A particle is moving in a circle of radius `(2)/( 3) m` and mass of the particle is 2kg. The kinetic energy of the particle dependes on distance 'S' travelled by the particle as K.E. = `4S^(4)`. The angle made by net acceleration with the radial acceleration when the particle rotate by `60^(@)` is

Options

A. `tan^(-1) ((3)/( pi ))`
B. `tan^(-1) ((6)/( pi ))`
C. `tan^(-1) ((1)/( pi ))`
D. `tan^(-1) ((4)/( pi ))`

Answer

To solve the problem step by step, we will analyze the motion of the particle and calculate the required angle made by the net acceleration with the radial acceleration.

### Step 1: Understand the given data
- Radius of the circular path, \( R = \frac{2}{3} \, \text{m} \)
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Kinetic energy of the particle, \( K.E. = 4S^4 \)

### Step 2: Calculate the velocity of the particle
The kinetic energy can also be expressed in terms of mass and velocity:
\[ K.E. = \frac{1}{2} m v^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ 4S^4 = \frac{1}{2} \cdot 2 \cdot v^2 \]
This simplifies to:
\[ 4S^4 = v^2 \]
Thus, we have:
\[ v^2 = 4S^4 \]

### Step 3: Determine the centripetal acceleration
Centripetal acceleration \( a_c \) is given by:
\[ a_c = \frac{v^2}{R} \]
Substituting \( v^2 \) from the previous step:
\[ a_c = \frac{4S^4}{\frac{2}{3}} = 6S^4 \]

### Step 4: Calculate the tangential acceleration
To find the tangential acceleration \( a_t \), we need to differentiate the expression for velocity with respect to time. The velocity \( v \) can be expressed as:
\[ v = 2S^2 \]
Now, the tangential acceleration is given by:
\[ a_t = \frac{dv}{dt} = \frac{d(2S^2)}{dt} = 4S \frac{dS}{dt} \]
Since \( \frac{dS}{dt} = v \):
\[ a_t = 4S \cdot 2S^2 = 8S^3 \]

### Step 5: Find the angle between net acceleration and radial acceleration
The net acceleration \( a \) is the vector sum of the tangential acceleration \( a_t \) and the centripetal acceleration \( a_c \). The angle \( \theta \) between the net acceleration and the radial acceleration can be found using:
\[ \tan \theta = \frac{a_t}{a_c} \]
Substituting the values we calculated:
\[ \tan \theta = \frac{8S^3}{6S^4} = \frac{4}{3S} \]

### Step 6: Calculate the distance \( S \) traveled when the particle rotates by \( 60^\circ \)
The distance \( S \) can be calculated using the formula:
\[ S = R \theta \]
where \( \theta \) in radians for \( 60^\circ \) is \( \frac{\pi}{3} \):
\[ S = \frac{2}{3} \cdot \frac{\pi}{3} = \frac{2\pi}{9} \]

### Step 7: Substitute \( S \) back into the equation for \( \tan \theta \)
Now substituting \( S = \frac{2\pi}{9} \):
\[ \tan \theta = \frac{4}{3 \cdot \frac{2\pi}{9}} = \frac{4 \cdot 9}{6\pi} = \frac{6}{\pi} \]

### Step 8: Find the angle \( \theta \)
Finally, we find \( \theta \) using the arctangent function:
\[ \theta = \tan^{-1}\left(\frac{6}{\pi}\right) \]

### Final Answer
Thus, the angle made by the net acceleration with the radial acceleration when the particle rotates by \( 60^\circ \) is:
\[ \theta = \tan^{-1}\left(\frac{6}{\pi}\right) \]