the extra 50N to lead to a 700N force given the weight is constant?

Nothing! Meaning that if you go fast enough that you require a net force of 700 N to maintain contact with the track and you only weight 650 N then you cannot maintain contact with the track. You're going too fast!

 

[jsmith613]

oh so I require a force of 700N to keep me in a circle IF i am moving at v^2 = (700*r)/m
if there is not enough force to provide this I will not stay in a circle and will follow a tagental line of motion

thanks!

 

[Doc Al]

Right!