a rain drop falls from rest in an atmosphere saturated with water vapour. As it falls, water vapour condenses on drop at rate of mass µ per second. If initial mass of drop is M0, how much distance the drop falls in time t?
Why would adding mass increase the drop speed (ignoring air friction)? Didn't Galileo demonstrate falling objects of different mass fall at the same rate?
d= 1/2 g t^2, it is mass independent (ignoring air friction).
To find the distance the raindrop falls in time t, we need to consider the forces acting on the raindrop.
When a raindrop falls through the atmosphere, it experiences two main forces: the gravitational force pulling it downward and the drag force acting in the opposite direction, which is related to air resistance. The drag force depends on the speed of the falling raindrop and can be calculated using Stokes' law.
However, in this case, we also need to consider the effect of water vapor condensing on the raindrop. As water vapor condenses onto the raindrop, its mass increases, which in turn affects the force of gravity acting on it.
Let's break down the problem step by step:
Step 1: Calculate the mass of the raindrop at any given time, t.
The rate of mass increase, µ, can be written as:
µ = dm/dt
Where dm/dt is the rate of change of mass with respect to time.
Therefore, dm = µ * dt
Integrating both sides gives us:
∫dm = ∫µ * dt
m - M0 = ∫µ * dt
m = M0 + ∫µ * dt
Step 2: Calculate the acceleration of the raindrop using Newton's second law.
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
F_net = m * a
In our case, the net force is the gravitational force minus the drag force:
F_net = mg - D
Where:
m = mass of the raindrop
g = acceleration due to gravity (approximately 9.8 m/s^2)
D = drag force
Step 3: Calculate the drag force using Stokes' law.
According to Stokes' law, the drag force acting on a spherical object in a viscous fluid is given by:
D = 6πηrν
Where:
η = dynamic viscosity of the fluid (air in this case)
r = radius of the raindrop
ν = velocity of the raindrop
We can approximate the velocity of the raindrop as v ≈ gt, assuming it has been falling for a sufficiently long time and has reached terminal velocity.
Step 4: Calculate the radius of the raindrop.
The volume of a raindrop can be calculated using:
V = (4/3)πr^3
The mass of the raindrop is related to its volume and density:
m = ρV
Where:
ρ = density of the raindrop (assumed to be constant)
Step 5: Combine the equations to calculate the distance the raindrop falls in time t.
Using the equations from steps 1 to 4, we can now calculate the mass of the raindrop at time t and the forces acting on it. From there, we can use kinematic equations to determine the distance it falls.
Please note that this calculation may involve assumptions and approximations. If you provide specific values for M0, µ, and any other necessary parameters, I can assist you further with the calculations.
To find the distance the raindrop falls in time t, we need to consider the forces acting on the raindrop as it falls and the change in its mass due to water vapor condensing.
Let's assume that the raindrop falls vertically downwards.
1. Initial situation:
At the beginning, the raindrop is at rest in the atmosphere saturated with water vapor (100% relative humidity). Its initial mass is M0.
2. Forces acting on the raindrop:
As the raindrop falls, it experiences two main forces: gravity and air resistance.
a) Gravity (weight): The raindrop is subject to the force of gravity, which pulls it downward. The force of gravity is given by F_gravity = m * g, where m is the mass of the raindrop and g is the acceleration due to gravity (approximately 9.8 m/s²).
b) Air resistance: As the raindrop falls through the atmosphere, it experiences air resistance, which opposes its motion. The exact calculation of air resistance is complex and depends on various factors such as the shape of the raindrop, its velocity, and the density of the air. However, we can assume that the effect of air resistance is negligible in this case, as the raindrop is relatively small compared to the forces of gravity and vapor condensation.
3. Vapor condensation:
As the raindrop falls through the atmosphere, water vapor condenses on its surface at a rate of mass µ per second. This means that the mass of the raindrop increases over time by µt (where t is the time in seconds).
4. Equation of motion:
Using Newton's second law of motion, we can write the equation of motion for the raindrop:
F_net = F_gravity + F_air_resistance = m * a
Since we assumed that the air resistance is negligible, the net force acting on the raindrop is equal to the force of gravity:
F_net ≈ F_gravity = m * g
5. Change in mass:
The mass of the raindrop changes due to the vapor condensation. The change in mass over time t is given by: ∆m = µt
6. Solving the equation of motion:
To find the distance the raindrop falls in time t, we need to solve the equation of motion for distance.
Using the equations from step 4 and the relationship between force, mass, and acceleration (F = m * a), we can substitute for force and rearrange the equation:
F_net = m * g
m * a = m * g
a = g
Since the acceleration is constant (equal to the acceleration due to gravity), the equation of motion becomes:
s = ut + (1/2) * a * t²
Where s is the distance the raindrop falls, u is the initial velocity (which is 0 since the raindrop starts from rest), and t is the time.
Substituting the values into the equation:
s = 0 * t + (1/2) * g * t²
s = (1/2) * g * t²
Therefore, the distance the raindrop falls in time t is (1/2) * g * t².