Spherical raindrop, mass, radius, and time

In summary: Not quite.M(t) = ρ(4/3)π*r3 at t=0, where ρ is the mass of the drop, r is its radius, and t is time.
  • #1
Raquel Aduriz
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Homework Statement



Separate variables and integrate to find an expression for r(t), given r0 at t=0

Homework Equations



M=ρ(4/3)πr3, thus V=(4/3)πr3

dM/dt=Cr3 where C is a constant

The Attempt at a Solution



∫dM=∫Cr3dt

M+constant=??

I have no idea how to integrate r because it's a function of t but we're not given the function. I don't think that integrating the left side this way will be very helpful either. Any advice would be much appreciated!
 
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  • #2
Raquel Aduriz said:

Homework Statement



Separate variables and integrate to find an expression for r(t), given r0 at t=0

Homework Equations



M=ρ(4/3)πr3, thus V=(4/3)πr3

dM/dt=Cr3 where C is a constant

The Attempt at a Solution



∫dM=∫Cr3dt

M+constant=??

I have no idea how to integrate r because it's a function of t but we're not given the function. I don't think that integrating the left side this way will be very helpful either. Any advice would be much appreciated!
You want to express M in terms of r. Once you do that, then dM(r) / dt = Cr3, and you can separate the variables to get r on one side and dt on the other.
 
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  • #3
Ok... I guess I don't really understand how to do that. Can I do that using just the M equation, or do I need something else?
 
  • #4
Raquel Aduriz said:
Ok... I guess I don't really understand how to do that. Can I do that using just the M equation, or do I need something else?
You gave an expression for the mass of the raindrop in terms of the radius right there in Section 2 of the template. It's the equation M = ...
 
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  • #5
Alright, so I have M=ρ(4/3)πr3. What exactly do I do with this? Take the derivative and set it equal to the other? Sorry, I'm sure I'm missing something really obvious.
 
  • #6
Raquel Aduriz said:
Alright, so I have M=ρ(4/3)πr3. What exactly do I do with this? Take the derivative and set it equal to the other? Sorry, I'm sure I'm missing something really obvious.
This problem is about solving a differential equation by separation of variables. Have you studied how to do this yet?
 
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  • #7
Yes, but a few years ago so maybe I've forgotten something key. So I have C*r*dt=ρ*4*π*dr. Is that correct?
 
  • #8
Raquel Aduriz said:
Yes, but a few years ago so maybe I've forgotten something key. So I have C*r*dt=ρ*4*π*dr. Is that correct?
Not quite.

Presumably, as the rain drop falls, it gets bigger as it collects more moisture; therefore, the radius of the drop grows as time passes.

Since the drop is spherical, M(t) = ρ(4/3)π*[r(t)]3, where M and r are written as functions of time.

You are also given the condition that the change in the mass of the drop, dM(t)/dt, at any given time is proportional to the cube of the radius of the drop, or dM(t)/dt = Cr3. At t = 0, r(t) = r(0) = r0.

You should check your differentiation of M(t) w.r.t. time. You might have to use the chain rule here, since M is a function of r, but r is a function of t.
 
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  • #9
I got it! Can't believe I forgot about the chain rule. Thank you so much for your help!
 

Related to Spherical raindrop, mass, radius, and time

1. What is a spherical raindrop?

A spherical raindrop is a perfectly round droplet of water that falls from the sky as part of precipitation.

2. How is the mass of a spherical raindrop determined?

The mass of a spherical raindrop is determined by its density and volume. The formula for calculating mass is: mass = density x volume.

3. Does the mass of a raindrop affect its size?

Yes, the mass of a raindrop is directly proportional to its size. This means that as the mass increases, so does the size of the raindrop.

4. How does the radius of a raindrop affect its fall time?

The radius of a raindrop has a significant influence on its fall time. According to the Stoke's law, larger raindrops fall faster than smaller raindrops due to their higher mass and surface area. Therefore, larger raindrops have a shorter fall time compared to smaller ones.

5. How long does it take for a spherical raindrop to reach the ground?

The time it takes for a spherical raindrop to reach the ground depends on various factors such as air resistance, wind speed, and the size of the raindrop. On average, it takes about 5-10 minutes for a raindrop to reach the ground.

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