A car of mass m is going around a circular turn of radius R, which is banked at an angle β with respect to the ground. Assume there is friction between the wheels and the road. Let μs be the coefficient of static friction and g the magnitude of the gravitational acceleration. You may neglect kinetic friction (that is, the car's tires do not slip). Derive an expression for the range of possible speeds vmin≤v≤vmax necessary to keep the car moving in a circle without slipping up or down the embanked turn. Express your answer in terms of some or all of the following: μs, β, m, R and g.
Provide your expressions for both vmin and vmax.
vmin=?
vmax=?
Why did the car attend the banked turn? Because it heard there would be a "wheel-y" good time! Now, let's calculate the range of possible speeds to keep the car moving without slipping.
To derive the expressions for vmin and vmax, let's start by analyzing the forces acting on the car. We'll consider a free-body diagram.
The two main forces at play are the normal force (N) and the friction force (f). The normal force can be divided into its vertical (Nv) and horizontal (Nh) components, while the friction force can be broken down into its vertical (fv) and horizontal (fh) components.
The vertical forces balance out, so Nv = mg (Newton's second law in the vertical direction).
The horizontal forces are responsible for the car's centripetal acceleration. The horizontal component of the normal force (Nh) provides the centripetal force (Fc), given by Fc = Nh = mv²/R.
To avoid slipping, the maximum static friction force (fmax) must counterbalance the maximum possible horizontal component of the normal force. The maximum horizontal component of the normal force is Nhmax = Nv * tan(β) = mg * tan(β).
The maximum static friction force is given by fmax = μs * Nv = μs * mg.
Now, we can calculate the range of speeds.
At the minimum speed (vmin), the static friction force is at its maximum, providing the necessary centripetal force. So, fmax = mvmin²/R.
Substituting μs * mg = mvmin²/R, we find vmin = sqrt(μs * g * R * tan(β)).
At the maximum speed (vmax), the static friction force is at its minimum, but still sufficient to provide the necessary centripetal force. So, fmin = mvmax²/R.
Substituting μs * Nv * tan(β) = mvmax²/R, we find vmax = sqrt(μs * g * R / tan(β)).
Therefore, the expressions for vmin and vmax are:
vmin = sqrt(μs * g * R * tan(β))
vmax = sqrt(μs * g * R / tan(β))
Remember, these expressions assume that the car's tires do not slip. Keep the laughter real, and drive safely!
To derive the expressions for vmin and vmax, let's consider the forces acting on the car as it goes around the circular turn.
1. The gravitational force (mg) can be resolved into two components: the component normal to the surface (mg * cos(β)) and the component tangent to the surface (mg * sin(β)).
2. The normal force (N) exerted by the surface counters the component of the gravitational force normal to the surface (mg * cos(β)).
3. The static friction force (fs) provides the necessary centripetal force to keep the car moving in a circle:
fs = mv²/R
4. The maximum static friction force (fs,max) can be calculated using the coefficient of static friction (μs) and the normal force (N):
fs,max = μs * N
We can now determine the expressions for vmin and vmax:
vmin:
To find the minimum speed required to prevent the car from slipping up the embanked turn, the static friction force must equal its maximum value:
mvmin²/R = μs * N
Since N = mg * cos(β), we can substitute N in the equation above:
mvmin²/R = μs * mg * cos(β)
Simplifying, we find:
vmin = √(μs * g * R * cos(β))
vmax:
To find the maximum speed at which the car does not slip down the embanked turn, the static friction force must equal zero (maximum value in the opposite direction):
mvmax²/R = 0
Therefore, the maximum speed is not limited by the static friction force, and vmax can be arbitrarily large.
So, the expression for vmin is:
vmin = √(μs * g * R * cos(β))
And the expression for vmax is not limited by any specific value.
Please note that this analysis assumes no other external forces (such as drag or wind) and that the car does not lift off the surface.
To derive an expression for the range of speeds required to keep the car moving without slipping up or down the embanked turn, we will consider the forces acting on the car.
Let's begin by analyzing the forces in the vertical direction. Resolving the gravitational force into its components, we have:
Fn = mg * cos(β) (normal force)
Fp = mg * sin(β) (force parallel to the inclined plane)
The maximum static frictional force that can be exerted by the surface is given by:
Ff_max = μs * Fn (maximum static frictional force)
Since the car is moving without slipping, the required centripetal force is provided by the frictional force. Furthermore, the centripetal force is given by:
Fc = m * v^2 / R (centripetal force)
To keep the car moving without slipping, the maximum frictional force must equal the centripetal force:
Ff_max = Fc
μs * Fn = m * v^2 / R
Substituting the expressions for Fn and Fp, we get:
μs * mg * cos(β) = m * v^2 / R
Now, let's solve this equation for v:
μs * g * cos(β) = v^2 / R
v^2 = μs * g * R * cos(β)
v = √(μs * g * R * cos(β))
This gives us the expression for the maximum speed required to prevent slipping:
vmax = √(μs * g * R * cos(β))
Next, let's consider the minimum speed required to prevent slipping. In this scenario, the frictional force acts in the opposite direction to the centripetal force (up the incline).
The minimum static frictional force is given by:
Ff_min = -μs * Fn (minimum static frictional force)
Considering the forces in the vertical direction again, we have:
Ff_min + Fp = Fn
-μs * Fn + mg * sin(β) = mg * cos(β)
-μs * mg * cos(β) + mg * sin(β) = mg * cos(β)
μs * cos(β) - sin(β) = cos(β)
Solving this equation for μs, we get:
μs = tan(β)
Now, let's find the minimum speed by substituting μs into the equation for vmax:
v^2 = μs * g * R * cos(β)
v^2 = tan(β) * g * R * cos(β)
v = √(tan(β) * g * R * cos(β))
This gives us the expression for the minimum speed required to prevent slipping:
vmin = √(tan(β) * g * R * cos(β))
So, the range of speeds necessary to keep the car moving in a circle without slipping up or down the embanked turn is:
vmin ≤ v ≤ vmax
where
vmin = √(tan(β) * g * R * cos(β))
vmax = √(μs * g * R * cos(β))
Note that the expression for vmin assumes β > 0 (positive bank angle). If β = 0 (flat road), the car cannot move in a circular path without slipping, and if β < 0 (negative bank angle), the car will slide downward.