A particle is to slide along a horizontal circular path on the inside of a funnel. The surface of the tunnel is friction less. How fast must the particle be moving (in terms of r and `theta`) if it is to execute this motion?

Answer

To solve the problem of a particle sliding along a horizontal circular path inside a frictionless funnel, we need to determine the speed of the particle in terms of the radius \( r \) and the angle \( \theta \) of the funnel. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Particle
The particle is moving in a circular path, which means it experiences centripetal acceleration. The forces acting on the particle are:
- The normal force \( N \) exerted by the surface of the funnel.
- The gravitational force \( mg \) acting downwards.

### Step 2: Resolve the Normal Force
The normal force can be resolved into two components:
- \( N \cos \theta \): This component acts towards the center of the circular path and provides the necessary centripetal force.
- \( N \sin \theta \): This component balances the weight of the particle.

### Step 3: Set Up the Equations of Motion
From the forces acting on the particle, we can write two equations:
1. For centripetal force:
\[ N \cos \theta = \frac{mv^2}{r} \quad \text{(1)} \]
2. For vertical force balance:
\[ N \sin \theta = mg \quad \text{(2)} \]

### Step 4: Divide the Equations
To eliminate \( N \), we can divide equation (1) by equation (2):
\[ \frac{N \cos \theta}{N \sin \theta} = \frac{\frac{mv^2}{r}}{mg} \]
This simplifies to:
\[ \cot \theta = \frac{v^2}{rg} \]

### Step 5: Solve for Velocity \( v \)
Rearranging the equation gives us:
\[ v^2 = rg \cot \theta \]
Taking the square root of both sides, we find:
\[ v = \sqrt{rg \cot \theta} \]

### Final Answer
Thus, the speed of the particle in terms of \( r \) and \( \theta \) is:
\[ v = \sqrt{rg \cot \theta} \]

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