A solid body rotates with deceleration about a stationary axis with an angular deceleration `|alpha|= ksqrt(omega)` where k is a constant and `omega` is the angular velocity of the body. If the initial angular velocity is `omega_(0)` then mean angular velocity of the body averaged over the whole time of rotation is

Options

A. `(7omega_(0))/(3)`
B. `(4omega_(0))/(3)`
C. `(omega_(0))/(3)`
D. `7omega_(0)`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the transcript and derive the mean angular velocity of the body.

### Step 1: Understand the relationship between angular deceleration and angular velocity
Given that the angular deceleration \(|\alpha| = k\sqrt{\omega}\), we can express this as:
\[ \frac{d\omega}{dt} = -k\sqrt{\omega} \]
Here, the negative sign indicates that the body is decelerating.

### Step 2: Rearranging the equation for integration
Rearranging the equation, we have:
\[ \frac{d\omega}{\sqrt{\omega}} = -k \, dt \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side will be integrated with respect to \(\omega\) from \(\omega_0\) to \(\omega\), and the right side will be integrated with respect to \(t\) from 0 to \(t\):
\[ \int_{\omega_0}^{\omega} \frac{d\omega}{\sqrt{\omega}} = -k \int_{0}^{t} dt \]

The left side integrates to:
\[ 2(\sqrt{\omega} - \sqrt{\omega_0}) = -kt \]
Thus, we can rearrange this to find \(\sqrt{\omega}\):
\[ \sqrt{\omega} = \sqrt{\omega_0} - \frac{kt}{2} \]

### Step 4: Find the time when \(\omega = 0\)
Setting \(\omega = 0\) to find the total time of rotation:
\[ 0 = \sqrt{\omega_0} - \frac{kt}{2} \]
This gives us:
\[ t = \frac{2\sqrt{\omega_0}}{k} \]

### Step 5: Calculate the average angular velocity
The average angular velocity \(\bar{\omega}\) is given by the formula:
\[ \bar{\omega} = \frac{1}{T} \int_{0}^{T} \omega(t) \, dt \]
To find \(\omega(t)\), we substitute \(t\) back into our equation:
\[ \omega(t) = \left(\sqrt{\omega_0} - \frac{kt}{2}\right)^2 \]

### Step 6: Integrate \(\omega(t)\) over the time period
Now we need to evaluate:
\[ \bar{\omega} = \frac{1}{T} \int_{0}^{T} \left(\sqrt{\omega_0} - \frac{kt}{2}\right)^2 dt \]
Where \(T = \frac{2\sqrt{\omega_0}}{k}\).

### Step 7: Solve the integral
Expanding the integrand:
\[ \left(\sqrt{\omega_0} - \frac{kt}{2}\right)^2 = \omega_0 - k\sqrt{\omega_0}t + \frac{k^2t^2}{4} \]
Now integrate term by term:
\[ \int_{0}^{T} \left(\omega_0 - k\sqrt{\omega_0}t + \frac{k^2t^2}{4}\right) dt \]
Calculating each term:
1. \(\int_{0}^{T} \omega_0 \, dt = \omega_0 T\)
2. \(\int_{0}^{T} k\sqrt{\omega_0}t \, dt = \frac{k\sqrt{\omega_0}T^2}{2}\)
3. \(\int_{0}^{T} \frac{k^2t^2}{4} \, dt = \frac{k^2T^3}{12}\)

### Step 8: Substitute \(T\) and simplify
Substituting \(T = \frac{2\sqrt{\omega_0}}{k}\) into the average angular velocity formula:
\[ \bar{\omega} = \frac{1}{T} \left(\omega_0 T - \frac{k\sqrt{\omega_0}T^2}{2} + \frac{k^2T^3}{12}\right) \]
After simplification, we find:
\[ \bar{\omega} = \frac{\omega_0}{3} \]

### Final Result
Thus, the mean angular velocity of the body averaged over the whole time of rotation is:
\[ \bar{\omega} = \frac{\omega_0}{3} \]