A point situated on a wheel decelerates obeying the relation `omega= omega_(0) - atheta` is angular displacement counted from t = 0. Which of the following graphs represents the angular acceleration of the wheel.

Options

A.
B.
C.
D.

Answer

To solve the problem, we need to analyze the given relation for angular velocity and find the corresponding angular acceleration. Let's break it down step by step.

### Step 1: Understand the given relation
The problem states that the angular velocity \(\omega\) of a point on the wheel is given by:
\[ \omega = \omega_0 - a\theta \]
where \(\omega_0\) is the initial angular velocity, \(a\) is a constant, and \(\theta\) is the angular displacement.

### Step 2: Relate angular velocity to angular displacement
We know that angular velocity \(\omega\) can also be expressed as the derivative of angular displacement with respect to time:
\[ \omega = \frac{d\theta}{dt} \]
Thus, we can rewrite the equation as:
\[ \frac{d\theta}{dt} = \omega_0 - a\theta \]

### Step 3: Rearranging the equation
To solve for \(\theta\) in terms of \(t\), we can rearrange the equation:
\[ d\theta = (\omega_0 - a\theta) dt \]

### Step 4: Integrate both sides
We will integrate both sides. The left side will be integrated with respect to \(\theta\) from \(0\) to \(\theta\), and the right side will be integrated with respect to \(t\) from \(0\) to \(t\):
\[ \int_0^\theta \frac{1}{\omega_0 - a\theta} d\theta = \int_0^t dt \]

### Step 5: Solve the integral
The left side integral can be solved using the natural logarithm:
\[ \log(\omega_0 - a\theta) - \log(\omega_0) = t \]
This can be simplified to:
\[ \log\left(\frac{\omega_0 - a\theta}{\omega_0}\right) = t \]

### Step 6: Exponentiate to eliminate the logarithm
Exponentiating both sides gives:
\[ \frac{\omega_0 - a\theta}{\omega_0} = e^{-at} \]
Rearranging gives:
\[ \omega_0 - a\theta = \omega_0 e^{-at} \]
So,
\[ a\theta = \omega_0(1 - e^{-at}) \]

### Step 7: Differentiate to find angular acceleration
Now, we need to find the angular acceleration \(\alpha\), which is the derivative of angular velocity with respect to time:
\[ \alpha = \frac{d\omega}{dt} \]
From our earlier expression for \(\omega\):
\[ \omega = \omega_0 e^{-at} \]
Differentiating gives:
\[ \alpha = -a \omega_0 e^{-at} \]

### Step 8: Analyze the result
The angular acceleration \(\alpha\) is negative, indicating deceleration, and it decreases exponentially with time. The graph of \(\alpha\) will start from a negative value and approach zero as \(t\) increases.

### Conclusion
The correct graph representing the angular acceleration of the wheel will show an exponential decay starting from \(-a\omega_0\) and approaching zero.