A particle of mass `m` describes a circle of radius `r` . The centripetal acceleration of the particle is `(4)/(r^(2))` . What will be the momentum of the particle ?

Options

A. `2(m)/(r)`
B. `2(m)/(sqrt(r))`
C. `4(m)/(sqrt(r))`
D. None of these

Explanation

To solve the problem step by step, we will use the relationship between centripetal acceleration, velocity, and momentum.

### Step 1: Understand the given information
We have a particle of mass \( m \) moving in a circular path of radius \( r \). The centripetal acceleration \( a_c \) is given as:
\[ a_c = \frac{4}{r^2} \]

### Step 2: Recall the formula for centripetal acceleration
The formula for centripetal acceleration is given by:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the linear velocity of the particle.

### Step 3: Set the two expressions for centripetal acceleration equal
Since we have two expressions for centripetal acceleration, we can set them equal to each other:
\[ \frac{v^2}{r} = \frac{4}{r^2} \]

### Step 4: Solve for velocity \( v \)
To find \( v^2 \), we can multiply both sides of the equation by \( r \):
\[ v^2 = \frac{4}{r^2} \cdot r = \frac{4}{r} \]
Now, take the square root of both sides to find \( v \):
\[ v = \sqrt{\frac{4}{r}} = \frac{2}{\sqrt{r}} \]

### Step 5: Calculate the momentum \( P \)
The momentum \( P \) of the particle is given by the product of its mass and velocity:
\[ P = m \cdot v \]
Substituting the value of \( v \) we found:
\[ P = m \cdot \frac{2}{\sqrt{r}} = \frac{2m}{\sqrt{r}} \]

### Conclusion
Thus, the momentum of the particle is:
\[ P = \frac{2m}{\sqrt{r}} \]