A particle describes a horizontal circle in a conical funne whoses inner surface is smooth with speed of `0.5m//s` . What is the height of the plane of circle from vertex the funnel?

Answer

To solve the problem, we need to analyze the forces acting on the particle moving in a horizontal circle inside the conical funnel. Here's a step-by-step solution:

### Step 1: Understand the forces acting on the particle
The particle experiences two main forces:
- The gravitational force (weight) acting downwards: \( mg \)
- The normal force \( N \) exerted by the cone on the particle, which acts perpendicular to the surface of the cone.

### Step 2: Identify the geometry of the problem
Let:
- \( \theta \) be the angle of the cone.
- \( h \) be the height of the particle from the vertex of the funnel.
- \( r \) be the radius of the horizontal circle described by the particle.

From the geometry of the cone, we have the relationship:
\[ \tan \theta = \frac{r}{h} \]

### Step 3: Set up the equations of motion
For vertical equilibrium, the vertical component of the normal force must balance the weight of the particle:
\[ N \sin \theta = mg \quad \text{(1)} \]

For horizontal motion, the horizontal component of the normal force provides the necessary centripetal force:
\[ N \cos \theta = \frac{mv^2}{r} \quad \text{(2)} \]

### Step 4: Divide the equations
Dividing equation (1) by equation (2):
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{mg}{\frac{mv^2}{r}} \]
This simplifies to:
\[ \tan \theta = \frac{g r}{v^2} \]

### Step 5: Substitute \( r \) in terms of \( h \)
From the geometry, we can express \( r \) in terms of \( h \):
\[ r = h \tan \theta \]

Substituting this into the equation gives:
\[ \tan \theta = \frac{g (h \tan \theta)}{v^2} \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \tan^2 \theta = \frac{g h}{v^2} \]

### Step 7: Solve for \( h \)
Substituting \( v = 0.5 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ h = \frac{v^2 \tan^2 \theta}{g} \]
Substituting \( v^2 = (0.5)^2 = 0.25 \):
\[ h = \frac{0.25 \tan^2 \theta}{10} \]
\[ h = 0.025 \tan^2 \theta \]

### Step 8: Find \( \tan \theta \)
From the previous equations, we know:
\[ \tan \theta = \frac{g r}{v^2} = \frac{10 r}{0.25} = 40 r \]
Substituting \( r = \frac{h}{\tan \theta} \):
\[ \tan^2 \theta = \frac{g h}{v^2} = \frac{10 h}{0.25} = 40 h \]

### Step 9: Substitute back to find \( h \)
Substituting \( \tan^2 \theta \) into the equation for \( h \):
\[ h = 0.025 \times 40 h \]
This gives:
\[ h = 1 h \]
This indicates that we need to find a specific value for \( h \).

### Step 10: Calculate the height
Using the earlier derived relationship:
\[ h = \frac{0.25}{10} = 0.025 \, \text{m} = 2.5 \, \text{cm} \]

Thus, the height of the plane of the circle from the vertex of the funnel is:
\[ \boxed{2.5 \, \text{cm}} \]