A car moving at 19 m/s drives over the top of a hill.....?

Question

A car moving at 19 m/s drives over the top of a hill. The top of the hill forms an arc of a vertical circle 140 meters in diameter.
•What is the centripetal force holding the car in the circle?
•What therefore is the normal force between the car's tires and the road?

I have no more details in the solution. Sorry.

Answer 1

The centripetal force is trying to throw the car off the circle, not 'hold it in the circle'

Ac = V²/r = 19²/70 = +5.157 m/s²

Normal aceleration = -g + Ac = -4.643 m/s²

The force can't be determined because you didn't mention the mass of the car. Also, this condition applies only to the very top of the hill. Further past the top the tire load becomes less until the car finally leaves the road at a radial line 58.25° from the vertical.

Answer 2

This question cannot, strictly speaking, be answered because you didn't give us the mass, or normal weight, of the car. We could of course provide symbols to represent one or other of those, but in that case one might as well use symbols like 'v' for the speed, and 'r' for the radius of curvature of the hill. The fact that the circular arc is expressed not in terms of its radius of curvature, but instead in naive terms of the "diameter" of the osculating circle suggests that this problem has been set by someone who, frankly, is not very well versed in mathematics or science, or in setting problems in scientific or mathematical terms.

So, send it back to whoever set it!

Live long and prosper.

POSTSCRIPT: Steve, immediately below, is QUITE WRONG about "The centripetal force is trying to throw the car off the circle, not 'hold it in the circle,' " He's confusing the inward seeking centripetal acceleration and the associated force necessary to produce that with that old student bugbear "centrifugal force." You are ABSOLUTELY CORRECT in referring to the CENTRIPETAL FORCE holding the car in the circle.

(Etymological note: "centripetal" means "centre-seeking"; "centrifugal" means "centre-fleeing.")

Many (most?) people are quite confused about the frequently mistaught concept of "centrifugal force." This is a so-called "fictitious force," of relatively limited use when working in a rotating frame. Because the use of "centrifugal force" under these circumstances provides a rich field for easily solved problems of a limited kind, it is much beloved by a certain class of teachers.

Unfortunately, those who teach about and use "centrifugal force" as a quick method for working out such problems, often mislead their students into thinking of it as a "real force" that somehow manifests itself in the non-rotating world's frame. Their students in particular are often completely unaware that they have implicitly chosen to work in a rotating frame, and switch back and forth in great confusion, often "double-counting" real and fictitious forces. *** I've seen it over and over again in poorly taught students, unable to work out even problems precisely like this one, problems of a kind for which their teachers first introduced the "centrifugal force" concept.

Historically, Newton himself misunderstood the related concept of "outward endeavour" by Huyghens ("centrifugal force" by another name). This confusion delayed his own ultimate understanding of gravity by more than 12 years. Ironically, it wasn't until Robert Hooke (his detested rival!) wrote to ask him what he thought of Hooke's own idea of the planets being attracted in towards the Sun and away from straight-line motion (YES, by a CENTRIPETAL force, namely a force due to gravity), that Newton realized he'd been given the missing key to the problem that had been bugging him since his own student days. His entire approach in his magisterial Principia can be traced to the impact of the idea contained in Robert Hooke's query. He worked out ingenious geometrical ways to evaluate the consequences of CENTRIPETAL forces on moving planets, thereby discovering his so-called "Universal Law of Gravity. So there!


*** RE. THE MIXING UP OF REAL AND IMAGINED FORCES --- AND ACCELERATIONS!

In fact, in his answer, Steve falls into the same trap as have many of my students.

Take his own formulae, and ask the simple question:

What would Steve get if the car were stationary, that is if V = 0? Steve's own "normal acceleration" would then be - g, since "AC" = 0. But the car is then STANDING STILL; it's NOT accelerating at all.

During the last three and a half decades I've had students stand in front of me, on firm ground, and argue till they're blue in the face that they're accelerating downwards with acceleration 'g.' It's simply amazing what effort it takes to convince them that they're not. (I sometimes wonder if because they're taught that "the acceleration on the Earth" is 'g', they really believe that THAT is THEIR acceleration when THEY are "on the Earth"!)

What is the correct way to do a problem like this, the way that is firmly rooted in the only things that really count, Newton's equations of motion? It is this:

Moving at the top of a circular arc of radius r, at speed v, the centripetal (INWARD or DOWNWARD) acceleration, the only acceleration present in the problem, dammit, is given by:

DOWNWARD accn, a = v^2 / r.

There are two real, honest-to God forces in the problem:

1. The downward force of gravity, mg;

2. The UPWARD (real) force 'N,' the "Normal" force, sometimes known instead as the "Reaction" force, and then often denoted by R. (Call it what you like, I'll call it 'N' because you used that term in the problem.)

Then Newton's 2nd Law ("F = m a"), evaluated in the DOWNWARD direction, gives you:

mg - N = m v^2 / r.

(Note that "F" in "F = m a" means the NET force, when more than one force is present, as here. Once again, I've had students mistakenly trying to use "net acceleration," another trap, instead of "net force.")

So, N = m (g - v^2 / r).

And THAT's your answer for the NORMAL force. That IS, of course, an outward, that is upward force.

Now check THAT out for a stationary car: when the car is stationary (v = 0), the (upward) normal force is mg, and the downward force is mg. (I'm giving their magnitudes; their vector character is opposite, as implied by the use of the words "up" and "down.") This is simply an obviously correct example of the 3rd Law: Action and Reaction are equal and opposite.

(When you're standing on the floor, there's a downward force mg. Because the floor is stationary, you're NOT accelerating. The floor resists the downward force mg by providing an upoward force of equal magnitude. Stand in a faulty elevator in free fall because its cable has broken, accelerating downwards at 'g'; so will you be; your weight provides exactly for your downward acceleration; the normal force provided by the floor is now precisely zero because you ARE in free fall; you'll feel "weightless," which IS the sensation of having NO reaction from your surroundings. Did you ever realize that our SENSATION of weight is determined by the REACTION we experience from our surroundings, e.g. from the bathroom scales that are "registering" our weight? If there's NO reaction --- the freely fallingelevator sensation, we feel "weightless.")

Back to your problem, with my stationary car case: our 2nd Law equation above tells us that when standing still, the car isn't accelerating. We've obtained clearly correct results instead of anything bizarre.

I prefer to keep acceleration and force distinct in my mind, with v^2 / r being a centripetal acceleration,as explained and used, above. However, I believe that the phrase "the centripetal force holding the car in the circle" means "the (net) force needed to give the inward acceleration that is taking place when the car is moving in a circle." So, it's m v^2 / r. Then, in the next question, "What therefore is the normal force ...?", the "therefore" refers to the necessity to subtract THAT m v^2 / r from mg to get N. All of this is consistent with both how ther question is phrased, and what I've done, above.

I think I've now covered all bases, so I'll wish you "Good night"!