Figure 5.1 is loop the loop. Ignoring the rotational energy of the ball, explain using equations, how you can determine the kinetic energy on the ball when it is at point A of the loop. R

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**Figure 5.1: Loop the Loop**

The diagram illustrates a ball moving through a vertical loop. The following components are labeled in the image:

- A sloped track leading to the loop, with the ball positioned at the top. The ball has a radius \( r \).
- A vertical height \( h \) from the horizontal base to the initial position of the ball.
- A loop with radius \( R \), with point \( A \) marked at the top of the loop.

**Task:**

Ignoring the rotational energy of the ball, explain using equations how you can determine the kinetic energy of the ball when it is at point \( A \) of the loop.

**Explanation:**

To solve this problem, apply the conservation of energy principle. Initially, the ball has potential energy due to its height \( h \). As the ball descends the slope and rises through the loop, its potential energy is converted into kinetic energy.

1. **Initial Potential Energy (\( PE_i \))**: 
   \[
   PE_i = mgh
   \]
   where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity, and \( h \) is the initial height.

2. **Potential Energy at Point \( A \) (\( PE_A \))**:
   \[
   PE_A = mg(2R)
   \]
   The height at point \( A \) is \( 2R \).

3. **Kinetic Energy at Point \( A \) (\( KE_A \))**:
   By the conservation of energy:
   \[
   PE_i = PE_A + KE_A
   \]
   \[
   mgh = mg(2R) + KE_A
   \]
   Solving for kinetic energy:
   \[
   KE_A = mgh - mg(2R)
   \]
   \[
   KE_A = mg(h - 2R)
   \]

Thus, the kinetic energy of the ball at point \( A \) is \( mg(h - 2R) \).
Transcribed Image Text:**Figure 5.1: Loop the Loop** The diagram illustrates a ball moving through a vertical loop. The following components are labeled in the image: - A sloped track leading to the loop, with the ball positioned at the top. The ball has a radius \( r \). - A vertical height \( h \) from the horizontal base to the initial position of the ball. - A loop with radius \( R \), with point \( A \) marked at the top of the loop. **Task:** Ignoring the rotational energy of the ball, explain using equations how you can determine the kinetic energy of the ball when it is at point \( A \) of the loop. **Explanation:** To solve this problem, apply the conservation of energy principle. Initially, the ball has potential energy due to its height \( h \). As the ball descends the slope and rises through the loop, its potential energy is converted into kinetic energy. 1. **Initial Potential Energy (\( PE_i \))**: \[ PE_i = mgh \] where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity, and \( h \) is the initial height. 2. **Potential Energy at Point \( A \) (\( PE_A \))**: \[ PE_A = mg(2R) \] The height at point \( A \) is \( 2R \). 3. **Kinetic Energy at Point \( A \) (\( KE_A \))**: By the conservation of energy: \[ PE_i = PE_A + KE_A \] \[ mgh = mg(2R) + KE_A \] Solving for kinetic energy: \[ KE_A = mgh - mg(2R) \] \[ KE_A = mg(h - 2R) \] Thus, the kinetic energy of the ball at point \( A \) is \( mg(h - 2R) \).
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