Required information A roller-coaster car of mass 307 kg (including passengers) travels around a horizontal curve of radius 38.2 m. Its speed is 22.7 m/s. If the track is frictionless, then what banking angle would ensure that the car does not slide off of the track? (The banking angle is the angle between the direction normal to the track and the vertical.)

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**Problem Statement:**

**Required Information:**
A roller-coaster car of mass 307 kg (including passengers) travels around a horizontal curve of radius 38.2 m. Its speed is 22.7 m/s.

If the track is frictionless, then what banking angle would ensure that the car does not slide off of the track? (The banking angle is the angle between the direction normal to the track and the vertical.)

**Solution:**

To solve this problem, we need to use the concept of banking angles in the context of circular motion. When a vehicle is moving on a curved path, banking helps to prevent it from slipping off due to the lack of friction. On a frictionless track, the banking angle θ is given by:

\[ \tan(\theta) = \frac{v^2}{r \cdot g} \]

where:
- \( v \) is the speed of the car (22.7 m/s),
- \( r \) is the radius of the curve (38.2 m),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

The banking angle \( \theta \) can be found by calculating the inverse tangent of the ratio above.

**Calculation Steps:**

1. Calculate \( \frac{v^2}{r \cdot g} \):
   - First, find \( v^2 \): \( (22.7)^2 = 515.29 \, \text{m}^2/\text{s}^2 \)
   - Then, calculate \( r \cdot g \): \( 38.2 \times 9.81 = 374.142 \, \text{m} \cdot \text{m/s}^2 \)
   - Divide: \( \frac{515.29}{374.142} \approx 1.377 \)

2. Calculate \( \theta \) using \( \tan^{-1}(1.377) \).

Use a calculator to find the angle in degrees. This angle is the required banking angle to prevent the roller-coaster from sliding on a frictionless track.

**Conclusion:**

The calculated banking angle θ will ensure that the roller-coaster car stays on the track without slipping, given the track is frictionless.

*Note: The page appears to be a part of an online educational platform, with navigation and resource buttons provided for enhanced learning.*
Transcribed Image Text:**Problem Statement:** **Required Information:** A roller-coaster car of mass 307 kg (including passengers) travels around a horizontal curve of radius 38.2 m. Its speed is 22.7 m/s. If the track is frictionless, then what banking angle would ensure that the car does not slide off of the track? (The banking angle is the angle between the direction normal to the track and the vertical.) **Solution:** To solve this problem, we need to use the concept of banking angles in the context of circular motion. When a vehicle is moving on a curved path, banking helps to prevent it from slipping off due to the lack of friction. On a frictionless track, the banking angle θ is given by: \[ \tan(\theta) = \frac{v^2}{r \cdot g} \] where: - \( v \) is the speed of the car (22.7 m/s), - \( r \) is the radius of the curve (38.2 m), - \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). The banking angle \( \theta \) can be found by calculating the inverse tangent of the ratio above. **Calculation Steps:** 1. Calculate \( \frac{v^2}{r \cdot g} \): - First, find \( v^2 \): \( (22.7)^2 = 515.29 \, \text{m}^2/\text{s}^2 \) - Then, calculate \( r \cdot g \): \( 38.2 \times 9.81 = 374.142 \, \text{m} \cdot \text{m/s}^2 \) - Divide: \( \frac{515.29}{374.142} \approx 1.377 \) 2. Calculate \( \theta \) using \( \tan^{-1}(1.377) \). Use a calculator to find the angle in degrees. This angle is the required banking angle to prevent the roller-coaster from sliding on a frictionless track. **Conclusion:** The calculated banking angle θ will ensure that the roller-coaster car stays on the track without slipping, given the track is frictionless. *Note: The page appears to be a part of an online educational platform, with navigation and resource buttons provided for enhanced learning.*
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