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 follow these instructions:

### Step 1: Write down the given values
We have:
- Mass of the particle, \( m \)
- Radius of the circle, \( r \)
- Centripetal acceleration, \( a_c = \frac{4}{r^2} \)

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

### Step 3: Substitute the given centripetal acceleration into the formula
We can substitute the given value of centripetal acceleration into the formula:
\[ \frac{4}{r^2} = \frac{v^2}{r} \]

### Step 4: Rearrange the equation to solve for \( v^2 \)
Multiplying both sides by \( r \) gives:
\[ \frac{4}{r^2} \cdot r = v^2 \]
This simplifies to:
\[ \frac{4}{r} = v^2 \]

### Step 5: Solve for \( v \)
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{4}{r}} = \frac{2}{\sqrt{r}} \]

### Step 6: Calculate the momentum
The momentum \( p \) of the particle is given by the formula:
\[ p = mv \]
Substituting the value of \( v \) we found:
\[ p = m \cdot \frac{2}{\sqrt{r}} = \frac{2m}{\sqrt{r}} \]

### Final Answer
Thus, the momentum of the particle is:
\[ p = 2m \sqrt{r} \]

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