On the application of a constant torque , a wheel is turned from rest through 400 radians in 10 s. The angular acceleration is :

Options

A. `8 rad s^(-2)`
B. `5 rad s^(-2)`
C. `6 rad s^(-2)`
D. `7 rad s^(-2)`

Answer

To find the angular acceleration of the wheel, we can use the equation of motion for rotational dynamics, which is analogous to the linear motion equations. The relevant equation is:

\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]

Where:
- \(\theta\) is the angular displacement (in radians),
- \(\omega_0\) is the initial angular velocity (in radians per second),
- \(t\) is the time (in seconds),
- \(\alpha\) is the angular acceleration (in radians per second squared).

### Step 1: Identify the known values
From the problem:
- \(\theta = 400\) radians (angular displacement),
- \(\omega_0 = 0\) radians/second (since the wheel starts from rest),
- \(t = 10\) seconds.

### Step 2: Substitute the known values into the equation
Since the initial angular velocity \(\omega_0\) is zero, the equation simplifies to:

\[ 400 = 0 \cdot 10 + \frac{1}{2} \alpha (10^2) \]

This simplifies to:

\[ 400 = \frac{1}{2} \alpha (100) \]

### Step 3: Solve for angular acceleration \(\alpha\)
Now, we can simplify the equation further:

\[ 400 = 50 \alpha \]

To find \(\alpha\), divide both sides by 50:

\[ \alpha = \frac{400}{50} = 8 \text{ radians/second}^2 \]

### Final Answer
The angular acceleration \(\alpha\) is \(8 \text{ radians/second}^2\).

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