In downhill speed skiing a skier is retarded by both the air drag force on the b

Question

In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. The equation for the air drag D is given by D = 1/2-C rhoA upsilon2 A is the effective cross-sectional area of a sky diver in a certain position, upsilon is the relative speed of the sky diver with respect to the air. and C is a drag coefficient. Suppose the slope angle is theta, the snow is dry snow with a coefficient of kinetic friction mu k, the mass of the skier and equipment is m (kg), the cross-sectional area of the (tucked) skier is A m2, the drag coefficient is C, and the air density is rho kg/m3. What is the terminal speed? Consider the balance of forces on the skier. Your answer should be in terms of all the variables given above. Note all the units are in SI so you can ignore the units for now. You can only do so when they are in same system. If a skier can vary the drag coefficient C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation (d upsilon) in the terminal speed? Answer this problem algebraically in the form of d upsilon = KdC and figure out what K is in terms of upsilon and C.

Explanation / Answer

a)balancing forces,
mgsin=mgcos(frictional force) + 1/2 CAv^2(air drag)

or v=(2mg(sin-cos)/CA)^1/2

b)v=(2mg(sin-cos)/A)^1/2*C^-1/2

or dv/dc=(-1/2)(2mg(sin-cos)/A)^1/2*C^-3/2

or dv/dc=(-1/2)(v/c)

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