A torque of 50 N-m acting on a wheel at rest rotates it through 200 rad in 5 sec. Its angular acceleration is -

Options

A. `0.16rad//s^(2)`
B. `1.6rad//s^(2)`
C. `16rad//s^(2)`
D. `20" "rad//s^(2)`

Explanation

To find the angular acceleration of the wheel, we can use the kinematic equation for rotational motion that is analogous to the linear motion equation \( s = ut + \frac{1}{2} a t^2 \). In this case, we will replace the linear variables with their rotational counterparts.

### Step-by-Step Solution:

1. **Identify the given values:**
- Torque (\( \tau \)) = 50 N-m
- Angular displacement (\( \theta \)) = 200 rad
- Time (\( t \)) = 5 s
- Initial angular velocity (\( \omega_0 \)) = 0 rad/s (since the wheel is at rest)

2. **Use the rotational kinematic equation:**
The equation we will use is:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Since the initial angular velocity (\( \omega_0 \)) is 0, the equation simplifies to:
\[ \theta = \frac{1}{2} \alpha t^2 \]

3. **Substitute the known values into the equation:**
\[ 200 = \frac{1}{2} \alpha (5^2) \]
\[ 200 = \frac{1}{2} \alpha (25) \]
\[ 200 = 12.5 \alpha \]

4. **Solve for angular acceleration (\( \alpha \)):**
\[ \alpha = \frac{200}{12.5} \]
\[ \alpha = 16 \text{ rad/s}^2 \]

5. **Final Answer:**
The angular acceleration (\( \alpha \)) is \( 16 \text{ rad/s}^2 \).