A bar of mass m resting on a smooth horizontal plane starts moving due to the force `F=mg//3` of constant magnitude. In the process of its rectilinear motion the angle `alpha` between the direction of this force and the horizontal varies as `alpha=as`, where a is a constant, and s is the distance traversed by the bar from its initial position. Find the velocity of the bar as a function of the angle `alpha`.

Answer

To solve the problem step by step, we will analyze the motion of the bar under the influence of the force \( F = \frac{mg}{3} \) and the relationship between the angle \( \alpha \) and the distance \( s \).

### Step 1: Understand the Forces Acting on the Bar
The bar of mass \( m \) is subjected to:
- Its weight \( mg \) acting downwards.
- The applied force \( F = \frac{mg}{3} \) at an angle \( \alpha \) with the horizontal.

### Step 2: Resolve the Applied Force into Components
The applied force can be resolved into two components:
- Horizontal component: \( F \cos \alpha \)
- Vertical component: \( F \sin \alpha \)

### Step 3: Write the Equations of Motion
Using Newton's second law, we can write the equations for the vertical and horizontal directions.

**Vertical Direction:**
\[ N + F \sin \alpha = mg \]
Where \( N \) is the normal force.

**Horizontal Direction:**
\[ F \cos \alpha = m a \]
Where \( a \) is the horizontal acceleration of the bar.

### Step 4: Substitute the Force
Substituting \( F = \frac{mg}{3} \) into the equations:
1. **Vertical Equation:**
\[ N + \frac{mg}{3} \sin \alpha = mg \]
Rearranging gives:
\[ N = mg - \frac{mg}{3} \sin \alpha = mg \left(1 - \frac{1}{3} \sin \alpha\right) \]

2. **Horizontal Equation:**
\[ \frac{mg}{3} \cos \alpha = m a \]
Simplifying gives:
\[ a = \frac{g \cos \alpha}{3} \]

### Step 5: Relate Acceleration to Velocity
We know that acceleration can also be expressed as:
\[ a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds} \]
Thus, we can write:
\[ v \frac{dv}{ds} = \frac{g \cos \alpha}{3} \]

### Step 6: Substitute the Angle Relation
Given that \( \alpha = as \), we can express \( \cos \alpha \) in terms of \( s \):
\[ \cos \alpha = \cos(as) \]
So, the equation becomes:
\[ v \frac{dv}{ds} = \frac{g \cos(as)}{3} \]

### Step 7: Integrate the Equation
We can separate variables and integrate:
\[ \int v \, dv = \frac{g}{3} \int \cos(as) \, ds \]
This gives:
\[ \frac{v^2}{2} = \frac{g}{3} \int \cos(as) \, ds \]

### Step 8: Solve the Integral
The integral \( \int \cos(as) \, ds = \frac{1}{a} \sin(as) \), so we have:
\[ \frac{v^2}{2} = \frac{g}{3} \cdot \frac{1}{a} \sin(as) + C \]
Assuming initial conditions where \( v = 0 \) when \( s = 0 \), we find \( C = 0 \).

### Step 9: Final Expression for Velocity
Thus, we can express \( v^2 \) as:
\[ v^2 = \frac{2g}{3a} \sin(as) \]
Taking the square root gives:
\[ v = \sqrt{\frac{2g}{3a} \sin(as)} \]

### Final Answer
The velocity of the bar as a function of the angle \( \alpha \) is:
\[ v = \sqrt{\frac{2g}{3a} \sin(a s)} \]