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 to that at its lowest point while completing vertical circular motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Forces at the Highest Point:**
At the highest point of the circular motion, the only forces acting on the particle are its weight (mg) acting downward and the tension in the string (T), which we will set to zero for the minimum speed required to maintain circular motion. Thus, the centripetal force required to keep the particle in circular motion is provided solely by the weight of the particle:
\[ mg = \frac{mv_H^2}{L} \]
where \( v_H \) is the velocity at the highest point and \( L \) is the length of the string.

2. **Finding the Velocity at the Highest Point:**
Rearranging the equation gives:
\[ v_H^2 = gL \]

3. **Understanding Forces at the Lowest Point:**
At the lowest point, the forces acting on the particle are its weight (mg) acting downward and the tension in the string (T) acting upward. The net force towards the center of the circle is:
\[ T + mg = \frac{mv_L^2}{L} \]
where \( v_L \) is the velocity at the lowest point.

4. **Applying Work-Energy Theorem:**
The work done by gravity as the particle moves from the highest point to the lowest point is equal to the change in kinetic energy. The height difference between the highest and lowest points is \( 2L \):
\[ W = mg \cdot 2L \]
This work done by gravity is equal to the change in kinetic energy:
\[ mg \cdot 2L = \frac{1}{2} m v_L^2 - \frac{1}{2} m v_H^2 \]

5. **Substituting for \( v_H \):**
Substitute \( v_H^2 = gL \) into the equation:
\[ mg \cdot 2L = \frac{1}{2} m v_L^2 - \frac{1}{2} m (gL) \]
Simplifying gives:
\[ 2gL = \frac{1}{2} v_L^2 - \frac{1}{2} gL \]
Multiplying through by 2:
\[ 4gL = v_L^2 - gL \]
Rearranging gives:
\[ v_L^2 = 5gL \]

6. **Calculating Kinetic Energies:**
The kinetic energies at the highest and lowest points are given by:
\[ KE_H = \frac{1}{2} mv_H^2 = \frac{1}{2} m(gL) = \frac{mgL}{2} \]
\[ KE_L = \frac{1}{2} mv_L^2 = \frac{1}{2} m(5gL) = \frac{5mgL}{2} \]

7. **Finding the Ratio of Kinetic Energies:**
The ratio of the kinetic energy at the highest point to that at the lowest point is:
\[ \frac{KE_H}{KE_L} = \frac{\frac{mgL}{2}}{\frac{5mgL}{2}} = \frac{1}{5} \]

### Final Answer:
The ratio of the kinetic energy of the particle at its highest point to that at its lowest point is \( \frac{1}{5} \).