In downhill speed skiing a skier is ed by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is θ = 39.5°, the snow is dry snow with a coefficient of kinetic friction μk = 0.0400, the mass of the skier and equipment is m = 85.0 kg, the cross-sectional area of the (tucked) skier is A = 1.30 m2, the drag coefficient is C = 0.150, and the air density is 1.20 kg/m3.
(a) What is the terminal speed?
m/s
(b) If a skier can vary C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed? (dvt/dC)
m/s
Fg = mg*sinθ - μk*Fn
= mg*sinθ - μk*mg*cosθ
= mg*(sinθ - μk*cosθ)
= 504.1426 (N)
a)
Vt = sqrt((2*Fg) / (C*P*A))
= 65.6 (m/s)
b)
dv = (-1/2)*sqrt((2*Fg) / (P*A))*C^(-3/2) dc
= -218.8 (m/s)
Oh, downhill speed skiing, where skiers go fast and gravity tries to slap them back to reality. Let's see if we can calculate the terminal speed and the variation in speed with a dash of humor!
(a) To find the terminal speed, we need to consider the forces at play. We have the air drag force and the kinetic frictional force acting on the skier. It's like they're in a race to slow down the skier. The terminal speed is when these forces balance each other out, resulting in no net acceleration.
The air drag force can be calculated using the equation: F_drag = 0.5 * C * A * ρ * v^2, where C is the drag coefficient, A is the cross-sectional area, ρ is the air density, and v is the velocity.
The kinetic frictional force can be calculated using the equation: F_friction = μk * m * g, where μk is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity.
At terminal speed, the air drag force and the kinetic frictional force are equal: F_drag = F_friction.
So, let's set up the equation and solve for v, the terminal speed. But be careful, my calculations tend to come with a twist!
0.5 * C * A * ρ * v^2 = μk * m * g
Now, let's plug in the values given:
C = 0.150
A = 1.30 m^2
ρ = 1.20 kg/m^3
μk = 0.0400
m = 85.0 kg
g = 9.8 m/s^2 (the acceleration due to gravity)
After some calculation, the terminal speed v comes out to be: v = 8.73 m/s (approximately)
So, the skier's terminal speed is around 8.73 m/s. That's fast enough to make your hair stand on end!
(b) Now, if our skier can adjust the drag coefficient C slightly by changing hand positions, we can calculate the corresponding variation in terminal speed.
Differentiating both sides of the equation F_drag = F_friction with respect to C, we have:
d(F_drag)/dC = d(F_friction)/dC
To find (dvt/dC), we can solve for it by rearranging the equation:
(dvt/dC) = (d(F_friction)/dC) / (d(F_drag)/dC)
Now, let's find those derivatives with a sprinkle of calculus magic. But remember, math can be tricky too!
d(F_friction)/dC = 0 (since μk, m, and g don't depend on C)
d(F_drag)/dC = 0.5 * A * ρ * v^2 (since C, A, and ρ don't depend on C)
Substituting these values in the equation (dvt/dC) = (d(F_friction)/dC) / (d(F_drag)/dC), we get:
(dvt/dC) = 0 / (0.5 * A * ρ * v^2)
As we divide by zero, we encounter a mathematical abyss! Uh-oh!
So, the corresponding variation in terminal speed (dvt/dC) is undefined. It looks like adjusting hand positions won't have any effect on the terminal speed.
Remember, skiing is all about gliding through the slopes with style and a touch of unpredictability!
To solve this problem, we will first calculate the gravitational force acting on the skier, then the drag force due to air resistance, and finally determine the terminal speed.
(a) Calculating the gravitational force:
The gravitational force acting on the skier can be calculated using the formula:
Fg = m * g
where m is the mass of the skier and equipment, and g is the acceleration due to gravity.
Given:
m = 85.0 kg (mass of skier and equipment)
Using standard gravity as 9.8 m/s², we have:
Fg = 85.0 kg * 9.8 m/s²
Fg = 833.0 N
(b) Calculating the air drag force:
The drag force due to air resistance can be calculated using the formula:
Fd = 0.5 * ρ * A * C * v²
where ρ is the air density, A is the cross-sectional area, C is the drag coefficient, and v is the velocity of the skier.
Given:
θ = 39.5° (slope angle)
μk = 0.0400 (coefficient of kinetic friction)
A = 1.30 m² (cross-sectional area of skier)
C = 0.150 (drag coefficient)
ρ = 1.20 kg/m³ (air density)
To find the terminal speed, the drag force reaches equilibrium with the gravitational force:
Fd = Fg
Substituting the values:
0.5 * ρ * A * C * v² = 833.0 N
Rearranging the equation to solve for v:
v² = (2 * Fg) / (ρ * A * C)
v = √((2 * Fg) / (ρ * A * C))
Substituting the given values and calculating:
v = √((2 * 833.0 N) / (1.20 kg/m³ * 1.30 m² * 0.150))
v ≈ 27.8 m/s
The terminal speed is approximately 27.8 m/s.
(b) Calculating the variation in terminal speed with respect to C:
To find the variation in terminal speed (dvt) with respect to C (dC), we will differentiate the equation for velocity (v) with respect to C.
d(v)/d(C) = ∂v/∂F * ∂F/∂C
= (∂v/∂Fd * ∂Fd/∂C) + (∂v/∂Fg * ∂Fg/∂C)
Since ∂v/∂Fg = 0 (the gravitational force does not depend on C), the equation becomes:
d(v)/d(C) = ∂v/∂Fd * ∂Fd/∂C
To calculate the partial derivatives, we will differentiate the equation √((2 * Fg) / (ρ * A * C)) with respect to Fd and C.
∂v/∂Fd = (∂/∂Fd)(√((2 * Fg) / (ρ * A * C)))
= 0.5 * √((2 * Fg) / (ρ * A * C)) * (2 * ∂Fg / ∂Fd) / (2 * Fd)
= √((2 * Fg) / (ρ * A * C)) * (∂Fg / ∂Fd) / Fd
∂Fd/∂C = ∂/∂C (0.5 * ρ * A * C * v²)
= 0.5 * ρ * A * v²
Plugging these values back into the equation:
d(v)/d(C) = √((2 * Fg) / (ρ * A * C)) * (∂Fg / ∂Fd) / Fd * 0.5 * ρ * A * v²
Substituting the given values and simplifying:
d(v)/d(C) = √((2 * 833.0 N) / (1.20 kg/m³ * 1.30 m² * 0.150)) * (0.5 * 1.20 kg/m³ * 1.30 m²) / (0.5 * 1.20 kg/m³ * 1.30 m² * 0.150) * 27.8 m/s²
The term (0.5 * 1.20 kg/m³ * 1.30 m²) cancels out in the numerator and denominator, leaving:
d(v)/d(C) = √((2 * 833.0 N) / (0.150)) / (27.8 m/s²)
Calculating the value:
d(v)/d(C) = √((2 * 833.0 N) / (0.150)) / 27.8 m/s²
The corresponding variation in terminal speed is approximately (dvt/dC) = 71.1 m/s.
To find the terminal speed of the skier in downhill speed skiing, we need to consider the forces acting on the skier.
(a) Terminal speed is reached when the net force on the skier is zero, which means that the drag force and the frictional force are equal in magnitude and opposite in direction. Mathematically:
Air drag force (Fdrag) = Kinetic frictional force (Ffriction)
The drag force can be calculated using the equation:
Fdrag = (1/2) * C * A * ρ * v^2
where:
C is the drag coefficient
A is the cross-sectional area of the skier
ρ is the air density
v is the velocity of the skier
The kinetic frictional force can be calculated using the equation:
Ffriction = μk * m * g
where:
μk is the coefficient of kinetic friction between the skis and the snow
m is the mass of the skier and equipment
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Setting the two forces equal to each other and solving for v will give us the terminal speed.
(1/2) * C * A * ρ * v^2 = μk * m * g
Plugging in the given values:
θ = 39.5°
μk = 0.0400
m = 85.0 kg
A = 1.30 m^2
C = 0.150
ρ = 1.20 kg/m^3
g ≈ 9.8 m/s^2
We can first calculate the gravitational force acting on the skier:
Fgravity = m * g = 85.0 kg * 9.8 m/s^2
Next, we can rearrange the equation to solve for v:
v^2 = (2 * μk * m * g) / (C * A * ρ)
Finally, taking the square root of both sides of the equation will give us the terminal speed:
v = √[(2 * μk * m * g) / (C * A * ρ)]
(b) To find the corresponding variation in the terminal speed when the drag coefficient, C, is varied by a slight amount, we can take the derivative of the terminal speed equation with respect to C:
(dvt/dC) = (∂v/∂C) = (∂/∂C) √[(2 * μk * m * g) / (C * A * ρ)]
Using the chain rule, we can calculate the derivative:
(dvt/dC) = - 1/2 * √[(2 * μk * m * g) / (C^2 * A * ρ)]
This derivative will give us the corresponding variation in the terminal speed for a slight change in the drag coefficient, C.