Suppose a gust of wind has carried a 52-μm-diameter dust particle to a height of 340 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 kg/m^3, the viscosity of 25∘C air is 2.0×10−5N⋅s/m^2, and you can assume that the falling dust particle reaches terminal speed almost instantly.
First, let's find the terminal speed of the dust particle. Terminal speed is reached when the force of gravity on a particle equals the drag force acting on it.
For a spherical particle falling through a fluid, we can use Stokes' Law to find the drag force. Stokes' Law states that the force of gravity on the particle is given by:
F = 6πηrv
Where F is the drag force, η is the fluid viscosity, r is the radius of the sphere, and v is the terminal speed of the particle.
We also know the force of gravity on the particle, given by:
F_gravity = mg
Where m is the mass of the particle and g is the acceleration due to gravity.
We can find the mass of the particle using its density:
m = Vρ
Where V is the volume of the particle and ρ is its density. For a sphere, the volume is given by:
V = (4/3)πr^3
So, the mass is:
m = (4/3)πr^3ρ
Now, substituting the mass back into the force of gravity equation, we get:
F_gravity = (4/3)πr^3ρg
Now, we set the force of gravity equal to the drag force and solve for the terminal speed:
6πηrv = (4/3)πr^3ρg
v = (2/9)(r^2ρg/η)
Now we can plug in the given values:
r = 52 μm/2 = 26 μm (convert diameter to radius)
ρ = 2700 kg/m³ (density of dust)
g = 9.81 m/s² (acceleration due to gravity)
η = 2.0 × 10⁻⁵ N⋅s/m² (viscosity of air)
v = (2/9)((26 × 10⁻⁶ m)² × (2700 kg/m³)(9.81 m/s²)/(2.0 × 10⁻⁵ N⋅s/m²))
v ≈ 0.0339 m/s
Now that we have the terminal speed, we can use the equation for speed to find the time it takes for the particle to fall:
distance = speed × time
time = distance/speed
Plug in the given distance (height) and calculated terminal speed:
time = (340 m)/(0.0339 m/s)
time ≈ 10028 s
It will take about 10028 seconds, or about 2.8 hours, for the dust particle to settle back to the ground.
To find the time it takes for the dust particle to settle back to the ground, we need to determine its terminal velocity first.
Terminal velocity is the constant speed reached by an object falling through a fluid, such as air, when the force of gravity is balanced by the drag force from the fluid.
The drag force acting on the dust particle can be calculated using Stokes' Law:
Fd = 6πηrv
where Fd is the drag force, η is the viscosity of the air, r is the radius of the dust particle, and v is the velocity of the particle.
The gravitational force acting on the dust particle can be calculated using its weight:
Fg = mg
where Fg is the force of gravity, m is the mass of the dust particle, and g is the acceleration due to gravity.
At terminal velocity, the drag force is equal to the gravitational force:
Fd = Fg
Substituting the expressions for the drag force and gravitational force:
6πηrv = mg
We can rearrange this equation to solve for v:
v = (mg) / (6πηr)
Now, let's calculate the mass of the dust particle:
The density of the dust is given as 2700 kg/m^3. The volume of the dust particle can be calculated using its diameter:
V = (4/3)πr^3
Substituting the diameter (52 μm) into the equation, we can find the radius:
r = 52 μm / 2 = 26 μm = 26 × 10^-6 m
Substituting the values of density and volume into the equation for mass:
m = ρV = 2700 kg/m^3 × (4/3)π(26 × 10^-6 m)^3
Now, we have all the values we need to calculate the terminal velocity:
v = (mg) / (6πηr)
Calculating the terminal velocity using the given values:
v = (2700 kg/m^3) × (4/3)π(26 × 10^-6 m)^3 × 9.8 m/s^2 / (6π(2.0 × 10^-5 N⋅s/m^2)(26 × 10^-6 m))
Simplifying the expression, we find:
v ≈ 0.005 m/s
Now that we know the terminal velocity, we can calculate the time it takes for the dust particle to settle back to the ground.
The distance the dust particle needs to cover is given as 340 m. The time it takes for an object to fall a certain distance can be calculated using the equation:
d = (1/2)gt^2
where d is the distance, g is the acceleration due to gravity, and t is the time.
Rearranging the equation to solve for t:
t = √(2d / g)
Substituting the given values:
t = √(2 × 340 m / 9.8 m/s^2)
Calculating the time, we find:
t ≈ 6.5 s
Therefore, it will take approximately 6.5 seconds for the dust particle to settle back to the ground.