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# Effects of rain on a train watch

1. Someone mentioned that they were asked the effects of rain on a train's speed. I spend a while thinking about it, and found it was very complicated.

I worked out that you have to model the rain as a continuous collision with the train, assuming that you reach an equilibrium mass of the train (from rain hitting the train and rain dripping off), and an equilibrium decelleration after rain has been falling fro long enough to make the coefficient of friction between the train and the rain to be independant of the decelleration.

Then I realised it would involve calculus because the rate of change of speed would be partially dependant on the current momentum of the train, which obviously is dependant on the speed.

So say the train and the rain's mass is 10 tonnes, and that the train at this mass is travelling at 30 m s^-1, and that 100kg of water lands on the train per second, (obviouisly this would imply 100kg leave the train, but this is irrelevant because this water has taken some of the train's kinetic energy, as it falls travelling at 30 m s^-1).

It's initial momentum is therefore 300 tonne m s^-1. Now every second imagine there is a collision with the train and 100kg of water with horizontal velocity 0 m s^-1. All we need now is a formula for the train's momentum in terms of t, time. Calling the train's momentum at one point p1, the train's momentum at a point p2, where the train has travelled for 1 more second, is equal to p1 = p2 + 100v, where 100v is the momentum of the second of water that has been transferred from the train.

Argh, so confusing. Can anyone help me, I've been thinking about this for ages, and I think going by momentum is the wrong way, but how else can you model the rain's effect on the horizontal speed of a train other than by collisions?
2. i THINK it is the only way to solve the problem in terms of momentum.
But it is not the collision with the rain droplets ahead but those arriving at the top of the train that matters.
When the droplets hit the top of the train, they are attached and the mass of the whole system, including the train and the droplets attached, increases. By conservation of momentum, the speed of the whole system, and thus the train, decreases
3. (Original post by keisiuho)
i THINK it is the only way to solve the problem in terms of momentum.
But it is not the collision with the rain droplets ahead but those arriving at the top of the train that matters.
When the droplets hit the top of the train, they are attached and the mass of the whole system, including the train and the droplets attached, increases. By conservation of momentum, the speed of the whole system, and thus the train, decreases
I already took the rain on top, but the rain doesn't just store up, as I have said there will be a point where the water dripping off the train will be the same as the water hitting the train.

THe problem comes when you find out the longer the train is being hit by rain, the lower its momentum, and the less momentum is transferred to the water, and the less the train slows down in the future. The train's speed loss decreases with time, but not enough so that the train does not stop.
4. yes, the mass of the water leaving the rain is equal to that arriving.
but the water leaving has virtual the same speed as the train and it means that it has already gained momentum from the train.
on the other hand, the water arriving has no momentum and thus it will also gain momentum.
therefore, although the mass seems to be conserved, the momentum keeps decreasing.

when the friction is also taken in to account, the train will eventually stop
5. Yeah that's what I thought. So the only way the train will not slow down is if the roof of the carridge is frictionless, in which case the water hits it and slides off having gained no momentum from the train.

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Updated: May 9, 2004
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