Question

A particle of mass m is moving in a horizontal circle of radius R under the centripetal force \( = -\frac{k}{R^2} \) where k is a constant. What is the total energy of the particle?

• A. \( \frac{k}{2R} \)
• B. \( -\frac{k}{2R} \)
• C. \( \frac{k}{R} \)
• D. \( -\frac{k}{R} \)

Hint:

For inverse square forces (like gravity), total energy in circular motion is always negative and equals half of potential energy.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The total energy of a particle in circular motion is the sum of its Kinetic Energy (KE) and Potential Energy (PE).
: Key Formula or Approach:
1. Centripetal Force \( F = \frac{mv^2}{R} \).
2. Kinetic Energy \( K = \frac{1}{2}mv^2 \).
3. Potential Energy \( U = -\int F \, dR \).
Step 2: Detailed Explanation:
Given the force magnitude \( F = \frac{k}{R^2} \).
This force provides the necessary centripetal force:
\[ \frac{mv^2}{R} = \frac{k}{R^2} \]
\[ mv^2 = \frac{k}{R} \]
Now, calculate the Kinetic Energy \( K \):
\[ K = \frac{1}{2}mv^2 = \frac{1}{2} \left( \frac{k}{R} \right) = \frac{k}{2R} \]
Next, calculate the Potential Energy \( U \). For an attractive central force \( F = -\frac{k}{R^2} \):
\[ U = -\int F \, dR = -\int \left( \frac{k}{R^2} \right) \, dR \]
(Note: Using the standard integration for conservative fields where \( F = -\frac{dU}{dR} \))
\[ U = -\frac{k}{R} \]
Total Energy \( E = K + U \):
\[ E = \frac{k}{2R} + \left( -\frac{k}{R} \right) \]
\[ E = \frac{k - 2k}{2R} = -\frac{k}{2R} \]
Step 3: Final Answer:
The total energy of the particle is \( -k/2R \).