The change of mass with respect to time of a rain drop falling though a cloud

In summary, the raindrop falls through the cloud and collects water. The rate of accretion is proportional to the cross-sectional area of the drop multiplied by the speed of fall. This equation implies that the acceleration of the raindrop is constant and equal to g/7.
  • #1
AaronKnight
10
0

Homework Statement


A raindrop is falling through a cloud, collecting water as it falls. If the raindrop has mass
m and velocity v, show that
mg = vdm/dt +mdv/dt.
Assume that the drop remains spherical and that the rate of accretion is proportional to the
cross-sectional area of the drop multiplied by the speed of fall. Show that this implies that
dm/dt = kvm2/3,
for some constant k. Using this equation and the assumption that the drop starts from rest
when it is infinitesimally small, find m as a function of x. Hence show that the acceleration
of the raindrop is constant and equal to g/7.


I have managed to show that mg = vdm/dt +mdv/dt easily, but I am at a lose for the second part where the question asks "Show that this implies that dm/dt = kvm2/3".
I am unsure where to begin, any help in pointing me in the right direction would be greatly appreciated.
Thank you.
 
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  • #2
hi aron, the rate of accretion is the rate at which mass is being put on the drop. so the rate of accretion is [itex]\frac{dm}{dt}[/itex]. The area of cross section of the sphere is [itex]\pi r^2 [/itex] , where r is the radius. So we are given that

[tex]\frac{dm}{dt} = k_1 v\; \pi r^2[/tex]

where k1 is just the constant of proportionality. now since the density of the drop remains constant, use that fact to write r2 in terms of density and the mass of the drop. whatever constants you encounter, just merge it and call it k.
 
  • #3
Thank you, it was the last step of writing r2 in terms of the mass and density that I was missing out.
 
  • #4
Hi. Sorry I know this thread has already been answered. Just one quick question: How do you know that the density of the drop remains constant?
 
  • #5
Well, the density of rainwater is constant.
 
  • #6
fair enough thanks.
 
  • #7
post full solution after you are done.
 

Related to The change of mass with respect to time of a rain drop falling though a cloud

1. How does the mass of a rain drop change as it falls through a cloud?

The mass of a rain drop does not change as it falls through a cloud. The drop is simply suspended in the cloud until it becomes heavy enough to overcome the air resistance and fall to the ground.

2. Why does the size of a rain drop decrease as it falls through a cloud?

The size of a rain drop decreases as it falls through a cloud due to evaporation. As the drop falls through the cloud, it encounters warmer and drier air which causes the water molecules in the drop to evaporate, resulting in a smaller drop.

3. Does the change in mass of a rain drop affect its speed of falling through a cloud?

No, the change in mass of a rain drop does not affect its speed of falling through a cloud. The drop's speed is determined by the force of gravity and air resistance, which are not affected by the drop's mass.

4. How does the temperature of a cloud affect the change in mass of a falling rain drop?

The temperature of a cloud does not directly affect the change in mass of a falling rain drop. However, a warmer cloud may cause the drop to evaporate more quickly, resulting in a larger decrease in mass.

5. Is the change in mass of a rain drop consistent for all clouds?

No, the change in mass of a rain drop can vary depending on the type and temperature of the cloud. Factors such as wind speed, humidity, and temperature can also affect the rate of evaporation and therefore the change in mass of a falling rain drop.

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