A particle of mass m is suspended from a ceiling through a string of length L. The particle moves in a horizontal circle of radius r. Find a. the speed of the particle and b. the tension in the string. Such a system is called a conical pendulum.

Options

A. `v=(2rsqrt(g))/((L^(2)-r^(2))^(1//2)) and T=(3mgL)/((L^(2)-r^(2))^(1//2))`
B. `v=(rsqrt(g))/((L^(2)-r^(2))^(1//4)) and T=(mgL)/((L^(3)-r^(3))^(1//3))`
C. `v=(3rsqrt(g))/((L^(2)-r^(2))^(1//2)) and T=(mgL)/((L^(2)-r^(2))^(1//2))`
D. `v=(rsqrt(g))/((L^(2)-r^(2))^(1//4)) and T=(mgL)/((L^(2)-r^(2))^(1//2))`

Answer

To solve the problem of a particle of mass \( m \) suspended from a ceiling through a string of length \( L \) moving in a horizontal circle of radius \( r \), we will follow these steps:

### Step 1: Understand the Geometry of the Problem
The particle moves in a horizontal circle, forming a cone shape with the string. The angle between the string and the vertical is \( \theta \). The vertical component of the tension \( T \) balances the weight of the particle, and the horizontal component provides the necessary centripetal force for circular motion.

### Step 2: Identify the Forces Acting on the Particle
1. The weight of the particle acting downwards: \( mg \)
2. The tension in the string \( T \), which can be resolved into two components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)

### Step 3: Set Up the Equations
From the vertical equilibrium (forces in the vertical direction):
\[ T \cos \theta = mg \quad \text{(1)} \]
From the horizontal motion (providing centripetal force):
\[ T \sin \theta = \frac{mv^2}{r} \quad \text{(2)} \]

### Step 4: Relate the Radius and the Length of the String
Using the geometry of the conical pendulum, we can relate \( r \), \( L \), and \( \theta \):
\[ L \cos \theta = h \quad \text{(height from the ceiling to the particle)} \]
\[ L \sin \theta = r \]
From these, we can express \( \cos \theta \) and \( \sin \theta \):
\[ \sin \theta = \frac{r}{L} \quad \text{and} \quad \cos \theta = \sqrt{1 - \left(\frac{r}{L}\right)^2} = \frac{\sqrt{L^2 - r^2}}{L} \]

### Step 5: Substitute into the Equations
Substituting \( \sin \theta \) and \( \cos \theta \) into equations (1) and (2):
From equation (1):
\[ T \cdot \frac{\sqrt{L^2 - r^2}}{L} = mg \quad \Rightarrow \quad T = \frac{mgL}{\sqrt{L^2 - r^2}} \quad \text{(3)} \]

From equation (2):
\[ T \cdot \frac{r}{L} = \frac{mv^2}{r} \quad \Rightarrow \quad T = \frac{mv^2L}{r^2} \quad \text{(4)} \]

### Step 6: Equate the Two Expressions for Tension
Setting equations (3) and (4) equal to each other:
\[ \frac{mgL}{\sqrt{L^2 - r^2}} = \frac{mv^2L}{r^2} \]
Cancelling \( m \) and \( L \) (assuming \( L \neq 0 \)):
\[ \frac{g}{\sqrt{L^2 - r^2}} = \frac{v^2}{r^2} \]
Rearranging gives:
\[ v^2 = \frac{gr^2}{\sqrt{L^2 - r^2}} \quad \Rightarrow \quad v = r \sqrt{\frac{g}{\sqrt{L^2 - r^2}}} \]

### Step 7: Solve for Tension
Substituting \( v^2 \) back into equation (4):
\[ T = \frac{m \cdot \frac{gr^2}{L^2 - r^2} \cdot L}{r^2} = \frac{mgL}{\sqrt{L^2 - r^2}} \]

### Final Answers
a. The speed of the particle \( v \) is:
\[ v = r \sqrt{\frac{g}{\sqrt{L^2 - r^2}}} \]

b. The tension in the string \( T \) is:
\[ T = \frac{mgL}{\sqrt{L^2 - r^2}} \]