A particle of mass m is moving in a horizontal circle of radius R under a centripetal force equal to `-A/r^(2)` (A = constant). The total energy of the particle is :-
(Potential energy at very large distance is zero)

Options

A. `A/R`
B. `- A/R`
C. `A/(2R)`
D. `- A/(2R)`

Answer

To find the total energy of a particle of mass \( m \) moving in a horizontal circle of radius \( R \) under a centripetal force given by \( -\frac{A}{r^2} \), we can follow these steps:

### Step 1: Identify the centripetal force
The centripetal force \( F_c \) required to keep a particle moving in a circle is given by:
\[ F_c = \frac{mv^2}{R} \]
According to the problem, the centripetal force is also given by:
\[ F_c = -\frac{A}{R^2} \]
For our calculations, we will consider the magnitude of the force, so we can write:
\[ \frac{mv^2}{R} = \frac{A}{R^2} \]

### Step 2: Solve for \( mv^2 \)
From the equation above, we can rearrange it to solve for \( mv^2 \):
\[ mv^2 = \frac{A}{R} \]

### Step 3: Calculate the kinetic energy (KE)
The kinetic energy \( KE \) of the particle is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting \( mv^2 \) from the previous step:
\[ KE = \frac{1}{2} \left(\frac{A}{R}\right) = \frac{A}{2R} \]

### Step 4: Find the potential energy (PE)
The potential energy \( U \) can be found using the relationship between force and potential energy:
\[ F = -\frac{dU}{dr} \]
Thus, we can express the potential energy as:
\[ U = -\int F \, dr \]
Substituting \( F = \frac{A}{r^2} \):
\[ U = -\int_{\infty}^{R} \frac{A}{r^2} \, dr \]
Calculating the integral:
\[ U = -\left[-\frac{A}{r}\right]_{\infty}^{R} = -\left(0 - \frac{A}{R}\right) = -\frac{A}{R} \]

### Step 5: Calculate the total energy (TE)
The total energy \( E \) of the particle is the sum of its kinetic energy and potential energy:
\[ E = KE + U \]
Substituting the values we found:
\[ E = \frac{A}{2R} - \frac{A}{R} \]
Combining the terms:
\[ E = \frac{A}{2R} - \frac{2A}{2R} = -\frac{A}{2R} \]

### Final Answer
Thus, the total energy of the particle is:
\[ E = -\frac{A}{2R} \]