Imagine water striking a water wheel. We can find the torque applied by calculating the force of the water and multiplying by the distance from the center.

To calculate force, we can use the formula F =$\ \frac{dp}{dt}$ .

However, I am confused about change in momentum. More specifically, when water hits the wheel, what will be the change in momentum? How can this be calculated?

To make the question easier, let's assume that a large drop of a certain mass m hits a stationary water wheel. Do we consider the water drop's final velocity to be zero at the point of impact? If not, how can we conceptually and mathematically determine the change in momentum, without experimenting?

Now, how about if the water wheel is already rotating. How can we conceptually imagine what the change in momentum would be (I assume lower than when the water wheel is stationary), and how can we mathematically calculate this if given real values?

It is possible that there are other factors, such as static friction of water wheel or mass of water wheel, however, I am not sure what is needed to be accounted for, so if anyone can improve this problem, please comment of edit.

Thanks for any help.

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    $\begingroup$ volumes have been written on the dynamic analysis of waterwheels and turbines like this. A good starting point is wikipedia. suggest also that you migrate this to the engineering stack exchange, lots of experts there. $\endgroup$ Aug 11 '18 at 5:33

The change in momentum is simply the difference between the momentum of the system before the event and the momentum of the system afterwards. Don't overthink it =)

The "correct" change in momentum depends on the exact shape of the water wheel, so your mileage may vary. However, a very reasonable first pass would be to assume the water "sticks" to the waterwheel. In this case, we can assume that the momentum of the water after the collision is exactly equal to the mass of the water multiplied by the tangental velocity of the wheel at the point where the water stuck.

However, given that you have a wheel rotating around a fixed point, it is easier to think in angular momentum. Angular momentum is also conserved, and the math is simpler. If you try to do this with just momentum, you'll have to consider how the forces are transmitted to the axel of the waterwheel. It'll yield the same answer, but using just linear momentum will require more math.

From this, we can put together a rough sketch of how a water wheel starts spinning. Start with the wheel at a standstill. One drop falls onto the wheel. Before the drop falls, the angular momentum of the drop is its mass multiplied by its downward velocity multiplied by its distance from the wheel (this is the definition of the angular momentum of a point mass), and the momentum of the waterwheel is zero. After the drop falls and "sticks" to the wheel, the momentum of the entire system is the angular velocity of the wheel after the collision multiplied by the moment of inertia of the waterdrop-waterwheel system. We can solve for this to see that the waterdroop-waterwheel system will start to turn ever so slightly, because of the collision.

As we have more collisions, the wheel starts to speed up. As the wheel speeds up, we start to find that the angular velocity of the water droplet before the collision and the angular velocity of the droplet after the collision start to get closer and closer together. This means the droplet will impart less and less angular momentum into the waterwheel. At some point, the wheel spins as fast as the droplets passing by it, and the wheel receives no angular momentum from the droplet, so it never accelerates faster.

Of course, this is only one effect. In a waterwheel like you have drawn, with no buckets, this sticking would only be momentary before the water rolls off. If there are buckets, we also have to consider that the water keeps acting on the waterwheel until it reaches the bottom. There are many ways this can be modeled, but energy may be the best way to do it. You can model the loss of gravitational potential energy as the bucket of water reaches the bottom of the wheel, and then look at how much kinetic energy is lost when you let the water go.

In a more complicated system, the water might splash or divert off at an angle. In such cases, you would need to take care to make sure it is modeled correctly.

  • $\begingroup$ Thanks for taking the time to answer! I have a better idea of the situation now. However, as the water wheel speeds up, would the angular momentum added to the wheel simply be the difference between the the water's tangential velocity and the wheel's tangential velocity (at point of impact), multiplied by the water mass and the distance? I think it would make sense because when the water falls as fast as the wheel rotates, the velocity difference is 0, thus the angular momentum added would also be 0 (as you described). $\endgroup$
    – F16Falcon
    Aug 11 '18 at 21:56
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    $\begingroup$ @f16falcon you have the idea correct. I phrased it in general terms of conservation of angular momentum, but if you do the math, you get exactly what you say. $\endgroup$
    – Cort Ammon
    Aug 11 '18 at 22:33

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