# B Relation between centrifugal force and speed

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1. Nov 29, 2017

### physicsnewbie101

Does centrifugal force generate in increase in overall speed? For example, if I am driving down a highway on-ramp which goes in a circular fashion, with my speed slowly increasing, does my overall speed increase because of the centrifugal force generated from the circular on-ramp? Is more force needed to control the car because of this?

2. Nov 29, 2017

### kuruman

No. It changes the direction of the motion but not the speed.
If your speed is slowly increasing it is because you are going downhill as you turn or because you are pushing on the accelerator or both.
I don't understand the question. What is "this" that might require "more force" to control the car?

3. Nov 29, 2017

### A.T.

That's what the centipetal froce does in the inertial frame. The centrufugal froce is what balances the centipetal force in the rotating frame, where the car is static.

4. Nov 29, 2017

### ZapperZ

Staff Emeritus
At some point, you need to learn about centripetal force, instead of "centrifugal force", which is only an "apparent force" when you are in a non-inertial reference frame. Centripetal force is "real", centrifugal force is nothing more than a reaction force to the centripetal force.

These are "central forces", meaning they act towards a center, and are always perpendicular to the direction of motion. If you've learned basic kinematics and basic vector addition, then a perpendicular force will not increase the speed or change the energy of motion. It merely changes the DIRECTION of motion. The force does no work.

Zz.

5. Nov 29, 2017

### kuruman

That is true. It wasn't clear to me whether OP distinguishes the difference between centrifugal and centripetal force.

6. Nov 30, 2017

### CWatters

Centrifugal and centripetal forces typically act sideways on a car so can't increase its forward speed.

The centripetal force required to make the car follow a circular path depends on the speed of the car (actually speed squared) so the faster you go the more friction is required to make the turn. However the available friction is limited and can vary due to changes in the road surface. One moment there might be enough friction for a turn at the speed you are going and the next there isn't. This can affect the ability of the car to make smooth turns or even retain control. Friction is also needed to slow down or stop, and if it's all been "used up" providing centripetal force there will be none left for braking. In short it's never a good idea to drive so fast you run out of friction when you need it

7. Nov 30, 2017

### pixel

With tight curves and/or high speeds, the road is banked so that the component of the force of gravity parallel to the road contributes to the centripetal force.

8. Nov 30, 2017

### jbriggs444

Huh? You can break gravity into components perpendicular to the pavement and parallel the pavement, sure. But gravity is still a downward force whose projection in the centripetal direction is zero regardless. This means that if gravity has a component parallel to the pavement that in turn has a projection in the centripetal direction that is positive, then it also has a component perpendicular to the pavement that has a projection in the centripetal direction that is negative. Gravity cancels out as a contribution to centripetal acceleration.

9. Nov 30, 2017

### FactChecker

The tilted pavement is a constraint that forces any motion in the direction of the centrifugal force to be associated with a gain in height. That gain in height is opposed by gravity.

10. Nov 30, 2017

### pixel

As you so kindly point out, my statement was not correct. I was thinking that with a banked road the axis about which the car is rotating is tilted and the centripetal force pointed along the road. Not so. That's what happens when one discusses something one hasn't thought about in decades.

But I think it's correct to say that gravity does play a role, as without it there would be no normal force, the horizontal component of which is providing the centripetal force. Angle needed to keep the car on the road surface and moving in a circle: tanθ = v2/gr .

Would what I originally said apply if you were going around a banked curve and also going downhill?

Last edited: Nov 30, 2017
11. Nov 30, 2017

### ZapperZ

Staff Emeritus
I have no idea why this is such a point of contention when one can easily and simply draw out the free-body diagram. In fact, that is what should have been done way in the beginning and would have saved a lot of discussion from the start.

This is basic physics. It should not be this painful to straighten out.

Zz.

12. Dec 1, 2017

### A.T.

Also, the Big Bang plays a role, as without it there would be no ....

13. Dec 1, 2017