A particle is moving in a circular orbit with a constant tangential acceleration. After a certain time t has elapsed after the beginning of motion, the between the total acceleration a and the direction along the radius r becomes equal to `45^(@)`. What is the angular acceleration of the particle.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have a particle moving in a circular orbit with a constant tangential acceleration. After a time \( t \), the angle between the total acceleration \( a \) and the radial direction \( r \) is \( 45^\circ \).

### Step 2: Define the accelerations
1. **Tangential Acceleration (\( a_t \))**: This is given by the formula:
\[ a_t = \alpha \cdot r \]
where \( \alpha \) is the angular acceleration and \( r \) is the radius of the circular path.

2. **Radial (Centripetal) Acceleration (\( a_r \))**: This is given by:
\[ a_r = \omega^2 \cdot r \]
where \( \omega \) is the angular velocity. Since the particle has a constant tangential acceleration, we can express \( \omega \) as:
\[ \omega = \alpha \cdot t \]
Therefore, substituting for \( \omega \):
\[ a_r = (\alpha \cdot t)^2 \cdot r = \alpha^2 \cdot t^2 \cdot r \]

### Step 3: Relate the accelerations using the angle
Given that the angle between the total acceleration \( a \) and the radial acceleration \( a_r \) is \( 45^\circ \), we can use the tangent of the angle:
\[ \tan(45^\circ) = 1 = \frac{a_t}{a_r} \]
This implies:
\[ a_t = a_r \]

### Step 4: Set the equations equal
From the definitions of tangential and radial accelerations:
\[ \alpha \cdot r = \alpha^2 \cdot t^2 \cdot r \]

### Step 5: Simplify the equation
Assuming \( r \neq 0 \), we can divide both sides by \( r \):
\[ \alpha = \alpha^2 \cdot t^2 \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ \alpha^2 \cdot t^2 - \alpha = 0 \]
Factoring out \( \alpha \):
\[ \alpha(\alpha \cdot t^2 - 1) = 0 \]

### Step 7: Solve for \( \alpha \)
This gives us two solutions:
1. \( \alpha = 0 \) (not applicable since the particle is accelerating)
2. \( \alpha \cdot t^2 - 1 = 0 \) which leads to:
\[ \alpha = \frac{1}{t^2} \]

### Final Answer
Thus, the angular acceleration of the particle is:
\[ \alpha = \frac{1}{t^2} \]

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