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 dependent 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)