Terminal velocity of a skier using a Momentum Balance

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SUMMARY

The discussion focuses on calculating the terminal velocity of a skier with a mass of 100 kg descending a slope at an angle of 30°, while considering factors such as air resistance and snow viscosity. The skier's terminal velocity is influenced by a drag coefficient of 0.2, air density of 1.2 kg/m³, and snow viscosity of 100 mPa*s. The momentum balance equations used include drag force and weight, leading to a complex analysis of forces acting on the skier. The participant identifies the need to incorporate viscous tension and normal force into the calculations for accurate results.

PREREQUISITES
  • Understanding of momentum balance equations
  • Knowledge of drag force calculations
  • Familiarity with fluid dynamics concepts, particularly viscosity
  • Basic physics principles related to forces and motion
NEXT STEPS
  • Calculate terminal velocity using the drag force equation for varying conditions
  • Explore the effects of viscosity on terminal velocity in different fluids
  • Investigate the role of normal force in inclined plane motion
  • Study the impact of varying drag coefficients on terminal velocity
USEFUL FOR

Physics students, engineers, and anyone interested in fluid dynamics and motion analysis, particularly in the context of sports and skiing.

Henrique Orlandini
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Homework Statement


[/B]
A skier (mass M = 100 kg) going down a slope with inclination θ = 30°, sliding in a fluid-like snow (viscosity μ = 100 mPa*s) of thickness h = 0.01 m, using a pair of skis, each one with a surface area of As = 0.15 m2, reaches terminal velocity vt after some distance. Air resistance (density ρ = 1.2 kg/m3) correspond to a drag coefficient (CD = 0.2) which refers to the crouched skier area AD.

a) Dermine the terminal velocity of the skier.
b) Dermine the terminal velocity of the skier without air resistance.
c) Dermine the terminal velocity of the skier without snow viscosity.

Additional info: g = 9.8 m/s2.

Homework Equations


[/B]
Drag force: ##F = C_D*ρ *(v^2/2)*A_D##
Weight: ##F = M*g##
Momentum balance: ##\dot {\mathbf L} = \sum \mathbf F_{ext} + \sum {\dot m} \mathbf v##

The Attempt at a Solution


[/B]
First I set the directions, with downward y and movement-forward x being positive. The I drew the FBD of the problem, with the weight of the skier going downward and drag force going against the movement, along the slope. Then I decomposed the drag force in x and y components, both negative according to my referential.

Using the momentum balance equation, I tried to solve the problem finding ##v_x## and ##v_y## so I could sum them to find ##v_t## in the slope direction.

For ##x## I found: ##\dot {\mathbf L} = -C_D*ρ *({v_x}^2/2)*A_D*sinθ##, in which ##\dot {\mathbf L} = 0## equals zero because we're at terminal velocity, so there is no acceleration.

I found something similar to ##y##, but with weight added: ##\dot {\mathbf L} = -C_D*ρ *({v_x}^2/2)*A_D*cosθ + M*g##, with the same conclusion ##\dot {\mathbf L} = 0##.

But only with those forces, I should get a zero speed in x and an absurdly high speed in y (about 255 m/s), which is clearly incorrect.

I also didn't use some parameters the problem gave me, such as h, μ and A_s, which all are related to "viscous tension", but from what I've read, drag force (and lift force) are the sum of pressure force and viscous tension, so putting tension in this FBD sounds wrong.

I think it's either this or ##\dot {\mathbf L}## isn't really zero, but I really cannot see a acceleration in a terminal velocity scenario.
 
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In your equations you have neglected normal force.
 

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