Centripetal acceleration of a particle of mass m moving in a circle of radius r is `4//r^(2)`. What is the angular momentum of the particle ?

Options

A. 4 m/r
B. `4 m// sqrt(r)`
C. `2 m sqrt(r)`
D. 2 m/r

Explanation

To find the angular momentum of a particle moving in a circle with a given centripetal acceleration, we can follow these steps:

### Step 1: Understand the relationship between centripetal acceleration and speed
The centripetal acceleration \( a_c \) of a particle moving in a circle is given by the formula:
\[ a_c = \frac{v^2}{r} \]
where \( v \) is the linear speed of the particle and \( r \) is the radius of the circle.

### Step 2: Set up the equation using the given centripetal acceleration
We are given that the centripetal acceleration is:
\[ a_c = \frac{4}{r^2} \]
Setting this equal to the formula for centripetal acceleration, we have:
\[ \frac{v^2}{r} = \frac{4}{r^2} \]

### Step 3: Solve for the speed \( v \)
To find \( v^2 \), we can rearrange the equation:
\[ v^2 = \frac{4}{r^2} \cdot r \]
This simplifies to:
\[ v^2 = \frac{4}{r} \]
Taking the square root of both sides gives us:
\[ v = \sqrt{\frac{4}{r}} = \frac{2}{\sqrt{r}} \]

### Step 4: Calculate the angular velocity \( \omega \)
The relationship between linear speed \( v \) and angular velocity \( \omega \) is given by:
\[ v = r \omega \]
Rearranging this gives:
\[ \omega = \frac{v}{r} \]
Substituting the expression for \( v \):
\[ \omega = \frac{2/\sqrt{r}}{r} = \frac{2}{r \sqrt{r}} = \frac{2}{r^{3/2}} \]

### Step 5: Find the moment of inertia \( I \)
For a point mass \( m \) at a distance \( r \) from the axis of rotation, the moment of inertia \( I \) is given by:
\[ I = m r^2 \]

### Step 6: Calculate the angular momentum \( L \)
The angular momentum \( L \) of a particle is given by the formula:
\[ L = I \omega \]
Substituting the expressions for \( I \) and \( \omega \):
\[ L = (m r^2) \left(\frac{2}{r^{3/2}}\right) \]
This simplifies to:
\[ L = 2m \frac{r^2}{r^{3/2}} = 2m r^{1/2} \]

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

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