# Homework Help: Centrifugal force vs Gravity

1. Oct 16, 2012

### lendav_rott

1. The problem statement, all variables and given/known data
The friction coefficient between the tires and the road is 0.6. What is the maximum velocity of the car without it sliding out of the turn. The radius of the curve is 80m. The curve is considered to be horizontal (as ridiculous as it may sound)

2. Relevant equations
F = ma

3. The attempt at a solution
Does the assignment imply that the centrifugal acceleration needs to overcome the gravitational acceleration?

If so then can I say that the centrifugal force has to be equal or less than the friction force between the tires and the road?

Fcf = ma = mv²/r
Ff = μmg cosα - but the road is horizontal so α is 0 and Ff = μmg

mv²/r = μmg
v²/r = μg => v = (μgr)^0.5 ≈ 21,7 (m/s)

Is this the correct assumption to be made? What exactly happens when the car starts sliding? Is that the same logic why water will stay in the bucket not fall out if you rotate the bucket quick enough?

This also made me think of the pilot training gadget - you know the centrifugal thing that creates overload. Does overload mean the centrifugal acceleration IS greater than the gravitational acceleration?
How would I determine the speed of the training device if I know the radius and that it would have to, say, create 5 times overload?
Does it mean that the person's weight is 5 times greater in relation to normal weight. eg. 5mg = m(g+a) ?

Last edited: Oct 16, 2012
2. Oct 16, 2012

### CWatters

Not sure why you say it's ridiculous? Many roads are flat and some even have have "adverse camber".

No it asks how fast the car can go before centrifugal force would exceed the maximium frictional force available.

You appear to have the right answer but this line..
mv²/r = μmg /m
mv²/r = μmg

When the car starts sliding μ might change. Typically μ is different for "static" and "dynamic" friction. Frequently μ falls when something starts sliding. Perhaps you have come across situations where it's hard to get something to move but once it's moving it becomes easier to keep it moving.

No it's not the same a the water in a bucket problem. In that case you need centrifugal force to exceed the force due to gravity (not friction) in order for the water to stay in the bucket.

3. Oct 16, 2012

### CWatters

Sounds like you are talking about one of these...

http://www.thespacereview.com/archive/402a.jpg

These simulate higher than normal g-force on the body. In the lump at the end of the rotating arm there is a seat that can tilt (eg lean over sideways). As it speeds up the seat normally leans over more and more so that the combination of normal gravity and centrifugal force allways acts downwards through the persons body.

If you were to lock the seat so it can't tilt it would feel like you are going around a corner very fast. Perhaps handy if you want to simulate the sideways forces on a racing car driver.

As regards to how fast it has to rotate to simulate 5g... You have to add (vector addition)together the centrifugal force and the regular gravitational force and arrange for the result to equal 5mg.

Note the machine cannot simulate less than 1g.

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4. Oct 16, 2012

### PeterO

On difficulty you are presenting to yourself is the use of the description "centrifugal acceleration" and "Centrifugal Force".

There is no such thing as "centrifugal force".

5. Oct 16, 2012

### lendav_rott

So is it referred to as Force of centrifugal acceleration in English?

6. Oct 16, 2012

### PeterO

There is no Centrifugal Acceleration either.

You are after Centripetal Force - a Force towards the centre of the circle, not some imaginary force outwards.
The acceleration is referred to as Centripetal Acceleration - and is inwards, not outwards.

You probably think it should be outward, since that seems to be the direction you would fall if you tried to stand on the tray of a truck while it went around a corner [unless you braced yourself to prevent falling].
But consider this: If you stood on the tray of a truck and it suddenly accelerated forwards, what direction would you fall [relative to the truck]?

7. Oct 16, 2012

### lendav_rott

Oh, right, I would fall opposite to the truck's direction and if it went around the corner the force would be pointed towards the center of the corner and I would fall in the middle of the street ...kind of maniacal examples, but I think I understand it - it's Newton's law of motion. The body will attempt to maintain its direction of moving or not-moving if influenced by other forces.

8. Oct 16, 2012

### PeterO

Much better way of thinking.

Fortunately/unfortunately you will get the correct sized, numerical answer if you assume there is a Centrifugal Force - but the direction will be wrong, and your descriptions will be full of rubbish.

9. Oct 16, 2012

### PeterO

As for the water in the bucket ...

There is nothing mystical about the size of the acceleration due to gravity. There is no barrier involved.

If you hold a stone in one hand, with your other hand beside that stone, then drop the stone, it is very easy to accelerate you other hand down at a much greater rate than the falling stone. In other words you can easily make something accelerate at a higher rate than gravity.

If you rotate a bucket of water in a vertical circle, you can easily do it such that the centripetal acceleration is greater than 9.8 ms-2.

If you were to have someone photograph you while the bucket is at its highest point, it would look like the water should fall out - since the bucket would appear to be stationary, and thus not accelerating.

If you held the bucket in that position, the water would indeed fall out. The bucket would be stationary while the water was accelerating down at a rate of 9.8 ms-2.

But while you spin the bucket, you may be forcing the bucket to accelerate [towards the centre] at a rate of 12 ms-2. The water in the bucket must also accelerate at that rate - or else the bucket will somehow overtake the water????

Only by having the bucket push on the water can the acceleration of the water be increased from the 9.8 ms-2 it would have if merely falling to the 12 ms-2 it must have to travel with the accelerating bucket. Only if the water stays in the bucket can the bucket apply that extra push needed.

Remember, most [all?]of the acceleration of the bucket of water has to do with a changing direction of velocity, not necessarily a change in magnitude of velocity.

10. Oct 16, 2012

### CWatters

Appologies for my very lax posts earlier that refer to centrifugal force. PeterO is correct, it's much better to think in terms of centripetal acceleration and that it allways acts towards the center.

11. Oct 16, 2012

### PeterO

That "which way would I fall" test is a handy way of reminding yourself of the direction an object is accelerating.