Finding the speed where friction is negligible.... please help!?

Question

A curve of radius 125 m is banked at an angle of 22°. At what speed can it be negotiated under icy conditions where friction is negligible?

Answer 1

Some hints to get you started:

1. If you drive around the curve at some speed "V", what is the amount of centripetal force that's acting on you (i.e. the force pushing you sideways toward the center of the circle). Write it as a formula that has "V" in it. (Hint: the formula will also have the car's mass "m" in it, and the radius "r" of the circle in it.)

2. When your car is on the track (tipped sideways at 22°), what are all the forces acting on the car? Okay, I'll tell you. Since they said their's no friction, there are just two forces: The force of gravity acting straight down (i.e. the car's weight); and the "normal force," which is the road pushing diagonally up (22° from vertical) on the car. These can be represented by two force vectors. (You should be drawing a diagram by now.)

3. Write a formula for magnitude of the downward (weight) vector (hint: it's got an "m" and a "g" in it.)

4. Break the "normal force" vector (the diagonal vector) into horizontal and vertical components (draw it!). How long must the vertical comonent be? (Hint: the car is not accelerating up or down.)

5. From the answer to #4, and knowning the tilt of 22°, you should be able to figure out how much the _horizontal_ component of the "normal force" vector is. Write a formula for it.

6. From #5, you should now have a formula for the sideways force pointing toward the center of the circle. But that is also what the formula in #1 shows. So set the #1 formula equal to the #5 formula.

7. Now you have an equation with V in it. Some of the other variables will cancel out, and whatever's left over are numbers that are given in the problem. Solve for V.

Answer 2

If there were friction no banking is necessary.

In order to make one move along a curve without depending upon friction the road is banked at angle θ.

The speed is given by the formula tan θ = v^2 /r

From this v = √ r tan θ = 7.1m/s


Only when the car either slows or speeds up this limit of 7.1 m/s, friction will come into action and accordingly the car move out or inside the arc of the curve.

Answer 3

You have to balance the vector forces parallel to that 22 deg angle and perpendicular to the 125m curve. Gravity will create a component of force down the slope, centrifugal force will create a component of force up the slope. When they are the same then the car will stay in the middle of the track.

Answer 4

N=mg/cos(θ)

Nsin(θ)=(mv^2)/r

v = root (Nrsin(θ)/m)
v = root (mgsin(θ)/m)
v = root [rgsin(θ)]
v = root [125sin(22)]
v = 6.84m/s

Answer 5

there is always friction with movement no matter what..you cant avoid it