A 400mm high open cylinder and 150mm in diameter is filled with water and rotated about its vertical axis at an angular speed of 33.5 rad/s. Determine (a) the depth of water in the cylinder when it is brought to rest, and (b) the volume of water that remains in the cylinder if the speed is doubled.

Problem Statement

An open cylinder 400 mm high and 150 mm in diameter is filled with water and rotated about its vertical axis at an angular speed of 33.5 rad/s.

Determine:

1. (a) The depth of water in the cylinder when it is brought to rest.
2. (b) The volume of water that remains in the cylinder if the speed is doubled.

Solution

1. Given:

   Diameter = 150 mm, so Radius, \( r = 0.075\,\text{m} \)
   Height, \( h = 0.4\,\text{m} \)
   Angular velocity, \( \omega = 33.5\,\text{rad/s} \)
   \( g = 9.81\,\text{m/s}^2 \)

2. Part (a): Depth of Water at Rest

   When the cylinder rotates, the free surface forms a paraboloid. The vertical depression at the center is given by \( z = \frac{r^2 \omega^2}{2g} \). Substituting the given values:

   \( z = \frac{(0.075)^2 \times (33.5)^2}{2 \times 9.81} \approx 0.32\,\text{m} \)

   The volume of water spilled (i.e., the volume of the paraboloid above the original level) is:

   \( V_{\text{spilled}} = \frac{1}{2}\pi r^2 z = \frac{1}{2}\pi \times (0.075)^2 \times 0.32 \approx 0.002827\,\text{m}^3 \)

   The original volume of water is:

   \( V_{\text{original}} = \pi r^2 h = \pi \times (0.075)^2 \times 0.4 \approx 0.007069\,\text{m}^3 \)

   Thus, the remaining volume of water is:

   \( V_r = V_{\text{original}} – V_{\text{spilled}} \approx 0.007069 – 0.002827 = 0.004242\,\text{m}^3 \)

   Let the depth of water at rest be \( d \) (so that \( V_r = \pi r^2 d \)). Then:

   \( 0.004242 = \pi \times (0.075)^2 \times d \)

   Solving for \( d \):

   \( d \approx 0.24\,\text{m} \)

3. Part (b): Volume of Water Remaining When Speed Is Doubled

   If the speed is doubled, then \( \omega = 2 \times 33.5 = 67\,\text{rad/s} \). The new free surface depression is:

   \( z = \frac{(0.075)^2 \times 67^2}{2 \times 9.81} \approx 1.287\,\text{m} \)

   The additional rise above the original water level is:

   \( z_1 = z – h = 1.287 – 0.4 = 0.887\,\text{m} \)

   To account for the spillage when the speed is doubled, a reduced effective radius \( r_1 \) is considered such that \( z_1 = \frac{r_1^2 \omega^2}{2g} \). Thus:

   \( 0.887 = \frac{r_1^2 \times 67^2}{2 \times 9.81} \)

   Solving for \( r_1 \), we obtain:

   \( r_1 \approx 0.062\,\text{m} \)

   The volume of water spilled when the speed is doubled is then the difference between the volumes of two paraboloids:

   V_spilled = \( \frac{1}{2}\pi (0.075)^2 \times 1.287 – \frac{1}{2}\pi (0.062)^2 \times 0.887 \approx 0.006016\,\text{m}^3 \)

   Therefore, the volume of water remaining in the cylinder is:

   \( V_{\text{left}} = V_{\text{original}} – V_{\text{spilled}} \approx 0.007069 – 0.006016 = 0.00105\,\text{m}^3 \)

Explanation

When the cylinder rotates, the free surface forms a paraboloid. In part (a), the depression \( z \) is computed using \( z = \frac{r^2 \omega^2}{2g} \). The volume spilled is the volume of the paraboloid above the original water level, and subtracting this from the original volume yields the remaining volume. The new depth \( d \) is then found from \( V_r = \pi r^2 d \).

Physical Meaning

This problem demonstrates the effect of rotation on the free surface of a liquid in an open cylinder. When the cylinder rotates, the liquid forms a parabolic shape and spills out, reducing the depth of water remaining. Doubling the rotational speed significantly increases the free surface depression, causing much more water to be lost.