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
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?
5 answers
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.
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)
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)
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.