Question

The model of a water wheel is desgined as shown. The rim of the wheel with a radius of R has N evenly spcaed compartments. When a compartment reaches the top position, a mass m is dropped into it with no initial velocity wrt ground. When the compartment reaches bpttom position, the mass falls out with no initialmvelo wrt the wheel. The wheel is light and all collisions are inelastic and there is no friction. find steady state angular velocity and thrust force on the wheel.

Answer 1

• Steady state angular velocity: $\displaystyle \omega=\sqrt{\frac{2g}{R}}$.
• Thrust force on the wheel: $\displaystyle F=\frac{Nm\,g}{\pi}$.

Answer 2

We will show one standard way to analyze the problem.

Let the wheel (of negligible mass) have a rim of radius $R$ with $N$ evenly spaced compartments. In one cycle a compartment:

1. At the top (angle $\theta=0$) a mass $m$ is dropped in. Dropped with zero velocity (ground frame) it suddenly has to move with the wheel at speed $\omega R$. Thus it “acquires” a tangential speed $\omega R$ and hence an angular momentum (about the wheel’s center) of

   $$\Delta L_{\text{in}} = mR^2\omega.$$

   This impulse is provided by the wheel (so the wheel loses this much angular momentum in an impulsive collision).

2. From top to bottom the mass stays in the rotating compartment. Under gravity, during its journey the tangential component of its weight $mg\sin\theta$ produces a torque. Writing the instantaneous torque as

   $$\tau = mgR \sin\theta,$$

   the angular momentum gained by the mass as it moves from the top ($\theta=0$) to the bottom ($\theta=\pi$) is

   $$\Delta L_{\text{grav}} = \int_{0}^{\pi} \tau\,dt = mgR\int_{0}^{\pi}\sin\theta\,\frac{d\theta}{\omega} =\frac{mgR}{\omega}\cdot(2)=\frac{2mgR}{\omega}.$$

3. At the bottom the mass is “dropped out” in such a way that it leaves with no speed relative to the wheel. (That is, no further impulse is exchanged at the bottom.)

In steady state the angular momentum given to the mass by gravity during its journey (from top to bottom) must exactly compensate the loss at the top when the mass is dropped into the compartment. Therefore we equate

$$mR^2\omega=\frac{2mgR}{\omega}.$$

Canceling $m$ and one factor of $R$ (and assuming nonzero $\omega$) gives

$$R\omega^2=\frac{2g}{\,}\quad\Longrightarrow\quad \omega^2=\frac{2g}{R}\quad\Longrightarrow\quad \omega=\sqrt{\frac{2g}{R}}.$$

Now, regarding the thrust force on the wheel: Each drop at the top imparts an impulse (change in momentum) of

$$\Delta p=mR\omega.$$

Since there are $N$ compartments, each making such a collision once per revolution, and the wheel rotates at angular speed $\omega$ (i.e. period $T=\frac{2\pi}{\omega}$), the average rate of collisions is

$$\text{collision rate}=\frac{N}{T}=\frac{N\omega}{2\pi}.$$

Thus the average thrust (force) is given by the total momentum change per unit time:

$$F=\Delta p\cdot\text{collision rate} = mR\omega\,\frac{N\omega}{2\pi}=\frac{N\,mR\omega^2}{2\pi}.$$

Substitute $\omega^2=\frac{2g}{R}$ to get

$$F=\frac{N\,mR\left(\frac{2g}{R}\right)}{2\pi}=\frac{N\,m\,2g}{2\pi}=\frac{Nm\,g}{\pi}.$$

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Explanation (minimal)

1. When a mass $m$ is dropped at the top (with zero ground speed) into a compartment moving at speed $\omega R$, it suddenly acquires angular momentum $mR^2\omega$.

2. From top to bottom, gravity gives the mass an angular momentum gain of $\frac{2mgR}{\omega}$. In steady state, these must balance.

3. This gives $\omega^2=\frac{2g}{R}$ so $\omega=\sqrt{\frac{2g}{R}}$.

4. The average thrust force is the impulse per drop times the frequency of drops: $F=\frac{N\,mR\,\omega^2}{2\pi}=\frac{Nm\,g}{\pi}$.

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Final Answer

$$\omega=\sqrt{\frac{2g}{R}},\qquad F=\frac{Nm\,g}{\pi}$$