ok if a car is going around a bank curve my book is telling me there is an ideal spead were no friction is required at all for it to complete the banked turn
it says that this occurs when the horizontal component of the normal force is equal to the radial acceleration
why?
I don't see how this relationship works
Fn sin theta = r^-1 m v^2
or more like why it works???
my text goes on to prove
tan theta = (rg)^-1 v^2
which I understand how to rearange I just don't get where the first relationship comes from
Fn sin theta = r^-1 m v^2
i get its the horizontal component of the normal force but I don't see why we can set it equal to the radial acceleration
If the radial acceleration to go around the curve is equal to the force of gravity trying to slide the car down towards the inside, then no friction is needed.
Looking at the horizonal:
centripetal acceleration=v^2/r and the amount of this force up the slope:
mv^2/r * cosTheta
Gravity down the slope: mgsintheta
set them equal
mg sinTheta=m v^2/r cosTheta
tan Theta= v^2/rg
I think I suggested to you once to take a look at getting Schaums Outline Series, college physics. You must have not done that, because there is an execellent drawn and explained problem on this.
Typo:
it says that this occurs when the horizontal component of the normal force is equal to the MASS TIMES THE radial acceleration
Now:
Force = mass times acceleration
To go in a circle you need a force toward the center of the circle equal to the mass times the centripetal acceleration.
For example if you swing a rock around you on a string, the horizontal component of the string tension keeps the rock going in a circle. If you let go, the rock leaves the circle and goes straight.
In this case we have no string.
The slope of the bank has to hold the car in the circle.
The bank can only exert a force normal to the surface, because there is no friction allowed.
Therefore the horizontal component of that force normal to the surface must give us that centripetal force.
If that force is Fn
and Ac = v^2/R
then Fn sin theta = m v^2/R
Now we need to get Fn
We know that the vertical component of Fn = mg, the weight
so
Fn cos theta = m g
or
Fn = m g/cos theta
so
(m g/cos theta) sin theta = m v^2/R
or
g tan theta = v^2/R
or
tan theta = v^2 / (R g)
When the car of mass m is turning at a radius of r at a speed of v, the centrifugal force that sends the car horizontally outwards is
mv²/r
The component that pushes the car upwards of the incline is
mv²cos&952;/r
For a banking of angle &952; (higher towards the outside), there is a component of the weight that pushes the car downwards. This amount is mg sin&952; where g is the acceleration due to gravity.
So equating the two gives the equation supplied.
Since m cancels on both sides of the equation, it can be simplified to
g sin&952; = v²cos&952;/r
tan&952;=v&supu2;/(rg)
replace &952; by θ.
To understand why the horizontal component of the normal force is equal to the radial acceleration in a banked turn, let's break it down step by step.
First, let's consider the situation where a car is driving around a curved banked road. When a vehicle turns, it experiences an acceleration towards the center of the circular path, known as the radial acceleration. This acceleration is necessary to keep the car moving in a circular path instead of moving in a straight line.
Next, let's understand the normal force. The normal force is the force exerted by a surface perpendicular to the surface itself. In this case, the road exerts a normal force on the car. When the car is traveling on a banked curve, the normal force has two components: the vertical component, which supports the weight of the car, and the horizontal component.
Now, let's consider the forces acting on the car. In a banked turn, there are three primary forces: the weight of the car acting downward, the normal force acting perpendicular to the surface of the road, and the frictional force between the road and the car's tires.
To maintain a circular path without slipping or skidding, the frictional force is not required as long as the horizontal component of the normal force is equal to the radial acceleration. When these two forces are equal, they cancel each other out, meaning there is no net force pulling or pushing the car towards the inside or outside of the curve. Hence, no friction is required to keep the car on the road.
This ideal speed at which the horizontal component of the normal force equals the radial acceleration is known as the "ideal speed" or "critical speed." If the car is traveling at speeds below this ideal speed, there will be a deficit of the radial acceleration supplied by the horizontal component of the normal force, causing the car to slip towards the inside of the turn. On the other hand, if the car is traveling at speeds above this ideal speed, there will be an excess of the radial acceleration, causing the car to slide towards the outside of the turn.
In summary, the horizontal component of the normal force is equal to the radial acceleration in a banked turn because it provides the necessary centripetal force, eliminating the need for friction to maintain a curved path.