Small, slowly moving spherical particles experience a drag force given by Stokes’ law: where r is the radius of the particle, is its speed, and ç is the coefficient of viscosity of the fluid medium. (a) Estimate the terminal speed of a spherical pollution particle of radius 1.00 x 10-5 m and density of 2000 kg/m3. (b) Assuming that the air is still and that ç is 1.80 x 0-5 N · s/m2, estimate the time it takes for such a particle to fall from a height of 100 m.
(a) To estimate the terminal speed of a spherical pollution particle, we need to determine the drag force acting on the particle using Stokes' law. The drag force is given by the equation:
F_drag = 6πηrv
Where F_drag is the drag force, π is a mathematical constant (approximately 3.14159), η is the coefficient of viscosity of the fluid medium, r is the radius of the particle, and v is the speed of the particle.
However, to determine the terminal speed, we equate the drag force to the weight (mg) of the particle, assuming the particle is in equilibrium and not accelerating. The weight of the particle is given by:
Weight = m * g = πr^3ρ * g
Where m is the mass of the particle, ρ is the density of the particle, and g is the acceleration due to gravity.
Equating the drag force to the weight gives us:
6πηrv = πr^3ρg
Simplifying the equation and solving for v, we have:
v = (2/9) * (r^2 * g) / η
Plugging in the given values:
r = 1.00 x 10^-5 m
ρ = 2000 kg/m^3
η = 1.80 x 10^-5 N · s/m^2
g = 9.8 m/s^2
v = (2/9) * [(1.00 x 10^-5 m)^2 * 9.8 m/s^2] / (1.80 x 10^-5 N · s/m^2)
Calculating the value of v gives us:
v ≈ 1.09 x 10^-3 m/s
Therefore, the estimated terminal speed of the spherical pollution particle is approximately 1.09 x 10^-3 m/s.
(b) To estimate the time it takes for the particle to fall from a height of 100 m, we can use the equation of motion:
v = u + at
Where v is the final velocity (terminal speed), u is the initial velocity (which is assumed to be zero because the particle starts from rest), a is the acceleration (gravity), and t is the time.
Rearranging the equation gives us:
t = (v - u) / a
Plugging in the given values:
v = 1.09 x 10^-3 m/s (terminal speed)
u = 0 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration due to gravity)
t = (1.09 x 10^-3 m/s - 0 m/s) / 9.8 m/s^2
Calculating the value of t gives us:
t ≈ 1.11 x 10^(-4) s
Therefore, it would take approximately 1.11 x 10^(-4) seconds for the particle to fall from a height of 100 m.
(a) To estimate the terminal speed of the pollution particle, we can use Stokes' law equation:
Drag force (F) = 6πηrv
Where:
F = Drag force
η = Coefficient of viscosity of the fluid medium
r = Radius of the particle
v = Speed of the particle (terminal speed)
Since we are looking for the terminal speed, we can assume that the drag force equals the gravitational force acting on the particle:
F = mg
Where:
m = Mass of the particle
g = Acceleration due to gravity (9.8 m/s^2)
We can calculate the mass of the particle using its density and radius:
m = (4/3)πr^3ρ
Substituting the equations together, we have:
6πηrv = (4/3)πr^3ρg
Simplifying the equation:
v = (2/9) (gr^2) (ρ/η)
Now we can plug in the given values:
g = 9.8 m/s^2
r = 1.00 x 10^-5 m
ρ = 2000 kg/m^3
η = 1.80 x 10^-5 N·s/m^2
Calculating the terminal speed:
v = (2/9) (9.8 m/s^2) ((1.00 x 10^-5 m)^2) ((2000 kg/m^3) / (1.80 x 10^-5 N·s/m^2))
v ≈ 2.71 x 10^-4 m/s