In vertical circular motion, the ratio of kinetic energy of a particle at highest point to that at lowest point is

Options

A. 5
B. 2
C. 0.5
D. 0.2

Explanation

To solve the problem of finding the ratio of the kinetic energy of a particle at the highest point to that at the lowest point during vertical circular motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- A particle is moving in a vertical circular path. At the highest point, the only forces acting on it are its weight (mg) and the tension in the string. For the particle to maintain circular motion at the highest point, the centripetal force must be provided by the weight of the particle.

2. **Finding Velocity at the Highest Point**:
- At the highest point, if we assume the tension in the string is zero (just enough to maintain circular motion), the centripetal force required is provided entirely by the weight of the particle.
- The centripetal force equation is given by:
\[ \frac{mv^2}{R} = mg \]
- Rearranging gives:
\[ v^2 = gR \quad \Rightarrow \quad v = \sqrt{gR} \]

3. **Calculating Kinetic Energy at the Highest Point**:
- The kinetic energy (KE) at the highest point (K.E. top) can be calculated using the formula:
\[ K.E. = \frac{1}{2} mv^2 \]
- Substituting the value of \(v\):
\[ K.E. \text{ (top)} = \frac{1}{2} m (\sqrt{gR})^2 = \frac{1}{2} mgR \]

4. **Finding Kinetic Energy at the Lowest Point**:
- To find the kinetic energy at the lowest point (K.E. bottom), we can use the work-energy theorem. The work done by gravity as the particle moves from the highest point to the lowest point is:
\[ W = -mg(2R) = -2mgR \quad (\text{since the displacement is downward}) \]
- The work done by tension is zero because the tension is perpendicular to the displacement.
- Thus, applying the work-energy theorem:
\[ W = K.E. \text{ (final)} - K.E. \text{ (initial)} \]
- Substituting the values:
\[ -2mgR = K.E. \text{ (bottom)} - \frac{1}{2} mgR \]
- Rearranging gives:
\[ K.E. \text{ (bottom)} = \frac{1}{2} mgR - 2mgR = \frac{1}{2} mgR + 2mgR = \frac{5}{2} mgR \]

5. **Finding the Ratio of Kinetic Energies**:
- Now, we can find the ratio of the kinetic energy at the highest point to that at the lowest point:
\[ \text{Ratio} = \frac{K.E. \text{ (top)}}{K.E. \text{ (bottom)}} = \frac{\frac{1}{2} mgR}{\frac{5}{2} mgR} \]
- Simplifying this gives:
\[ \text{Ratio} = \frac{1}{5} = 0.2 \]

### Final Answer:
The ratio of the kinetic energy of a particle at the highest point to that at the lowest point is \( \frac{1}{5} \) or \( 0.2 \).

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