# Understanding centripetal force

1. Sep 15, 2011

### jsmith613

1. The problem statement, all variables and given/known data
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.

thanks

2. Relevant equations

I know the formula for acceleration (v^2/r) and Force = m(v^2/r).
3. 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.

2. Sep 15, 2011

### PeterO

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 lets 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.

3. Sep 15, 2011

### 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?

4. Sep 15, 2011

### jsmith613

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

5. Sep 15, 2011

### jsmith613

6. Sep 15, 2011

### 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

7. Sep 15, 2011

### Staff: Mentor

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'.

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.

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)

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?)

8. Sep 15, 2011

### 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

9. Sep 15, 2011

### Staff: Mentor

What working are you referring to? Which post?

10. Sep 15, 2011

### jsmith613

post 6

11. Sep 15, 2011

### Staff: Mentor

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

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.)

12. Sep 15, 2011

### jsmith613

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

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

13. Sep 15, 2011

### jsmith613

Last edited: Sep 15, 2011
14. Sep 16, 2011

### jsmith613

post #12

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

15. Sep 16, 2011

### Staff: Mentor

16. Sep 16, 2011

### Staff: Mentor

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

17. Sep 16, 2011

### jsmith613

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

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

18. Sep 16, 2011

### Staff: Mentor

This is correct.

This is not correct.

Take the correct equation and solve for R.

19. Sep 16, 2011

### jsmith613

sorry
R = mg - mv^2/r

20. Sep 16, 2011

### Staff: Mentor

Good. Now how would you answer the question?