speedway turn, with radius of curvature R, is banked at an angle θ above the horizontal.
(a) What is the optimal speed at which to take the turn if the track's surface is iced over (that is, if there is very little friction between the tires and the track)? (Use any variable or symbol stated above along with the following as necessary: g.)
vzero friction =
(b) If the track surface is ice-free and there is a coefficient of friction μs between the tires and the track, what are the maximum and minimum speeds at which this turn can be taken? (Use any variable or symbol stated above along with the following as necessary: g.)
vmin =
vmax =
(c) Evaluate the results of parts (a) and (b) for R = 365 m, θ = 45°, and μs = 0.64.
vzero friction =
vmin =
vmax =
a) you want the resultant of the force vectors mg (vertical) and mv^2/r (horizontal) to be normal to Theta.
TanTheta=v^2/rg Or alternative, you want the component of force mg down the track equal to the sliding force centripetal force.
mv^2/r *cosTheta=mgsinTheta
and then tantheta=v^2/rg
b) Now one has an additional force, Parallel to the track,friction.
so, mgSinTheta-mu*mgCosTheta=mv^2/r cosTheta
or g Tantheta= mu*g+v^2/r
solve for v, in this case vmax
Now, vmin means friction is preventing it going down, or g TanTheta=-mu*g+v^2/r
is this how we solve for v for part a and b .
a ) v=sqrt ( g r / tan theta)
b) v min= sqrt ( r tan theta + r mu g )
v max= sqrt ( r tan theta - r mu g )
The book gives an answer for part a below.
v = sqrt (rg*cos theta*sin theta)
I assume that's incorrect?
Hey Bob, I think you need to check your algebra. As the student substituted 2 posts up vmin is bigger than vmax.
To find the optimal speed at which to take the turn when the track's surface is iced over and there is very little friction, we need to consider the forces acting on the car.
(a) First, we need to analyze the forces acting on the car. There are two main forces to consider: the gravitational force (mg) and the normal force (N) exerted by the track surface on the car.
The component of the gravitational force perpendicular to the track surface is given by mg * cos(θ), where θ is the angle at which the track is banked. This force provides the necessary centripetal force for the car to take the turn.
The net force acting on the car is given by the difference between the centripetal force and the frictional force (which is negligible in this case due to the ice).
Therefore, we have:
mg * cos(θ) = mv²/R,
where m is the mass of the car, v is the velocity of the car, R is the radius of curvature of the turn, and g is the acceleration due to gravity.
To find the optimal speed, we can solve this equation for v:
v = √(R * g * cos(θ)).
Substituting the given values for R = 365 m, θ = 45°, and g ≈ 9.8 m/s², we can calculate the optimal speed when there is no friction:
vzero friction = √(365 * 9.8 * cos(45°)).
(b) When there is a coefficient of friction (μs) between the tires and the track, we need to consider the maximum and minimum speeds at which the turn can be taken.
The maximum speed occurs when the frictional force (f) just reaches its maximum value before the car starts sliding off the track. The maximum frictional force is given by μs * N, where N is the normal force exerted by the track surface on the car. The normal force is equal to mg * cos(θ), as before.
For the maximum speed:
μs * N = mvmax²/R,
Substituting N = mg * cos(θ) and solving for vmax:
vmax = √(μs * R * g * cos(θ)).
The minimum speed occurs when the frictional force is at its minimum value, which is zero. The car can still take the turn at this minimum speed without sliding off due to the banking of the track.
For the minimum speed, the car experiences the gravitational force (mg) as the only force acting to provide the necessary centripetal force:
mg * cos(θ) = mvmin²/R,
Solving for vmin:
vmin = √(R * g * cos(θ)/μs).
(c) Substituting R = 365 m, θ = 45°, and μs = 0.64 into the formulas derived in parts (a) and (b), we can calculate the results:
vzero friction = √(365 * 9.8 * cos(45°)),
vmin = √(365 * 9.8 * cos(45°)/0.64),
vmax = √(0.64 * 365 * 9.8 * cos(45°)).
Evaluating these expressions will give us the values for vzero friction, vmin, and vmax.