A paticle of mass `m` is executing uniform circular motion on a path of radius `r`. If `p` is the magnitude of its linear momentum, then the radial force acting on the particle is

Options

A. `p m r`
B. `r m//p`
C. `m p^(2)//r`
D. `p^(2) //r m`

Explanation

To solve the problem, we need to find the radial force acting on a particle of mass \( m \) executing uniform circular motion with a radius \( r \), given that \( p \) is the magnitude of its linear momentum.

### Step-by-Step Solution:

1. **Understanding Linear Momentum**:
The linear momentum \( p \) of the particle is given by the formula:
\[ p = mv \]
where \( m \) is the mass of the particle and \( v \) is its linear velocity.

2. **Relating Linear Velocity to Momentum**:
From the above equation, we can express the linear velocity \( v \) in terms of momentum \( p \):
\[ v = \frac{p}{m} \]

3. **Centripetal Force in Circular Motion**:
For an object moving in a circle of radius \( r \) with a constant speed \( v \), the centripetal (radial) force \( F \) required to keep the object in circular motion is given by:
\[ F = \frac{mv^2}{r} \]

4. **Substituting Velocity in Terms of Momentum**:
Now, we will substitute \( v \) from step 2 into the centripetal force equation:
\[ F = \frac{m\left(\frac{p}{m}\right)^2}{r} \]

5. **Simplifying the Expression**:
Simplifying the expression gives:
\[ F = \frac{m \cdot \frac{p^2}{m^2}}{r} = \frac{p^2}{mr} \]

6. **Final Result**:
Therefore, the radial force acting on the particle is:
\[ F = \frac{p^2}{mr} \]