Let's break down and solve the problem step by step. --- ## **Given Data** - **Diameter of container (D):** 20 cm ⇒ **Radius (R):** 10 cm = .1 m - **Height of container:** 60 cm = .6 m - **Height of liquid:** 50 cm = .5 m - **Density of liquid:** ρ = 850 kg/m³ (Not needed for the calculation) - **Acceleration due to gravity (g):** 9.81 m/s² - **Formula:** \[ z_s = h_ - \frac{\omega^2}{4g}(R^2 - 2r^2) \] --- ## **Step 1: Understanding the Formula** - \( z_s \) = height of liquid surface at radius \( r \) - \( h_ \) = initial height of liquid without rotation = .5 m - \( R \) = radius of the cylinder = .1 m - \( r \) = radial distance from the center We want the liquid to just start **spilling** from the edge, i.e., the height at \( r = R \) reaches the top of the container. So, - At \( r = R \), \( z_s = \) height at edge \( = \) **.6 m** (top of container) --- ## **Step 2: Substitute Values** Plug into formula at \( r = R \): \[ z_s = h_ - \frac{\omega^2}{4g}(R^2 - 2R^2) \] \[ z_s = h_ - \frac{\omega^2}{4g}(-R^2) \] \[ z_s = h_ + \frac{\omega^2 R^2}{4g} \] Set \( z_s = .6 \) m and \( h_ = .5 \) m: \[ .6 = .5 + \frac{\omega^2 (.1)^2}{4 \times 9.81} \] --- ## **Step 3: Solve for \( \omega \)** \[ .6 - .5 = \frac{\omega^2 \times .01}{39.24} \] \[ .1 = \frac{\omega^2 \times .01}{39.24} \] \[ .1 \times 39.24 = \omega^2 \times .01 \] \[ 3.924 = \omega^2 \times .01 \] \[ \omega^2 = \frac{3.924}{.01} \] \[ \omega^2 = 392.4 \] \[ \omega = \sqrt{392.4} \] \[ \omega \approx 19.81 \ \text{rad/s} \] --- ## **Final Answer** ### **The rotational speed at which the liquid will start spilling from the edges of the container is:** \[ \boxed{19.8 \ \text{rad/s}} \] (Rounded to one decimal place) --- ### **Summary of Steps** 1. Used the formula at the edge (\( r = R \)), set height to .6 m. 2. Solved for \( \omega \) using the given heights and radius. 3. **Final result:** \( \omega \approx 19.8 \) rad/s.