4. A 1.45-kg stone is tied to a 1.18-m long string and whirled in a vertical circle at a constant speed of 5.60 m/s. What is the tension in the string when the stone passes by the highest point? N

Physics for Scientists and Engineers with Modern Physics
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Chapter6: Circular Motion And Other Applications Of Newton's Laws
Section: Chapter Questions
Problem 33AP: The pilot of an airplane executes a loop-the-loop maneuver in a vertical circle. The speed of the...
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### Problem Statement:

A 1.45-kg stone is tied to a 1.18-m long string and whirled in a vertical circle at a constant speed of 5.60 m/s. What is the tension in the string when the stone passes by the highest point?

### Solution:

To solve this problem, we need to analyze the forces acting on the stone when it is at the highest point of its circular path. At this point, the forces acting on the stone are:

1. The gravitational force (weight) acting downward.
2. The tension in the string acting towards the center of the circle.

#### Step-by-Step Solution:

1. **Calculate the gravitational force (weight) acting on the stone:**
   - Weight, \( W \) = mass \( m \) * gravitational acceleration \( g \)
   - Here, \( m = 1.45 \) kg and \( g = 9.81 \) m/s².
   - Therefore, \( W = 1.45 \text{ kg} * 9.81 \text{ m/s}^2 = 14.2245 \text{ N} \).

2. **Determine the centripetal force required to keep the stone moving in a circular path:**
   - Centripetal force, \( F_c \) = \( \frac{m v^2}{r} \)
   - Here, \( m = 1.45 \) kg (mass of the stone), \( v = 5.60 \) m/s (speed of the stone), and \( r = 1.18 \) m (length of the string).
   - Therefore, \( F_c = \frac{1.45 \text{ kg} * (5.60 \text{ m/s})^2}{1.18 \text{ m}} = 38.277 \text{ N} \).

3. **Relate the tension in the string to the forces acting on the stone at the highest point:**
   - At the highest point, both the tension force \( T \) and the weight \( W \) act downward.
   - To provide enough centripetal force to keep the stone moving in a circular path, the tension in the string must satisfy:
     \[
     F_c = T + W 
     \]
     Therefore,
     \[
     T = F_c - W 
     \]
Transcribed Image Text:### Problem Statement: A 1.45-kg stone is tied to a 1.18-m long string and whirled in a vertical circle at a constant speed of 5.60 m/s. What is the tension in the string when the stone passes by the highest point? ### Solution: To solve this problem, we need to analyze the forces acting on the stone when it is at the highest point of its circular path. At this point, the forces acting on the stone are: 1. The gravitational force (weight) acting downward. 2. The tension in the string acting towards the center of the circle. #### Step-by-Step Solution: 1. **Calculate the gravitational force (weight) acting on the stone:** - Weight, \( W \) = mass \( m \) * gravitational acceleration \( g \) - Here, \( m = 1.45 \) kg and \( g = 9.81 \) m/s². - Therefore, \( W = 1.45 \text{ kg} * 9.81 \text{ m/s}^2 = 14.2245 \text{ N} \). 2. **Determine the centripetal force required to keep the stone moving in a circular path:** - Centripetal force, \( F_c \) = \( \frac{m v^2}{r} \) - Here, \( m = 1.45 \) kg (mass of the stone), \( v = 5.60 \) m/s (speed of the stone), and \( r = 1.18 \) m (length of the string). - Therefore, \( F_c = \frac{1.45 \text{ kg} * (5.60 \text{ m/s})^2}{1.18 \text{ m}} = 38.277 \text{ N} \). 3. **Relate the tension in the string to the forces acting on the stone at the highest point:** - At the highest point, both the tension force \( T \) and the weight \( W \) act downward. - To provide enough centripetal force to keep the stone moving in a circular path, the tension in the string must satisfy: \[ F_c = T + W \] Therefore, \[ T = F_c - W \]
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