Calculate the magnitude and direction of total linear acceleration of a particle moving in a circle of radius 0.4 m having an instantaneous angular velocity of `2 "rad" s^(-1)` and angular acceleration 5"rad"s^(-2)

Explanation

To solve the problem of calculating the magnitude and direction of the total linear acceleration of a particle moving in a circle, we will follow these steps:

### Step 1: Identify the given values
- Radius of the circle (R) = 0.4 m
- Instantaneous angular velocity (ω) = 2 rad/s
- Angular acceleration (α) = 5 rad/s²

### Step 2: Calculate the radial acceleration (A_r)
Radial acceleration (A_r) is given by the formula:
\[ A_r = R \cdot \omega^2 \]
Substituting the values:
\[ A_r = 0.4 \, \text{m} \cdot (2 \, \text{rad/s})^2 \]
\[ A_r = 0.4 \cdot 4 \]
\[ A_r = 1.6 \, \text{m/s}^2 \]

### Step 3: Calculate the tangential acceleration (A_t)
Tangential acceleration (A_t) is given by the formula:
\[ A_t = R \cdot \alpha \]
Substituting the values:
\[ A_t = 0.4 \, \text{m} \cdot 5 \, \text{rad/s}^2 \]
\[ A_t = 2.0 \, \text{m/s}^2 \]

### Step 4: Calculate the total linear acceleration (A_net)
The total linear acceleration (A_net) is the vector sum of radial and tangential accelerations. We can calculate it using the Pythagorean theorem:
\[ A_{net} = \sqrt{A_r^2 + A_t^2} \]
Substituting the values:
\[ A_{net} = \sqrt{(1.6 \, \text{m/s}^2)^2 + (2.0 \, \text{m/s}^2)^2} \]
\[ A_{net} = \sqrt{2.56 + 4} \]
\[ A_{net} = \sqrt{6.56} \]
\[ A_{net} \approx 2.56 \, \text{m/s}^2 \]

### Step 5: Determine the direction of total linear acceleration
The direction of the total linear acceleration can be found using the tangent of the angle θ:
\[ \tan(\theta) = \frac{A_r}{A_t} \]
Substituting the values:
\[ \tan(\theta) = \frac{1.6}{2.0} \]
Calculating θ:
\[ \theta = \tan^{-1}(0.8) \]
Using a calculator:
\[ \theta \approx 38.46^\circ \]

### Final Result
- Magnitude of total linear acceleration: **2.56 m/s²**
- Direction of total linear acceleration: **38.46° with respect to the tangential direction**

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