A fluid of mass $M$ fills a cylindrical glass to a height $H$. The glass is rotated about its axis at such an angular velocity that the fluid rises to a height of $2H$ at the glass wall (see the figure).

a) What is the speed of the fluid at the glass wall?

b) What is the kinetic energy of the fluid?

c) What is the minimum work required to accelerate the fluid?

![](https://cdn.mathpix.com/cropped/2024_05_02_997f923a4691fb7eedf3g-1.jpg?height=330&width=282&top_left_y=407&top_left_x=911)
In a resting fluid, the surface is perpendicular to the resultant of the forces acting on it. In a rotating reference frame with the fluid, the fluid is at rest. The surface of the water is acted upon by a centrifugal force of magnitude $\omega^{2} x$ and a gravitational force of magnitude $g$, where the distance from the axis of rotation is denoted by $x$, and the angular velocity of the water by $\omega$.

![](https://cdn.mathpix.com/cropped/2024_05_02_3d815240d8d8b0a00437g-1.jpg?height=230&width=198&top_left_y=308&top_left_x=953)

If the angle between the tangent of the equilibrium surface and the horizontal is $\alpha$, then based on the figure,

$$
\operatorname{tg} \alpha=\frac{\omega^{2} x}{g}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_3d815240d8d8b0a00437g-1.jpg?height=266&width=255&top_left_y=810&top_left_x=924)

Furthermore, $\operatorname{tg} \alpha$ is the slope of the curve $y(x)$ in the plane section at point $x$, which is the derivative of $y$ with respect to $x$, thus

$$
\frac{d y}{d x}=\frac{\omega^{2} x}{g}
$$

It is known that such a curve is a parabola:

$$
y(x)=\frac{\omega^{2} x^{2}}{2 g}+y_{0}
$$

where $y_{0}$ is the height of the water surface at the axis of rotation. Additionally, we know that

$$
y(R)=2 H
$$

The volume of the rotating fluid can be determined by integration. Divide the radius $R$ into segments of length $\Delta x_{i}=x_{i}-x_{i-1}=\frac{R}{n}$ (for simplicity, equal) $(i=1,2, \ldots, n)$. The volume of the rotating fluid can be approximated by the sum of the volumes of cylindrical shells of thickness $\Delta x_{i}$ and height $y\left(x_{i}\right)$, specifically the

$$
\sum_{i=1}^{n} y\left(x_{i}\right) 2 \pi x_{i} \Delta x_{i}
$$

sum; its limit as $n \rightarrow \infty$ is given by the

$$
\int_{0}^{R} y(x) 2 \pi x \mathrm{~d} x
$$

integral. Based on (1),

$$
\int_{0}^{R} y(x) 2 \pi x \mathrm{~d} x=\frac{\pi \omega^{2}}{g} \int_{0}^{R} x^{3} \mathrm{~d} x+2 \pi y_{0} \int_{0}^{R} x \mathrm{~d} x=\frac{\pi \omega^{2} R^{4}}{4 g}+\pi y_{0} R^{2}
$$

The volume of the fluid is constant, so

$$
\frac{\pi \omega^{2} R^{4}}{4 g}+\pi y_{0} R^{2}=\pi R^{2} H
$$

thus

$$
\frac{\omega^{2} R^{2}}{4 g}+y_{0}=H
$$

From the equality of (1), (2), and (3),

$$
y_{0}=0, \quad \omega=\frac{2}{R} \sqrt{g H}, \quad y(x)=2 H \frac{x^{2}}{R^{2}}
$$

so the height of the water at the axis of rotation is 0.

The speed at the edge of the cup is

$$
v=R \cdot \omega=2 \sqrt{g H}
$$

The kinetic energy of the water can also be determined by integration:

$$
E_{\mathrm{kin}} \approx \sum_{i=1}^{n} \frac{1}{2} m_{i} v_{i}^{2}=\sum_{i=1}^{n} \frac{1}{2} 2 \pi x_{i} \cdot \Delta x_{i} \cdot y\left(x_{i}\right) \cdot \varrho \cdot\left(x_{i} \omega\right)^{2}
$$

thus

$$
E_{\text {kin }}=\pi \varrho \omega^{2} \int_{0}^{R} x^{3} y(x) \mathrm{d} x
$$

where $\varrho=\frac{M}{R^{2} \pi H}$. Evaluating the integral, we get

$$
E_{\mathrm{kin}}=\frac{4}{3} M g H
$$

During acceleration, at least as much work must be done as the increase in the water's energy:

$$
W_{\min }=\Delta E_{\mathrm{kin}}+\Delta E_{\mathrm{pot}}
$$

The change in kinetic energy is $\frac{4}{3} M g H$. The potential energy initially was $\frac{1}{2} M g H$, and after acceleration, it is

$$
E_{\mathrm{pot}}=\sum_{i=1}^{n} m_{i} g h_{i}=\int g \frac{1}{2} y(x) \varrho \mathrm{d} V
$$

Using the expressions for $y(x)$, $\varrho$, and $\mathrm{d} V$, we get

$$
E_{\mathrm{pot}}=\frac{2}{3} M g H
$$

Thus, the minimum work done is

$$
W_{\min }=\left(\frac{2}{3}+\frac{4}{3}-\frac{1}{2}\right) M g H=\frac{3}{2} M g H
$$

This much work would need to be done if we ignore all losses (viscosity, the mass of the cup, etc.).