2. A conical pendulum consists of a pendulum bob of mass M at the end of a "mass-less" string of length L. The pendulum bob travels in a horizontal circle at a constant speed. The string is oriented at a constant angle from vertical. Give answers in terms of L, 0, M, and/or g. a. Draw the free body diagram of forces acting on the pendulum bob. Use Newton's 2nd law and circular motion dynamics to determine the tension of the string. Show all steps in the FBD method. b. C. What is the speed of the bob? O 0

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### Conical Pendulum Analysis

Consider a conical pendulum system, which consists of a pendulum bob of mass \( M \) at the end of a "mass-less" string of length \( L \). This pendulum moves in a horizontal circle at a constant speed, with the string making a constant angle \( \theta \) with the vertical axis. The analysis involves examining the system in terms of \( L, \theta, M, \) and/or \( g \) (acceleration due to gravity).

1. **Free Body Diagram (FBD):**

   Firstly, draw the free body diagram of the forces acting on the pendulum bob. The forces that need to be represented are:
   - The gravitational force \( Mg \) acting vertically downward.
   - The tension \( T \) in the string acting along the string's direction, making an angle \( \theta \) with the vertical.

   The corresponding sketch should show two forces on the bob:
   - \( Mg \) vertically downward.
   - \( T \), directed along the string, inclined at an angle \( \theta \) to the vertical.

   ![Diagram](URL_to_the_FBD_image) — sketch your free body diagram here.

2. **Using Newton’s Second Law and Circular Motion:**

   To determine the tension in the string, we analyze the forces in both vertical and horizontal directions.

   - **Vertical Direction:**

     The vertical component of the tension \( T \) must balance the gravitational force:
     \[
     T \cos\theta = Mg \quad \text{(Equation 1)}
     \]

   - **Horizontal Direction:**

     The horizontal component of the tension provides the centripetal force required for circular motion:
     \[
     T \sin\theta = M \left(\frac{v^2}{r}\right) \quad \text{(Equation 2)}
     \]
     Where:
     \[
     r = L \sin\theta
     \]
     (the radius of the horizontal circle described by the bob).

     Substituting \( r \) in the equation:
     \[
     T \sin\theta = M \left(\frac{v^2}{L \sin\theta}\right)
     \]

   Solving Equations 1 and 2 together to find \( T \):
   \[
   T \cos\theta = Mg \quad \
Transcribed Image Text:### Conical Pendulum Analysis Consider a conical pendulum system, which consists of a pendulum bob of mass \( M \) at the end of a "mass-less" string of length \( L \). This pendulum moves in a horizontal circle at a constant speed, with the string making a constant angle \( \theta \) with the vertical axis. The analysis involves examining the system in terms of \( L, \theta, M, \) and/or \( g \) (acceleration due to gravity). 1. **Free Body Diagram (FBD):** Firstly, draw the free body diagram of the forces acting on the pendulum bob. The forces that need to be represented are: - The gravitational force \( Mg \) acting vertically downward. - The tension \( T \) in the string acting along the string's direction, making an angle \( \theta \) with the vertical. The corresponding sketch should show two forces on the bob: - \( Mg \) vertically downward. - \( T \), directed along the string, inclined at an angle \( \theta \) to the vertical. ![Diagram](URL_to_the_FBD_image) — sketch your free body diagram here. 2. **Using Newton’s Second Law and Circular Motion:** To determine the tension in the string, we analyze the forces in both vertical and horizontal directions. - **Vertical Direction:** The vertical component of the tension \( T \) must balance the gravitational force: \[ T \cos\theta = Mg \quad \text{(Equation 1)} \] - **Horizontal Direction:** The horizontal component of the tension provides the centripetal force required for circular motion: \[ T \sin\theta = M \left(\frac{v^2}{r}\right) \quad \text{(Equation 2)} \] Where: \[ r = L \sin\theta \] (the radius of the horizontal circle described by the bob). Substituting \( r \) in the equation: \[ T \sin\theta = M \left(\frac{v^2}{L \sin\theta}\right) \] Solving Equations 1 and 2 together to find \( T \): \[ T \cos\theta = Mg \quad \
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