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Very small objects, such as dust particles, experience a
Chapter 6, Problem 66P(choose chapter or problem)
Very small objects, such as dust particles, experience a linear drag force, \(\vec{D}=\)( \(b v\), direction opposite the motion), where \(b\) is a constant. That is, the quadratic model of drag of Equation 6.16 fails for very small particles. For a sphere of radius \(R\), the drag constant can be shown to be \(b=6 \pi \eta R\), where \(\eta\) is the viscosity of the gas.
a. Find an expression for the terminal speed \(v_{\text {tem }}\) of a spherical particle of radius \(R\) and mass \(m\) falling through a gas of viscosity \(\eta\).
b. Suppose a gust of wind has carried a 50- \(\mu\) m-diameter dust particle to a height of \(300 \mathrm{~m}\). If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density of \(2700 \mathrm{~kg} / \mathrm{m}^{3}\), the viscosity of \(25^{\circ} \mathrm{C}\) air is \(2.0 \times 10^{-5} \mathrm{Ns} / \mathrm{m}^{2}\), and you can assume that the falling dust particle reaches terminal speed almost instantly.
Equation Transcription:
Text Transcription:
vec{D}= ( b v
b
R
b=6 pi eta R
eta
v_{\text {tem }}
R
m
eta
mu
300 m
2700 kg/}^3
25^circ C
2.0 X 10^{-5} Ns/m^2
Questions & Answers
QUESTION:
Very small objects, such as dust particles, experience a linear drag force, \(\vec{D}=\)( \(b v\), direction opposite the motion), where \(b\) is a constant. That is, the quadratic model of drag of Equation 6.16 fails for very small particles. For a sphere of radius \(R\), the drag constant can be shown to be \(b=6 \pi \eta R\), where \(\eta\) is the viscosity of the gas.
a. Find an expression for the terminal speed \(v_{\text {tem }}\) of a spherical particle of radius \(R\) and mass \(m\) falling through a gas of viscosity \(\eta\).
b. Suppose a gust of wind has carried a 50- \(\mu\) m-diameter dust particle to a height of \(300 \mathrm{~m}\). If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density of \(2700 \mathrm{~kg} / \mathrm{m}^{3}\), the viscosity of \(25^{\circ} \mathrm{C}\) air is \(2.0 \times 10^{-5} \mathrm{Ns} / \mathrm{m}^{2}\), and you can assume that the falling dust particle reaches terminal speed almost instantly.
Equation Transcription:
Text Transcription:
vec{D}= ( b v
b
R
b=6 pi eta R
eta
v_{\text {tem }}
R
m
eta
mu
300 m
2700 kg/}^3
25^circ C
2.0 X 10^{-5} Ns/m^2
ANSWER:
Step 1 of 5
a.)
We have to find an expression for the terminal speed \(v_{\text {term }}\) of a spherical particle of radius \(R\) and mass \(m\) falling through a gas of viscosity \(\eta\).
The speed at which the exact balance between the upward drag force and the downward gravitational force causes an object to fall without acceleration is called the terminal speed
\(v_{\text {term }}\).
\(D=F_{G}\)
Where,
\(D=\) Drag force \(F_{G}=\) gravitational force