8. At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, their backs against the outer wall. At one instant the outer wall moves at a speed of 3.70 m/s, and an 87.0-kg person feels a 565-N force pressing against his back. What is the radius of the chamber? m sf60 50 $$ sfo 60 s ssfor

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### Physics Problem: Amusement Park Ride Dynamics

**Problem Statement:**

At an amusement park, there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, with their backs against the outer wall. At one instant, the outer wall moves at a speed of \(3.70 \, \text{m/s}\), and an \(87.0 \, \text{kg}\) person feels a \(565 \, \text{N}\) force pressing against his back. What is the radius of the chamber?

**Solution:**

Given data:
- Speed of the outer wall (\(v\)) = \(3.70 \, \text{m/s}\)
- Force experienced by the person (\(F\)) = \(565 \, \text{N}\)
- Mass of the person (\(m\)) = \(87.0 \, \text{kg}\)

We need to find the radius of the chamber (\(r\)).

Using the formula for centripetal force:
\[ F = \frac{m v^2}{r} \]

Rearranging to solve for \(r\):
\[ r = \frac{m v^2}{F} \]

Substitute the given values:
\[ r = \frac{87.0 \, \text{kg} \times (3.70 \, \text{m/s})^2}{565 \, \text{N}} \]
\[ r = \frac{87.0 \times 13.69}{565} \]
\[ r = \frac{1191.03}{565} \]
\[ r \approx 2.11 \, \text{m} \]

**Answer:**
The radius of the chamber is approximately \(2.11 \, \text{m}\).
Transcribed Image Text:### Physics Problem: Amusement Park Ride Dynamics **Problem Statement:** At an amusement park, there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats facing the axis, with their backs against the outer wall. At one instant, the outer wall moves at a speed of \(3.70 \, \text{m/s}\), and an \(87.0 \, \text{kg}\) person feels a \(565 \, \text{N}\) force pressing against his back. What is the radius of the chamber? **Solution:** Given data: - Speed of the outer wall (\(v\)) = \(3.70 \, \text{m/s}\) - Force experienced by the person (\(F\)) = \(565 \, \text{N}\) - Mass of the person (\(m\)) = \(87.0 \, \text{kg}\) We need to find the radius of the chamber (\(r\)). Using the formula for centripetal force: \[ F = \frac{m v^2}{r} \] Rearranging to solve for \(r\): \[ r = \frac{m v^2}{F} \] Substitute the given values: \[ r = \frac{87.0 \, \text{kg} \times (3.70 \, \text{m/s})^2}{565 \, \text{N}} \] \[ r = \frac{87.0 \times 13.69}{565} \] \[ r = \frac{1191.03}{565} \] \[ r \approx 2.11 \, \text{m} \] **Answer:** The radius of the chamber is approximately \(2.11 \, \text{m}\).
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