Let's break down the problem step-by-step: ### **Step 1: Understanding the Physics** - **Radius of cylinder**: \( r = 2.5\,\text{m} \) - **Coefficient of static friction**: \( \mu_s = 0.45 \) - **Goal**: - Find the minimum frequency \( f \) (in revolutions per second) for which people don’t fall when the floor drops. - Calculate the "g-force" experienced by the riders at this frequency. --- ### **Step 2: Free-Body Diagram and Forces** When the floor drops: - **Centrifugal force** (outward, due to rotation): \( F_c = m \omega^2 r \) - **Normal force** (from wall): \( N = m \omega^2 r \) - **Frictional force** (upward, opposes gravity): \( f_{\text{fric}} = \mu_s N \) - **Gravity** (downward): \( mg \) For the person **not to fall**, friction must balance gravity: \[ f_{\text{fric}} \geq mg \] \[ \mu_s N \geq mg \] \[ \mu_s m \omega^2 r \geq mg \] \[ \mu_s \omega^2 r \geq g \] \[ \omega^2 \geq \frac{g}{\mu_s r} \] \[ \omega \geq \sqrt{\frac{g}{\mu_s r}} \] --- ### **Step 3: Plug in Values** - \( g = 9.8\,\text{m/s}^2 \) - \( \mu_s = 0.45 \) - \( r = 2.5\,\text{m} \) \[ \omega \geq \sqrt{\frac{9.8}{0.45 \times 2.5}} \] \[ \omega \geq \sqrt{\frac{9.8}{1.125}} \] \[ \omega \geq \sqrt{8.711} \] \[ \omega \geq 2.953\,\text{rad/s} \] --- ### **Step 4: Convert Angular Speed to Frequency** \[ \omega = 2\pi f \implies f = \frac{\omega}{2\pi} \] \[ f = \frac{2.953}{2\pi} \] \[ f = \frac{2.953}{6.283} \] \[ f \approx 0.47\,\text{Hz} \] --- ### **Step 5: Find the "g-force"** "g-force" here is the ratio of the centripetal acceleration to \( g \): \[ a_c = \omega^2 r \] \[ \text{g-force} = \frac{a_c}{g} \] \[ a_c = (2.953)^2 \times 2.5 = 8.711 \times 2.5 = 21.78\,\text{m/s}^2 \] \[ \text{g-force} = \frac{21.78}{9.8} \approx 2.22 \] --- ## **Final Answers** ### **1. Minimum Frequency:** **The floor can drop without people falling out if the cylinder rotates at a frequency above** \[ \boxed{0.47\,\text{Hz}} \] (or about 28 revolutions per minute) ### **2. "g-force" on Victims:** **The g-force on the victims at this frequency is** \[ \boxed{2.2\,g} \] --- **Summary Table:** | Quantity | Value | |-----------------|---------| | Minimum frequency | \( 0.47\,\text{Hz} \) | | "g-force" | \( 2.2\,g \) |

The minimum rotation frequency required to prevent victims from falling is when the static friction force balances gravity, calculated as \(f \approx 0.47\,\text{Hz}\). At this frequency, the victims experience approximately 2.2 times Earth's gravity ("g-force").