A particle is moving in a circular path of radius a under the action of an attractive potential `U=-(k)/(2r^(2))`. Its total energy is :

Options

A. `-(3)/(2)(k)/(a^(2))`
B. `-(k)/(4a^(2))`
C. `(k)/(2a^(2))`
D. Zero

Answer

To find the total energy of a particle moving in a circular path under the influence of the given potential \( U = -\frac{k}{2r^2} \), we will follow these steps:

### Step 1: Determine the Force from the Potential Energy
The force \( F \) associated with a potential energy \( U \) is given by:
\[ F = -\frac{dU}{dr} \]
Given \( U = -\frac{k}{2r^2} \), we differentiate it with respect to \( r \):
\[ \frac{dU}{dr} = \frac{d}{dr}\left(-\frac{k}{2r^2}\right) = -\frac{k}{2} \cdot \frac{d}{dr}(r^{-2}) = -\frac{k}{2} \cdot (-2r^{-3}) = \frac{k}{r^3} \]
Thus, the force is:
\[ F = -\frac{dU}{dr} = -\frac{k}{r^3} \]

### Step 2: Relate the Force to Centripetal Force
For a particle moving in a circular path of radius \( a \), the centripetal force required to keep the particle in circular motion is given by:
\[ F_{centripetal} = \frac{mv^2}{a} \]
Setting the attractive force equal to the centripetal force, we have:
\[ \frac{mv^2}{a} = \frac{k}{a^3} \]
Rearranging gives:
\[ mv^2 = \frac{k}{a^2} \]

### Step 3: Calculate the Kinetic Energy
The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2}mv^2 \]
Substituting \( mv^2 \) from the previous step:
\[ KE = \frac{1}{2} \cdot \frac{k}{a^2} = \frac{k}{2a^2} \]

### Step 4: Calculate the Potential Energy
The potential energy \( U \) at radius \( a \) is:
\[ U = -\frac{k}{2a^2} \]

### Step 5: Calculate the Total Energy
The total energy \( E \) of the system is the sum of kinetic energy and potential energy:
\[ E = KE + U \]
Substituting the values we found:
\[ E = \frac{k}{2a^2} - \frac{k}{2a^2} = 0 \]

### Final Result
Thus, the total energy of the particle is:
\[ \boxed{0} \]