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

Concept: For an inverse square central force: \[ F = -\frac{k}{r^2} \] Potential energy: \[ U = -\frac{k}{r} \] For circular motion: \[ \frac{mv^2}{r} = \frac{k}{r^2} \]

Step 1: Find kinetic energy.
\[ mv^2 = \frac{k}{r} \Rightarrow v^2 = \frac{k}{mr} \] \[ K = \frac{1}{2}mv^2 = \frac{1}{2}\cdot \frac{k}{r} = \frac{k}{2r} \]

Step 2: Find potential energy.
\[ U = -\frac{k}{r} \]

Step 3: Total energy.
\[ E = K + U = \frac{k}{2r} - \frac{k}{r} = -\frac{k}{2r} \] Putting \( r = R \): \[ E = -\frac{k}{2R} \]