What is the Greatest Speed a Car Can Drive Over a Hill Without Leaving the Road?

[spyroarcher]

Homework Statement

A person drives a car (without negative lift) over the top of a hill, the corss section of which can be approximated by a circle of radius 250m. What is the greatest speed at which he can drive without the car leaving the road at the top of the hill?

Homework Equations

F=(mv^2)/R (I think, there probably is a more)

The Attempt at a Solution

I'm sorry that I can't provide a visual, but imagine a hill, and the curve of that hill is 250m and a car is going up it. So I know that I'm trying to find the greatest speed of the car, just so it wouldn't "jump" after the top. Now I'm guessing that I need to find a velocity where F(g) or F(n) would be infinitesimally small, but I'm not even sure where to start...
Thanks in advance.

 

[fzero]

Consider the force diagram for the car. What equation must be satisfied for the car to remain in circular motion?

 

[spyroarcher]

I actually don't know too many equations, my teacher barely teaches, so I practically have to rely on my physics textbook, and I don't completely understand it...

 

[fzero]

Do you know any of the forces (vertical or whatever) on the car? Do you know what the equation F=(mv^2)/R describes?

 

[spyroarcher]

The only forces I am aware of on the car is friction (I'm not sure if it is incorporated into this problem), gravity and the normal force. Also from what I can understand, that equation just represents the magnitude of centripetal force, wait so would the magnitude be the Force of gravity and normal since the car is on top of the hill at that moment? But then again I'm not sure why that is significant...

 

[fzero]

The only forces I am aware of on the car is friction (I'm not sure if it is incorporated into this problem), gravity and the normal force. Also from what I can understand, that equation just represents the magnitude of centripetal force, wait so would the magnitude be the Force of gravity and normal since the car is on top of the hill at that moment? But then again I'm not sure why that is significant...

Well some forces act horizontally to the ground, such as friction and the acceleration from the engine. We can assume that the car's engine is being used to accelerate against the friction force to keep the speed of the car constant.

The question is asking about the speed at which the car will leave the ground, so the relevant forces are vertical to the ground. These are gravity and the normal force. For circular motion, the normal force is equal in magnitude to the centripetal force. Can you put these forces together into an equation for the vertical acceleration?

 

[spyroarcher]

So would I begin with F(net)= F(N)+F(G)+F(centripetal)=0 and try solving from there?

 

[fzero]

So would I begin with F(net)= F(N)+F(G)+F(centripetal)=0 and try solving from there?

It's a little bit confusing, which is why I suggested setting the magnitude of the normal force equal to the centripetal force. The force law should be the sum of gravitational + centripetal force, with appropriate signs, is equal to the vertical component of acceleration. If there is net acceleration away from the ground, the car will leave the road.

 

[spyroarcher]

Just to clarify, I want the net acceleration to be 0 for the car not to leave the road?
Also explain to me why would you want the magnitude of normal force and centripetal force equal? (I don't understand the concept)
Also thanks for bearing with me.

 

[fzero]

Just to clarify, I want the net acceleration to be 0 for the car not to leave the road?

The car will not leave the road if the vertical acceleration is either zero or pointed toward the ground. If it's being accelerated away from the ground it will leave the road.

Also explain to me why would you want the magnitude of normal force and centripetal force equal? (I don't understand the concept)
Also thanks for bearing with me.

Maybe it's not a good idea to think of it that way because it is confusing. The important thing to know is the centripetal force m v²/r is the inward force required to keep the object in circular motion. For zero acceleration we can set F(grav) = F(cent) and the car will travel in a circle (not leave the ground). But because the ground is there, if F(grav) > F(cent), the car will still remain on the ground. If F(grav) < F(cent), there is not enough gravitational force to hold the car in a circle and it will leave the ground.

The formula I suggested for the acceleration reproduces the physics precisely, which is the main reason I suggested it.

 

[spyroarcher]

I finally got it!
The steps I did:
ma=F(cent)+F(g), a=0
F(cent)=-F(g)
(mv^2)/R=-mg
Then after the cancel with the mass and some value plugging I get 49.5m/s.
However I am a bit shaky on the whole idea of the F(N) and F(cent), but I generally get it since both F(cent) and F(N) are inward forces in this case.
Once again sorry for the trouble I caused you, combining Newton's second law and centripetal force+friction, confuses me a lot and thanks a lot.

 

[ttdb]

Sorry to interrupt the conservation :)
can i know where the normal force that the ground acts on the car gone to?
i just know that in this question, the weight of the car provides the centripetal force acts on the car to keep it travels in a circular path.
but when i consider a free body diagram which consists of vertical forces only, what i found out is the normal force of the ground on the car is
N = F(g) + F (centripetal)
is this statement correct?

 

[Mindscrape]

Now to test your understanding, try to figure out the minimum speed a car would have to go to stay on a hill with a height given by h(x)=dcos(kx).

P.S. Don't actually try this, I was only joking. It's a tough problem in reality.

 

[fzero]

Sorry to interrupt the conservation :)
can i know where the normal force that the ground acts on the car gone to?
i just know that in this question, the weight of the car provides the centripetal force acts on the car to keep it travels in a circular path.
but when i consider a free body diagram which consists of vertical forces only, what i found out is the normal force of the ground on the car is
N = F(g) + F (centripetal)
is this statement correct?

The best way to write Newton's 2nd law would be

$$N - F_c = m a_y,$$

where we use the 3rd law to equate

$$N = | F_g |.$$

This still isn't great because the centripetal force can only appear if the motion is circular, so that equation still doesn't make sense.

I like to think of centripetal force in this kind of problem as a limit, similar to the way we use static friction. The car stays on the ground as long as

$$-F_g + F_c \leq 0,$$

similar to the way an object subject to static friction remains at rest unless the other forces exceed the static friction.

 

[ttdb]

hmm... then we aren't going to use a free body diagram here?

 

[fauxbeast]

I finally got it!
The steps I did:
ma=F(cent)+F(g), a=0
F(cent)=-F(g)
(mv^2)/R=-mg
Then after the cancel with the mass and some value plugging I get 49.5m/s.
However I am a bit shaky on the whole idea of the F(N) and F(cent), but I generally get it since both F(cent) and F(N) are inward forces in this case.
Once again sorry for the trouble I caused you, combining Newton's second law and centripetal force+friction, confuses me a lot and thanks a lot.

F(cent) is the same thing as Fnet, except for circular motion. it is NOT a separate force.