A particle starts rotating from rest. Its angular displacement is expressed by the following equation `theta=0.025t^(2)-0.1t` where `theta` is in radian and t is in seconds. The angular acceleration of the particle is

Options

A. 0.5 rad/sec at the end of 10 sec
B. 0.3 rad/sec at the end of 2 sec
C. 0.05 rad/sec at the end of 1 sec
D. Constant 0.05 rad/sec

Answer

To find the angular acceleration of the particle given the angular displacement equation \( \theta = 0.025t^2 - 0.1t \), we will follow these steps:

### Step 1: Differentiate the angular displacement to find angular velocity
The angular velocity \( \omega \) is the first derivative of angular displacement \( \theta \) with respect to time \( t \).

\[ \omega = \frac{d\theta}{dt} = \frac{d}{dt}(0.025t^2 - 0.1t) \]

### Step 2: Calculate the derivative
Using the power rule of differentiation:

\[ \omega = 0.025 \cdot 2t - 0.1 = 0.05t - 0.1 \]

### Step 3: Differentiate angular velocity to find angular acceleration
The angular acceleration \( \alpha \) is the derivative of angular velocity \( \omega \) with respect to time \( t \).

\[ \alpha = \frac{d\omega}{dt} = \frac{d}{dt}(0.05t - 0.1) \]

### Step 4: Calculate the derivative
Again, applying the power rule:

\[ \alpha = 0.05 \]

### Step 5: State the result
The angular acceleration of the particle is constant and equal to:

\[ \alpha = 0.05 \, \text{radians per second squared} \]