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

VISCOSITY: Dropping Solid Cylinder

PROBLEM STATEMENT:
A solid cylinder of diameter D, length L, density ρ_s falls due to gravity inside a tube of diameter D₀. The clearance, D₀ - D << D, is filled with a film of viscous fluid. Derive a formula for terminal fall velocity and apply to SAE 30 oil at 20° C for a steel cylinder with D=2 cm, D₀ = 2.04 cm, and L = 15 cm. Neglect the effect of any air in the tube.

SOURCE:
unknown

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PROCESS:

1) Convert all givens to SI units

D = 0.02 m
L = 0.15 m
D₀ = 0.0204 m

2) Look up material properties

ρ_steel = 7859 kg/m³ [source]
ρ_{SAE 30 oil} = 8.815×10^-4 kg/m³ [source]
μ_{SAE 30 oil} = 0.310 N⋅s/m² [source]

3) Analyze the problem & draw a picture

So it asks for us to derive a formula for the terminal fall velocity.
"Terminal velocity is the highest velocity attainable by an object as it falls through a fluid. It occurs when the sum of the drag force and the buoyancy is equal to the downward force of gravity."

4) Pick out appropriate equations & formulas

Shear Stress ≡ τ = μ⋅dv/dy = F/A
Area_circle ≡ A_ο = πr²
Circumference_circle ≡ C_ο = 2πr
Volume_cylinder ≡ V_cyl = πr²⋅L
Mass ≡ m = ρ⋅V
Force ≡ F = m⋅a

5) Plug, Chug, Solve

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

→ Area_{steel cylinder} ≡ A_s = πD²/4 j^

Shear Stress: τ = μ⋅dv/dy
dv/dy = V(y)/clearance → dv/dy = V(y)/(½⋅(D₀ - D))

→ dv/dy = 2⋅V(y)/(D₀ - D)
→ τ = μ⋅2⋅V(y)/(D₀ - D)

Drag: τ = μ⋅dv/dy = F/A
F_drag = ∫_A τ dA
dA refers to the surface area around the outside of the steel cylinder 
dA = circumference⋅dy = 2π(1/2 D)⋅dy

→ dA = πD⋅dy

plugging equations for dA, dv/dy, & τ into F_drag equation
→ F_drag = ∫₀^L μ⋅2⋅V(y)/(D₀ - D)⋅πD⋅dy

→ F_drag = [2π⋅μ⋅D⋅L⋅V(y)]⋅[D₀ - D]^-1​

Volume of steel cylinder: V = πr²⋅L

→ V_s = πLD²/4​

Mass of steel cylinder: m = ρ⋅V

→ m_s = ρ_s⋅πLD²/4​

Gravitational force on steel cylinder: F = ma

→ F_g = ρ_s⋅πLD²/4 ⋅ g​

Sum of forces: +ΣF_y = 0: 0 = F_drag - F_g
→ F_drag = F_g
→ [2π⋅μ_oil⋅D⋅L⋅V(y)]⋅[D₀ - D]^-1 = ρ_s⋅πLD²/4 ⋅ g

⇒ V = ρ_s⋅D⋅g⋅(D₀ - D)/(8μ)

6) Plug in actual values

V = (7859 kg/m³)⋅(0.02 m)⋅(9.81 m/s²)⋅[(0.0204 m) - (0.02 m)]/[8⋅(0.310 N⋅s/m²)]

⇒ V = 0.2487 m/s     [WolframAlpha Solution]

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SOLUTION:
 V = ρ_s⋅D⋅g⋅(D₀ - D)/(8μ)
V = 0.2487 m/s

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TOPICS COVERED: Viscosity, Solid Cylinder, Oil, Density, Terminal Velocity, Shear Stress, Force, Chapter 10