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, we need to find the angle made by the net acceleration with the radial acceleration when the particle rotates by \(60^\circ\). We will follow these steps:

### Step 1: Understand the given data
- Radius of the circle, \( R = \frac{2}{3} \, \text{m} \)
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Kinetic energy as a function of distance \( S \):
\[ K.E. = 4S^4 \]

### Step 2: Find the expression for velocity
The kinetic energy can also be expressed as:
\[ K.E. = \frac{1}{2} mv^2 \]
Equating the two expressions for kinetic energy:
\[ \frac{1}{2} mv^2 = 4S^4 \]
Substituting \( m = 2 \):
\[ \frac{1}{2} \cdot 2 \cdot v^2 = 4S^4 \implies v^2 = 8S^4 \]

### Step 3: Find radial acceleration \( A_c \)
The centripetal (radial) acceleration is given by:
\[ A_c = \frac{v^2}{R} \]
Substituting \( v^2 = 8S^4 \) and \( R = \frac{2}{3} \):
\[ A_c = \frac{8S^4}{\frac{2}{3}} = 12S^4 \]

### Step 4: Find tangential acceleration \( A_t \)
The tangential acceleration can be found from the change in velocity. We know:
\[ A_t = \frac{d(v^2)}{dt} \cdot \frac{1}{2v} \]
To find \( \frac{d(v^2)}{dt} \), we can use:
\[ v^2 = 8S^4 \implies \frac{d(v^2)}{dt} = 32S^3 \cdot \frac{dS}{dt} \]
Since \( \frac{dS}{dt} = v \), we have:
\[ A_t = \frac{32S^3 \cdot v}{2v} = 16S^3 \]

### Step 5: Calculate the angle \( \alpha \)
The net acceleration \( A_{net} \) can be found using the Pythagorean theorem:
\[ A_{net} = \sqrt{A_c^2 + A_t^2} \]
The angle \( \alpha \) made by the net acceleration with the radial acceleration is given by:
\[ \tan(\alpha) = \frac{A_t}{A_c} \]
Substituting the values:
\[ \tan(\alpha) = \frac{16S^3}{12S^4} = \frac{4}{3S} \]

### Step 6: Find \( S \) when the particle has rotated \( 60^\circ \)
The distance \( S \) traveled when the particle rotates by \( 60^\circ \) is given by:
\[ S = R \cdot \theta = \frac{2}{3} \cdot \frac{\pi}{3} = \frac{2\pi}{9} \]

### Step 7: Substitute \( S \) into the angle equation
Substituting \( S = \frac{2\pi}{9} \):
\[ \tan(\alpha) = \frac{4}{3 \cdot \frac{2\pi}{9}} = \frac{4 \cdot 9}{6\pi} = \frac{6}{\pi} \]

### Step 8: Find \( \alpha \)
Thus, the angle \( \alpha \) is:
\[ \alpha = \tan^{-1}\left(\frac{6}{\pi}\right) \]

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