A particle is attached to a string and the other end of the string is attached to a fixed support. In order to just complete the vertical circular motion, what should be the ratio of the kinetic energy of the particle at its highest point of that at its highest point to that at its lowest point?

Answer

To find the ratio of the kinetic energy of a particle at its highest point (K_A) to that at its lowest point (K_B) in a vertical circular motion, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Conditions for Vertical Circular Motion:**
- For a particle to complete a vertical circular motion, it must have sufficient speed at the highest point to ensure that the tension in the string does not become zero. This means that the gravitational force must provide enough centripetal force.

2. **Setting Up the Forces at the Highest Point:**
- At the highest point, the forces acting on the particle are the gravitational force (mg) acting downwards and the tension (T) in the string acting downwards as well. The centripetal force required for circular motion is provided by these forces.
- The condition for just completing the circular motion is given by:
\[ mg = \frac{mv^2}{L} \]
- Here, \(v\) is the speed at the highest point, and \(L\) is the length of the string.

3. **Finding the Speed at the Highest Point:**
- Rearranging the equation gives:
\[ v^2 = gL \]
- Therefore, the speed at the highest point (V_A) is:
\[ V_A = \sqrt{gL} \]

4. **Using Energy Conservation:**
- We can apply the conservation of mechanical energy between the highest and lowest points.
- The potential energy at the highest point is maximum, and the kinetic energy is minimum. At the lowest point, the potential energy is minimum, and the kinetic energy is maximum.

5. **Calculating the Potential Energy Change:**
- The height difference between the highest and lowest points is \(2L\). The work done by gravity when moving from the highest to the lowest point is:
\[ W = mg \cdot (2L) = 2mgL \]

6. **Setting Up the Energy Equation:**
- The total mechanical energy at the highest point (K_A + Potential Energy at A) equals the total mechanical energy at the lowest point (K_B + Potential Energy at B).
- At the lowest point, the potential energy is zero, so:
\[ K_A + mg(2L) = K_B \]

7. **Expressing Kinetic Energies:**
- The kinetic energy at the highest point (K_A) is:
\[ K_A = \frac{1}{2} m V_A^2 = \frac{1}{2} m (gL) \]
- The kinetic energy at the lowest point (K_B) can be expressed as:
\[ K_B = K_A + 2mgL \]

8. **Substituting Values:**
- Substitute \(K_A\) into the equation:
\[ K_B = \frac{1}{2} m (gL) + 2mgL \]
- Simplifying gives:
\[ K_B = \frac{1}{2} mgL + 2mgL = \frac{1}{2} mgL + \frac{4}{2} mgL = \frac{5}{2} mgL \]

9. **Finding the Ratio of Kinetic Energies:**
- Now, we can find the ratio of kinetic energies:
\[ \frac{K_A}{K_B} = \frac{\frac{1}{2} mgL}{\frac{5}{2} mgL} = \frac{1}{5} \]

### Final Answer:
The ratio of the kinetic energy of the particle at its highest point to that at its lowest point is:
\[ \frac{K_A}{K_B} = \frac{1}{5} \]