A wheel starting from rest is uniformly accelerated at `2 red/s^(2)` for 20 seconds. It is allowed to rotate uniformly for the next 10 seconds and is finally brought to rest in next 20 seconds. The total angle rotated by the wheel (in radian) is :-

Options

A. 600
B. 1200
C. 1800
D. 300

Answer

To solve the problem step by step, we will break it down into three phases of the wheel's motion:

1. **Phase 1: Uniform Acceleration**
2. **Phase 2: Uniform Motion**
3. **Phase 3: Uniform Deceleration**

### Step 1: Calculate the final angular velocity after acceleration

The wheel starts from rest, so the initial angular velocity (\( \omega_0 \)) is 0. The angular acceleration (\( \alpha \)) is given as \( 2 \, \text{rad/s}^2 \) and the time (\( t_1 \)) for this phase is 20 seconds.

Using the formula for final angular velocity:
\[ \omega_f = \omega_0 + \alpha t_1 \]
Substituting the known values:
\[ \omega_f = 0 + (2 \, \text{rad/s}^2)(20 \, \text{s}) = 40 \, \text{rad/s} \]

### Step 2: Calculate the angle rotated during acceleration

Next, we calculate the angle (\( \theta_1 \)) rotated during the acceleration phase using the formula:
\[ \theta_1 = \omega_0 t_1 + \frac{1}{2} \alpha t_1^2 \]
Substituting the values:
\[ \theta_1 = 0 \cdot 20 + \frac{1}{2} \cdot 2 \cdot (20)^2 \]
\[ \theta_1 = 0 + \frac{1}{2} \cdot 2 \cdot 400 = 400 \, \text{radians} \]

### Step 3: Calculate the angle rotated during uniform motion

In the second phase, the wheel rotates uniformly for 10 seconds at the final angular velocity (\( \omega_f = 40 \, \text{rad/s} \)). The angle (\( \theta_2 \)) rotated during this phase can be calculated as:
\[ \theta_2 = \omega_f \cdot t_2 \]
Where \( t_2 = 10 \, \text{s} \):
\[ \theta_2 = 40 \, \text{rad/s} \cdot 10 \, \text{s} = 400 \, \text{radians} \]

### Step 4: Calculate the angle rotated during deceleration

In the third phase, the wheel comes to rest in 20 seconds. The initial angular velocity for this phase is \( \omega_f = 40 \, \text{rad/s} \) and the final angular velocity (\( \omega_f' \)) is 0. The angular deceleration (\( \alpha' \)) can be calculated using:
\[ \alpha' = \frac{\Delta \omega}{t_3} = \frac{0 - 40}{20} = -2 \, \text{rad/s}^2 \]

Now, we can calculate the angle (\( \theta_3 \)) rotated during this deceleration phase:
\[ \theta_3 = \omega_f t_3 + \frac{1}{2} \alpha' t_3^2 \]
Substituting the values:
\[ \theta_3 = 40 \cdot 20 + \frac{1}{2} \cdot (-2) \cdot (20)^2 \]
\[ \theta_3 = 800 - \frac{1}{2} \cdot 2 \cdot 400 = 800 - 400 = 400 \, \text{radians} \]

### Step 5: Calculate the total angle rotated

Finally, we sum up all the angles rotated in each phase:
\[ \theta_{total} = \theta_1 + \theta_2 + \theta_3 \]
\[ \theta_{total} = 400 + 400 + 400 = 1200 \, \text{radians} \]

### Final Answer
The total angle rotated by the wheel is \( 1200 \, \text{radians} \).

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