The potential energy of a particle of mass ? at a distance ? from a fixed point ? is given by `V(r)=kr^(2)//2`, where ? is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius ? about the point ?. If ? is the speed of the particle and ? is the magnitude of its angular momentum about ?, which of the following statements is (are) true?

Options

A. `v=sqrt(k/(2m))R`
B. `v=sqrt(k/m)R`
C. `L=sqrt(mkR^(2))`
D. `L=sqrt((mk)/2)R^(2)`

Explanation

To solve the problem step by step, we will analyze the potential energy given, derive the necessary expressions for the motion of the particle, and then evaluate the statements regarding the speed and angular momentum.

### Step 1: Understand the Potential Energy Function
The potential energy \( V(r) \) of the particle is given by:
\[ V(r) = \frac{kr^2}{2} \]
where \( k \) is a positive constant and \( r \) is the distance from the fixed point.

**Hint:** The potential energy function indicates that the force acting on the particle is derived from this potential energy.

### Step 2: Calculate the Force from Potential Energy
The force \( F \) acting on the particle can be found using the relation:
\[ F = -\frac{dV}{dr} \]
Calculating the derivative:
\[ F = -\frac{d}{dr}\left(\frac{kr^2}{2}\right) = -kr \]
This force acts towards the center of the circular orbit, indicating that it is a restoring force.

**Hint:** Remember that the negative sign indicates that the force is attractive towards the center.

### Step 3: Apply the Centripetal Force Condition
For a particle moving in a circular orbit, the centripetal force required to keep the particle in circular motion is given by:
\[ F_{\text{centripetal}} = \frac{mv^2}{r} \]
Setting the centripetal force equal to the force derived from the potential energy:
\[ kr = \frac{mv^2}{r} \]
Multiplying both sides by \( r \):
\[ kr^2 = mv^2 \]

**Hint:** This equation relates the force acting on the particle to its mass and velocity.

### Step 4: Solve for the Speed \( v \)
From the equation \( kr^2 = mv^2 \), we can solve for the speed \( v \):
\[ v^2 = \frac{kr^2}{m} \]
Taking the square root gives:
\[ v = \sqrt{\frac{kr^2}{m}} = r\sqrt{\frac{k}{m}} \]

**Hint:** This expression shows how the speed depends on the radius and the constants \( k \) and \( m \).

### Step 5: Calculate the Angular Momentum \( L \)
The angular momentum \( L \) of the particle about the center is given by:
\[ L = mvr \]
Substituting the expression for \( v \):
\[ L = m \left(r\sqrt{\frac{k}{m}}\right) r = mr^2\sqrt{\frac{k}{m}} \]
Simplifying this gives:
\[ L = r^2\sqrt{km} \]

**Hint:** Angular momentum is a measure of the rotational motion of the particle.

### Conclusion
We have derived the expressions for speed \( v \) and angular momentum \( L \):
- Speed: \( v = r\sqrt{\frac{k}{m}} \)
- Angular Momentum: \( L = r^2\sqrt{km} \)

Now, we can evaluate which statements regarding these quantities are true based on the derived equations.