Car on circular hill: what happens when speed exceeds maximum?

[titansarus]

Homework Statement

The Actual Question is that:

A car is moving on a circular hill with radius R. what is the maximum speed it can have at the apex of the hill such that it doesn't jump of from the hill. I know the solution but I want to know how will be the equations if the velocity is greater than that.

The Attempt at a Solution

[/B]
$mg - N = m v^2 /r$ the maximum is at $N=0$ so $v = \sqrt{rg}$.

I want to know what will happen to the equation if v gets higher than that? Why the car will take off from the ground? what will be the acceleration of it when it gets off from the ground?

Sorry for my English.

 

[.Scott]

We are assuming that there is no friction with the air or anything else.

The car will follow a parabolic path that starts out tangent to the road surface and ends soon after that with a collision with the road.
Of course, in the extreme case, the velocity will exceed escape velocity and the hill will bump the car into space never to return.

 

[titansarus]

We are assuming that there is no friction with the air or anything else.

The car will follow a parabolic path that starts out tangent to the road surface and ends soon after that with a collision with the road.
Of course, in the extreme case, the velocity will exceed escape velocity and the hill will bump the car into space never to return.

I want to know how can I say that when $v$ gets higher than that, the car will take off from the ground. $N$ will get negative but what does that mean? Why this means the car is not binded to the ground anymore?

 

[.Scott]

I don't know what your $N$ is.

The maximum downward acceleration will be from gravity $g=9.8m/s^2$.
At the top of the hill with radius r, there will be no vertical velocity.

Let's put the origin (x,y=(0,0)) at the apex of the hill.
So without the hill, the car would free-fall, following a parabolic part: $x=vt$ and $y=-gt^2/2$ where $t$ is time.
$y$ as a function of $x$ would be $y=-gx^2/2v^2$.
But as you mentioned, the radius of the hill is $r$ while the radius of this parabola is $v^2/a$.
So when $v^2/g < r$, the car follows the surface of the hill: $y=-(r^2-x^2)^{1/2}$
Otherwise, the car becomes airborne: $y=-gx^2/2v^2$

 

[CWatters]

v=√rgv=rgv = \sqrt{rg}.

I want to know what will happen to the equation if v gets higher than that? Why the car will take off from the ground?

Inertia causes objects to move in a straight line if there is no force acting on them. So the car will travel in a straight line (tangentially to the hill) unless there is a downwards force acting on it. As it happens there is a downwards force (mg due to gravity) acting on the car. So the car will initially try to move in a circle of radius r which we can calculate...

The force required to move with radius r is mv²/r. The available force is mg. So...

mv²/r = mg
m cancels
r = v²/g

If r = R or r < R the car will stay on the road.

If r > R the car will leave the road and will initially follow a path with a greater radius than R.

Edit: I think Scott may have better maths but I like the approach that an object moves in a circle with a radius that depends on the net force acting on it. eg change the force and the radius changes.

 

[kuruman]

$N$ will get negative but what does that mean? Why this means the car is not binded to the ground anymore?

I believe by $N$ you mean the normal force exerted by the surface on the car. When this force goes to zero, it's as if the surface is not there. Something that does not exist exerts zero force. Conversely, something that exerts zero force does not exist as far as the system's motion is concerned because it cannot influence that motion. $N$ is positive when the force is away from the surface. It cannot be negative because this will mean that the direction of $N$ is into the surface. Surfaces push but not pull, so negative $N$ is meaningless.