A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius r. Find an expression for the ball's angular speed ω.
Express your answer in terms of the variables  L, r, and the free-fall acceleration g.

To find the angular speed \( \omega \) of the ball in a conical pendulum, we need to analyze the forces acting on the ball and relate them to the motion in circular motion.

### Step-by-Step Derivation:

1. **Forces Acting on the Ball:**
   - The tension \( T \) in the string has two components:
     - The vertical component, \( T \cos \theta \), balances the weight of the ball \( mg \).
     - The horizontal component, \( T \sin \theta \), provides the centripetal force for the circular motion.

2. **Trigonometry in the Conical Pendulum:**
   The ball moves in a horizontal circle with radius \( r \), and the string makes an angle \( \theta \) with the vertical. Using geometry, we have:
   \[
   \cos \theta = \frac{\sqrt{L^2 - r^2}}{L}
   \quad \text{and} \quad
   \sin \theta = \frac{r}{L}
   \]

3. **Vertical Force Balance:**
   The vertical component of the tension balances the gravitational force:
   \[
   T \cos \theta = mg
   \]

4. **Horizontal Force (Centripetal Force):**
   The horizontal component of the tension provides the centripetal force necessary for circular motion:
   \[
   T \sin \theta = m \frac{v^2}{r}
   \]
   where \( v \) is the linear speed of the ball.

5. **Substitute for \( T \):**
   From the vertical force equation, solve for \( T \):
   \[
   T = \frac{mg}{\cos \theta}
   \]
   Substituting this into the horizontal force equation:
   \[
   \frac{mg}{\cos \theta} \sin \theta = m \frac{v^2}{r}
   \]
   Simplifying:
   \[
   g \tan \theta = \frac{v^2}{r}
   \]

6. **Angular Speed \( \omega \):**
   The angular speed \( \omega \) is related to the linear speed \( v \) by \( v = \omega r \). Substituting \( v = \omega r \) into the previous equation:
   \[
   g \tan \theta = \omega^2 r
   \]

7. **Relating \( \tan \theta \) to \( r \) and \( L \):**
   From the geometry of the pendulum:
   \[
   \tan \theta = \frac{r}{\sqrt{L^2 - r^2}}
   \]
   Substituting this into the equation \( g \tan \theta = \omega^2 r \):
   \[
   g \frac{r}{\sqrt{L^2 - r^2}} = \omega^2 r
   \]
   Simplifying by dividing both sides by \( r \) (assuming \( r \neq 0 \)):
   \[
   g \frac{1}{\sqrt{L^2 - r^2}} = \omega^2
   \]

8. **Final Expression for \( \omega \):**
   Solving for \( \omega \):
   \[
   \omega = \sqrt{\frac{g}{\sqrt{L^2 - r^2}}}
   \]

Thus, the angular speed \( \omega \) of the ball is:
\[
\boxed{\omega = \sqrt{\frac{g}{\sqrt{L^2 - r^2}}}}
\]

### Do you need any details or have any questions?

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Here are 5 related questions:
1. How would the angular speed \( \omega \) change if the radius \( r \) increased while keeping \( L \) constant?
2. What would happen to \( \omega \) if the length of the string \( L \) increased?
3. How does the mass of the ball affect the angular speed in this conical pendulum?
4. Can you derive an expression for the tension in the string in terms of \( L \), \( r \), and \( g \)?
5. How would the conical pendulum behave if the system were placed on a planet with a different gravitational acceleration \( g \)?

**Tip:** In conical pendulum problems, always check the geometric relationships between the string length, radius, and the angle the string makes with the vertical to simplify your calculations.

A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius r. Find an expression for the ball's angular speed ω. Express your answer in terms of the variables L, r, and the free-fall acceleration g.

Discover how to derive the angular speed ω for a conical pendulum in this detailed guide. We explore key physics concepts such as tension, centripetal force, and geometric relationships, providing a step-by-step solution to express ω in terms of L, r, and g. Perfect for Grades 11-12 students.

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