Problem Statement
A cylindrical tank is spun at 300 rpm with its axis vertical. The tank is 0.6 m high and 45 cm in diameter and is completely filled with water before spinning.
Calculate:
- (a) The speed at which the water surface will just touch the top rim and the center bottom of the tank.
- (b) The level to which the water will return when the tank stops spinning and the amount of water lost.
Solution
-
Given:
Diameter = 45 cm, so Radius \( r = 0.225\,\text{m} \)
Tank Height \( h = 0.6\,\text{m} \)
\( g = 9.81\,\text{m/s}^2 \) -
Part (a): Critical Speed
For the water surface to just touch the top rim and the center bottom, the vertical difference must equal the tank height, so we set \( z = 0.6\,\text{m} \).
The free surface depression is given by \( z = \frac{r^2 \omega^2}{2g} \). Therefore,
\( 0.6 = \frac{(0.225)^2\,\omega^2}{2 \times 9.81} \)Solving for \( \omega \):
\( \omega \approx 15.25\,\text{rad/s} \)Converting angular velocity to rpm:
\( N = \frac{60\,\omega}{2\pi} \approx \frac{60 \times 15.25}{2\pi} \approx 146\,\text{rpm} \) -
Part (b): Water Level After Rest and Water Lost
The original volume of water in the tank is:
\( V_{\text{original}} = \pi r^2 h = \pi \times (0.225)^2 \times 0.6 \approx 0.0954\,\text{m}^3 \)The volume of water spilled (the volume of the paraboloid above the original water level) is:
\( V_{\text{spilled}} = \frac{1}{2}\pi r^2 z = \frac{1}{2}\pi \times (0.225)^2 \times 0.6 \approx 0.0477\,\text{m}^3 \)Hence, the remaining volume of water is:
\( V_r = V_{\text{original}} – V_{\text{spilled}} \approx 0.0954 – 0.0477 = 0.0477\,\text{m}^3 \)When the tank is at rest, the remaining water occupies a depth \( d \) such that
\( V_r = \pi r^2 d \)Therefore:
\( 0.0477 = \pi \times (0.225)^2 \times d \)Solving for \( d \) gives:
\( d \approx 0.3\,\text{m} \)
Explanation
In part (a), the free surface of the spinning tank forms a paraboloid. The condition for the water surface to just touch both the top rim and the center bottom is met when the vertical difference \( z = \frac{r^2 \omega^2}{2g} \) equals the tank height (0.6 m). This yields an angular velocity of approximately 15.25 rad/s, which converts to about 146 rpm.
In part (b), the water lost during spinning is equal to the volume of the paraboloid above the original water level. Subtracting this spilled volume from the original volume gives the remaining water volume. When the tank is at rest, this remaining water occupies a depth of approximately 0.3 m.
Physical Meaning
This problem illustrates the impact of centrifugal forces on the free surface of water in a rotating tank. At high rotational speeds, the water is forced outward, causing the center to drop and the edges to rise. The critical speed is reached when the surface touches the top rim at the edge and the bottom at the center. Once the tank stops spinning, the water returns to a lower level, and the difference in volume represents the water lost during rotation.
