# Rotating Cylinder with Water: Step-by-Step Solution Let's solve the problem step by step: ## **Given Data** - Cylinder diameter, \( D = 50\, \text{cm} = 0.5\, \text{m} \) - Cylinder height, \( H = 100\, \text{cm} = 1.0\, \text{m} \) - Initial water height, \( h_0 = 70\, \text{cm} = 0.7\, \text{m} \) - The cylinder is rotated about its axis with angular speed \( \omega \). - Water just reaches the rim (i.e., the parabolic surface touches the rim). --- ## **1. Shape of the Water Surface (Paraboloid Equation)** When the cylinder rotates at angular speed \( \omega \): - The free surface of the liquid forms a **paraboloid**. - The height of the surface at radius \( r \) is: \[ z(r) = z_0 + \frac{\omega^2 r^2}{2g} \] where: - \( z_0 \) = height of water at the axis (lowest point), - \( r \) = distance from axis (from 0 to \( R = D/2 \)), - \( g \) = acceleration due to gravity (use \( 9.81\, \text{m/s}^2 \)). --- ## **2. Condition: Water Just Touches the Rim** At \( r = R \), the water surface just reaches the rim, i.e., \[ z(R) = H \] Also, the **volume of water stays the same** (no spill). --- ## **3. Volume Conservation** **Initial volume:** \[ V_0 = \pi R^2 h_0 \] **Volume after rotation (under paraboloid):** \[ V = \int_0^R 2\pi r \left[z_0 + \frac{\omega^2 r^2}{2g}\right] dr \] Calculate: \[ V = 2\pi \int_0^R z_0 r\, dr + 2\pi \int_0^R \frac{\omega^2 r^3}{2g} dr \] \[ = 2\pi z_0 \left[\frac{r^2}{2}\right]_0^R + 2\pi \frac{\omega^2}{2g} \left[\frac{r^4}{4}\right]_0^R \] \[ = 2\pi z_0 \frac{R^2}{2} + 2\pi \frac{\omega^2}{2g} \frac{R^4}{4} \] \[ = \pi R^2 z_0 + \frac{\pi \omega^2 R^4}{4g} \] Set this equal to \( V_0 \): \[ \pi R^2 z_0 + \frac{\pi \omega^2 R^4}{4g} = \pi R^2 h_0 \] \[ z_0 + \frac{\omega^2 R^2}{4g} = h_0 \] \[ z_0 = h_0 - \frac{\omega^2 R^2}{4g} \] --- ## **4. Rim Condition** At rim: \( z(R) = H \): \[ z(R) = z_0 + \frac{\omega^2 R^2}{2g} = H \] Substitute \( z_0 \): \[ h_0 - \frac{\omega^2 R^2}{4g} + \frac{\omega^2 R^2}{2g} = H \] \[ h_0 + \frac{\omega^2 R^2}{2g} - \frac{\omega^2 R^2}{4g} = H \] \[ h_0 + \frac{\omega^2 R^2}{4g} = H \] \[ \frac{\omega^2 R^2}{4g} = H - h_0 \] \[ \omega^2 = \frac{4g}{R^2} (H - h_0) \] --- ## **5. Insert Values** - \( R = D/2 = 0.25\, \text{m} \) - \( H = 1.0\, \text{m} \) - \( h_0 = 0.7\, \text{m} \) - \( g = 9.81\, \text{m/s}^2 \) \[ \omega^2 = \frac{4 \times 9.81}{(0.25)^2} (1.0 - 0.7) \] \[ = \frac{39.24}{0.0625} \times 0.3 \] \[ = 627.84 \times 0.3 \] \[ = 188.35 \] \[ \omega = \sqrt{188.35} \approx 13.73\, \text{rad/s} \] --- ## **6. Lowest Height Reached** \[ z_0 = h_0 - \frac{\omega^2 R^2}{4g} \] But \(\frac{\omega^2 R^2}{4g} = H - h_0 = 0.3\, \text{m}\): \[ z_0 = h_0 - (H - h_0) = 2h_0 - H \] \[ = 2 \times 0.7 - 1.0 = 0.4\, \text{m} \] --- ## **7. **Pressure Distribution** ### **At the Side Walls:** - The pressure at a point with radius \( r \) and depth \( h \) below the surface is: \[ P = P_0 + \rho g (z_\text{surface at } r - z) \] ### **At the Bottom:** - The bottom is deepest at the wall, shallowest at the center. - **At the center**: depth below surface = \( z_0 \) - **At the wall**: depth = \( H - H = 0 \) (surface reaches rim) ### **Pressure is highest at the center of the bottom, lowest at the rim.** - **Pressure at center (bottom):** \[ P_\text{center} = P_0 + \rho g z_0 \] - **Pressure at rim (bottom):** \[ P_\text{rim} = P_0 \] --- ## **Summary Table** | Quantity | Value | |----------|-------| | **Angular speed (\(\omega\))** | \(\boxed{13.7\, \text{rad/s}}\) | | **Lowest height (center)** | \(\boxed{0.4\, \text{m}}\) | | **Height at rim** | \(1.0\, \text{m}\) (full) | --- ## **Pressure Distribution** - **Bottom**: Max in center (\(P_0 + \rho g \times 0.4\)), min at rim (\(P_0\)) - **Sides**: Increases with depth below the rotating paraboloid surface, not a simple linear hydrostatic distribution. --- ## **Final Answers** ### **1. The angular speed required:** \[ \boxed{\omega = 13.7\ \text{rad/s}} \] ### **2. The lowest height the water reaches:** \[ \boxed{0.4\ \text{m}} \] ### **3. Pressure distribution:** - **Sides:** Pressure increases with depth below the paraboloid surface. - **Bottom:** Maximum at center, minimum at rim, linear from center to rim. --- **Let me know if you'd like to see a graph or a more detailed explanation of the pressure at a specific location!**