A flywheel revolves at 100 rev/min, a torque is applied to the flywheel for 10 s If the torque increases the speed to 200 rev/min, then the angular acceleration of the flywheel will be

Options

A. `(pi)/(6)rad//s^(2)`
B. `(pi)/(5)rad//s^(2)`
C. `(pi)/(4)rad//s^(2)`
D. `(pi)/(3)rad//s^(2)`

Explanation

To find the angular acceleration of the flywheel, we can follow these steps:

### Step 1: Convert revolutions per minute (rev/min) to radians per second (rad/s)
The initial angular velocity \( \omega_1 \) is given as 100 rev/min. We need to convert this to rad/s.

\[ \omega_1 = 100 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{100 \times 2\pi}{60} \, \text{rad/s} \]

Calculating this gives:

\[ \omega_1 = \frac{200\pi}{60} \approx 10.47 \, \text{rad/s} \]

### Step 2: Convert the final angular velocity (200 rev/min) to rad/s
The final angular velocity \( \omega_2 \) is given as 200 rev/min. We convert this similarly:

\[ \omega_2 = 200 \, \text{rev/min} \times \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{s}} = \frac{200 \times 2\pi}{60} \, \text{rad/s} \]

Calculating this gives:

\[ \omega_2 = \frac{400\pi}{60} \approx 20.94 \, \text{rad/s} \]

### Step 3: Calculate the change in angular velocity
The change in angular velocity \( \Delta \omega \) is:

\[ \Delta \omega = \omega_2 - \omega_1 \]

Substituting the values we found:

\[ \Delta \omega = 20.94 \, \text{rad/s} - 10.47 \, \text{rad/s} = 10.47 \, \text{rad/s} \]

### Step 4: Calculate the angular acceleration
Angular acceleration \( \alpha \) is defined as the change in angular velocity over time:

\[ \alpha = \frac{\Delta \omega}{\Delta t} \]

Given that the torque is applied for \( \Delta t = 10 \, \text{s} \):

\[ \alpha = \frac{10.47 \, \text{rad/s}}{10 \, \text{s}} = 1.047 \, \text{rad/s}^2 \]

### Final Answer
The angular acceleration of the flywheel is approximately \( 1.05 \, \text{rad/s}^2 \).

---