VISCOSITY: Air-Hockey Puck

PROBLEM STATEMENT:
An air-hockey puck has a mass of 50 g and is 9 cm in diameter. When placed on the air table, a 20° C air film, of 0.12-mm thickness, forms under the puck. The puck is struck with an initial velocity of 10 m/s. Assuming a linear velocity distribution in the air film, how long will it take the puck to 
(a) slow down to 1 m/s?
(b) stop completely?
Also, (c) how far along this extremely long table will the puck have traveled in condition (a)?

SOURCE:
unknown

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PROCESS:

1) Convert all givens to SI units

m = 0.05 kg
D = 0.09 m
T = 1.2×10^-4 m
V(t₀ = 0) = 10 m/s

2) Look up material properties​

ρ_air = 1.205 kg/m³ [source]
μ_air = 1.81×10^-5 N⋅s/m² [source]

3) Analyze the problem & draw a picture​

At some arbitrary time t, there is a linear velocity profile V(t) between the bottom of the puck and the surface of the table. The only force acting on the puck while it is in motion is the drag force. Since this force is against the direction of motion, it is considered negative. This force can be determined by using the shear force equation.

4) Pick out appropriate equations & formulas​

Shear Stress ≡ τ = μ⋅dv/dy = F/A
Area_circle ≡ A_ο = πr²
Force ≡ F = m⋅a

5) Plug, Chug, Solve​

Cross sectional area of hockey puck: (in j^ direction)
A_ο = πr² 
r = D/2

→ Area_puck ≡ A_p = πD²/4 j^​

Shear Stress: τ = μ⋅dv/dy
dv/dy = V(t)/thickness → dv/dy = V(t)/T

→ dv/dy = V(t)/T
→ τ = μ⋅V(t)/T​

Drag: F = τA= μ⋅dv/dy ⋅A

→ F_drag = μ⋅V(t)/T ⋅ πD²/4​

Sum of forces: +ΣF_y = m⋅a_x: 
m⋅a_x = m⋅dV/dt = -F_drag
→ m⋅dV/dt = -(μ⋅V(t)/T ⋅ πD²/4)
→ dV/V(t) = -(μ⋅πD²)/(4mT) dt

integrate...
ln(V(t)) = -(μ⋅πD²)/(4mT)⋅t +C

→V(t) = C⋅e^{-(μ⋅πD2)/(4mT)⋅t}​

V(0) = V₀ ... ∴ C = V₀

⇒ V(t) = V₀⋅e^{-(μ⋅πD2)/(4mT)⋅t}​

Now that velocity function has been derived, the remaining is simple kinematics

(a) Time until V(t) = 1 m/s​

V(t_a) = 1 m/s = V₀⋅e^{-(μ⋅πD2)/(4mT)⋅t_a}

⇒ t_a = [-1⋅ln(1 m/s / V₀)⋅4mT] / [πμD²]​

(b) Time until V(t) = 0 m/s​

V(t_b) = 0 m/s = V₀⋅e^{-(μ⋅πD2)/(4mT)⋅t_b}

⇒ t_b = [-1⋅ln(0 m/s / V₀)⋅4mT] / [πμD²]​

(c) Distance traveled in part (a)​

V = dx/dt
dx = V⋅dt
→ x(t) = ∫V⋅dt
→ x(t) = (-4⋅m⋅T⋅V₀)/(μ⋅π⋅D²)⋅e^{-(μ⋅π⋅D2)/(4mT)⋅t} + C
if x(0) = 0 m.... 0 m = (-4⋅m⋅T⋅V₀)/(μ⋅π⋅D²) + C
→ C = (4⋅m⋅T⋅V₀)/(μ⋅π⋅D²)

⇒ x(t) = [1-e^{-(μ⋅π⋅D2)/(4mT)⋅t}]⋅(4⋅m⋅T⋅V₀)/(μ⋅π⋅D²)​

→ x(t_a) = [1-e^{-(μ⋅π⋅D2)/(4mT)⋅[-1⋅ln(1 m/s / V₀)⋅4mT] / [πμD2]}]⋅(4⋅m⋅T⋅V₀)/(μ⋅π⋅D²)
= (1-e^{ln(1 m/s / V₀)})⋅(4⋅m⋅T⋅V₀)/(μ⋅π⋅D²) 
= (1-(1 m/s)/( V₀)⋅(4⋅m⋅T⋅V₀)/(μ⋅π⋅D²) 

⇒ x(t_a) = [V₀ - (1 m/s)]⋅(4⋅m⋅T)/(μ⋅π⋅D²)​

6) Plug in actual values​

(a)
t_a = [-1⋅ln[1 m/s / (10 m/s)]⋅4⋅(0.05 kg)⋅(1.2×10^-4 m)] / [π⋅(1.81×10^-5 N⋅s/m²)⋅(0.09 m)²]

⇒ t_a = 120 s [WolframAlpha Computation]​

(b)
t_b = [-1⋅ln[0 m/s / (10 m/s)]⋅4⋅(0.05 kg)⋅(1.2×10^-4 m)] / [π⋅(1.81×10^-5 N⋅s/m²)⋅(0.09 m)²]

⇒ t_b = ∞ [WolframAlpha Computation]​

Since this is a logarithmic function, it will never actually reach zero. However, if the settling time is considered, such as that defined by harmonic motion, t_settling = time it takes for velocity to reach 2% of initial velocity
(2%)⋅V₀ = 0.2 m/s
Therefore, the time required for it to stop comlpetely can be approimated to about 203.9 s. [WolframAlpha Computation]

(c)
x(t_a) = [(10 m/s) - (1 m/s)]⋅(4⋅(0.05 kg)⋅(1.2×10^-4 m))/(π⋅(1.81×10^-5 N⋅s/m²)⋅(0.09 m)²)

⇒ x(t_a) = 469 m [WolframAlpha Computation]​

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SOLUTION:
(a)
 t_a = [-1⋅ln(1 m/s / V₀)⋅4mT] / [πμD²]
 t_a = 120 s 

(b)
t_b = [-1⋅ln(0 m/s / V₀)⋅4mT] / [πμD²]​
 t_b = ∞

(c)
x(t_a) = [V₀ - (1 m/s)]⋅(4⋅m⋅T)/(μ⋅π⋅D²)​
x(t_a) = 469 m
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TOPICS COVERED: Viscosity, Puck, Air, Shear Stress, Force, Chapter 10, Velocity, Function