What is the effect of centripetal force on velocity?

[olivermsun]

Perhaps you are talking about the lobbed ball and bat ... I see that error. The force has to go across time and will go in the direction of the hit. I agree that was a poorly thought out example.

But turning a car takes no work? The lateral force does no work? Somehow there is something I am not seeing.

If the lateral force is truly lateral, that is, across the path of the car rather than along, then it is not doing work.

If the car is front wheel drive, and the wheels are turned, all the force is now directed along the direction of the forward rotation of the wheels. The car was going straight, and the force was in that direction, exactly balancing the resistance.

The resistance obviously resists the forward motion of the car whether it is turning or not, and hence resistance makes the problem more complicated than just the problem of turning. In that case, then the answer is yes, the engine is doing work to maintain forward speed. However, I think you should first work out your questions about the idealized case (no friction/resistance), in which case you don't need to consider the forward force exerted by the engine.

 

[rcgldr]

What causes a car to turn is that the front tires and rear tires don't point in the same direction.

That is very true for conventional cars but you can still achieve cornering with all your wheels being steered ...

My statement was the front and rear tires don't point in the same direction, this could be front wheel steer, rear wheel steer, all wheel steer. My next statement mentioned front wheel steer, so to generalize it, if you were to extend the axis of the front wheels and rear wheels, the intersection of where those extended axis cross is the center of the circle that the car tries to turn. The tires flex, so the actual turn radius is a bit larger than that. This assumes an idealized case where Ackerman like steering setup is used on the steered wheels so that the extended axis of all wheels cross at the same point.

 

[sophiecentaur]

But turning a car takes no work? The lateral force does no work? Somehow there is something I am not seeing.

No in principle. The 'work done' is strictly the centripetal force times the change in radius (which is zero, round the bend). [Edit: I didn't put that quite right but I can't think of an instant correction there] That will be zero but that's the same old 'work done on' paradox that applies to many mechanics situations and which gets people going so much. The fact is that there are many components of forces times distances that are involved in a real case because there is motion in the direction of several of the forces involved in a real tyre. One of the reasons that steel wheels and rails are so successful is that the situation on bends is much nearer the ideal.
I could go so far as to say that the 'work done on' the car is irrelevant and not worth getting bothered about. The work that is done is on distorting the tyres and the road surface etc. etc..

 

[votingmachine]

No in principle. The 'work done' is strictly the centripetal force times the change in radius (which is zero, round the bend). [Edit: I didn't put that quite right but I can't think of an instant correction there] That will be zero but that's the same old 'work done on' paradox that applies to many mechanics situations and which gets people going so much. The fact is that there are many components of forces times distances that are involved in a real case because there is motion in the direction of several of the forces involved in a real tyre. One of the reasons that steel wheels and rails are so successful is that the situation on bends is much nearer the ideal.
I could go so far as to say that the 'work done on' the car is irrelevant and not worth getting bothered about. The work that is done is on distorting the tyres and the road surface etc. etc..

For some reason, my intuition leads me to completely different thinking with a train on rails, and a car. I see that error now.

The force (torque of the tires) seems directed by the tire's direction. And then the force has to be divided into the component parallel to the forward direction of the car, and the sideways direction of the car.

No one needs to correct that ... it was an explanation of my intuitive thinking, not a statement of the physics.

 

[jbriggs444]

What torque on the tires? In the ideal case (no wind resistance and a car coasting at a constant speed) there is no torque at the tires.

 

[votingmachine]

What torque on the tires? In the ideal case (no wind resistance and a car coasting at a constant speed) there is no torque at the tires.

The OP was specifically about a real case, not ideal. With the accelerator applying constant gas/torque, and the net resistance equal to that, and the speed constant.

EDIT: That leads to confusion as people have shifted between "real" and "ideal". I've been trying to stick with "real". With net resistance and net forward forces equal and constant forward speed. Then a turn of the steering wheel to a new, steady angle.

 

[sophiecentaur]

different thinking with a train on rails, and a car.

It brings to mind what Newton's Third Law tells us. Resistance is very low with rails and steel wheels and Work is Force times Distance. The force (at constant speed) in the direction of the rails cannot be very great - or the train would be accelerating - so the work can't be much. So you have large (lateral) force and no distance and small (tangential) force in the direction where there is distance.
Some modern trams use rubber wheels because, perhaps, low noise is more important than efficiency.

 

[rcgldr]

The 'work done' is strictly the centripetal force times the change in radius (which is zero, round the bend). [Edit: I didn't put that quite right but I can't think of an instant correction there]

Work done would be a radial force times change in radius. While the radius is changing, the path spirals inwards or outwards, a component of the radial force is in the direction of that path (so it's not a centripetal "only" force), so work is done. A centripetal force is always perpendicular to the path, so doesn't perform work.

 

[sophiecentaur]

Work done would be a radial force times change in radius. While the radius is changing, the path spirals inwards or outwards, a component of the radial force is in the direction of that path (so it's not a centripetal "only" force), so work is done. A centripetal force is always perpendicular to the path, so doesn't perform work.

Yes, of course. That makes perfect sense! Cheers

 

[sophiecentaur]

What torque on the tires? In the ideal case (no wind resistance and a car coasting at a constant speed) there is no torque at the tires.

I don't think that's right because the direction in which the 'footprint' points is not in the plane of the wheel. The process of compressing the front of the footprint and releasing it at the back involves hysteresis and the forces (torques) do not cancel. This will produce a torque which 'fights' against the applied torque on the steering wheel.

 

[A.T.]

I've been trying to stick with "real". With net resistance and net forward forces equal and constant forward speed. Then a turn of the steering wheel to a new, steady angle.

Addressed here:
https://www.physicsforums.com/threa...-force-on-velocity.885192/page-2#post-5570267

 

[jbriggs444]

I don't think that's right because the direction in which the 'footprint' points is not in the plane of the wheel. The process of compressing the front of the footprint and releasing it at the back involves hysteresis and the forces (torques) do not cancel. This will produce a torque which 'fights' against the applied torque on the steering wheel.

Be careful. The steering wheel does not apply any propulsive torque, that's the engine's job. The torque that the steering wheel fights is due to the rearward offset of the contact patch from the steering axis, i.e. caster.

 

[Rupert Young]

This would be complicated. It's a combination of factors, cornering load, deformation of the tires due to a lateral load, and the hysteresis factor of the tire compound.

Can we simplify things by considering free floating bodies orbiting a central point? E.g. the planets, but with equal sized objects.

Would those nearer the centre (on a path of higher curvature) move at a slower speed?

 

[rcgldr]

This would be complicated. It's a combination of factors, cornering load, deformation of the tires due to a lateral load, and the hysteresis factor of the tire compound.

Can we simplify things by considering free floating bodies orbiting a central point? E.g. the planets, but with equal sized objects. Would those nearer the centre (on a path of higher curvature) move at a slower speed?

The loss in speed is due to tires consuming energy (conversion into heat). The amount of energy consumed by tires is increased when cornering. In a scenario where there are no energy losses, then cornering would not reduce speed (this is a self fulfilling statement, no energy lost means that kinetic energy and the related speed don't change).

 

[A.T.]

considering free floating bodies orbiting a central point? E.g. the planets, but with equal sized objects. Would those nearer the centre (on a path of higher curvature) move at a slower speed?

No, faster:
https://en.wikipedia.org/wiki/Orbital_speed#Mean_orbital_speed