A uniform rod AB of mass m and length 2a is falling freely without rotation under gravity with AB horizontal. Suddenly the end A is fixed when the speed of the rod is v. The angular speed which the rod begains to rotate is

Options

A. `(v)/(2a)`
B. `(4v)/(3a)`
C. `(v)/(3a)`
D. `(3v)/(4a)`

Answer

To solve the problem, we will use the principle of conservation of angular momentum. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a uniform rod AB of mass \( m \) and length \( 2a \) that is falling freely under gravity with its horizontal position. The end A of the rod is suddenly fixed when the rod has a speed \( v \).

### Step 2: Identify the Initial Angular Momentum
Before the end A is fixed, the rod is falling with a linear velocity \( v \). The center of mass of the rod is at a distance \( a \) from point A (the midpoint of the rod). The initial angular momentum \( L_i \) about point A can be calculated using the formula:
\[ L_i = m \cdot v \cdot r \]
where \( r \) is the distance from point A to the center of mass, which is \( a \). Therefore, we have:
\[ L_i = m \cdot v \cdot a \]

### Step 3: Identify the Final Angular Momentum
Once point A is fixed, the rod will start to rotate about point A. The moment of inertia \( I \) of the rod about point A is given by:
\[ I = \frac{1}{3} m L^2 \]
where \( L = 2a \) (the length of the rod). Thus,
\[ I = \frac{1}{3} m (2a)^2 = \frac{4}{3} m a^2 \]

The final angular momentum \( L_f \) when the rod starts rotating with angular speed \( \omega \) is given by:
\[ L_f = I \cdot \omega = \frac{4}{3} m a^2 \cdot \omega \]

### Step 4: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
Substituting the expressions we found:
\[ m \cdot v \cdot a = \frac{4}{3} m a^2 \cdot \omega \]

### Step 5: Simplify and Solve for Angular Speed \( \omega \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ v \cdot a = \frac{4}{3} a^2 \cdot \omega \]
Now, divide both sides by \( a \) (assuming \( a \neq 0 \)):
\[ v = \frac{4}{3} a \cdot \omega \]
Rearranging gives:
\[ \omega = \frac{3v}{4a} \]

### Final Answer
Thus, the angular speed \( \omega \) that the rod begins to rotate with is:
\[ \omega = \frac{3v}{4a} \]