A uniform rod of mass `2 kg` and length `1 m` lies on a smooth horizontal plane. A particle of mass 1 kg moving at a speed of `2ms^-1` perpendicular to the length of the rod strikes it at a distance `(1)/(4)` m from the centre and stops . Find the angular velocity of the rod about its centre just after the collision (in rad/s)

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given data
- Mass of the rod, \( m_{rod} = 2 \, \text{kg} \)
- Length of the rod, \( l = 1 \, \text{m} \)
- Mass of the particle, \( m_{particle} = 1 \, \text{kg} \)
- Speed of the particle, \( v = 2 \, \text{m/s} \)
- Distance from the center of the rod where the particle strikes, \( d = \frac{1}{4} \, \text{m} \)

### Step 2: Calculate the initial momentum of the system
The initial momentum of the particle before the collision is given by:
\[ p_{initial} = m_{particle} \cdot v = 1 \, \text{kg} \cdot 2 \, \text{m/s} = 2 \, \text{kg m/s} \]

### Step 3: Apply the conservation of linear momentum
After the collision, the particle comes to a stop, and the rod starts moving. Let \( V \) be the velocity of the center of mass of the rod after the collision. According to the conservation of linear momentum:
\[ p_{initial} = p_{final} \]
\[ 2 \, \text{kg m/s} = m_{rod} \cdot V \]
Substituting the mass of the rod:
\[ 2 = 2 \cdot V \implies V = 1 \, \text{m/s} \]

### Step 4: Calculate the moment of inertia of the rod
The moment of inertia \( I \) of a uniform rod about its center is given by:
\[ I = \frac{1}{12} m_{rod} l^2 = \frac{1}{12} \cdot 2 \cdot (1)^2 = \frac{1}{6} \, \text{kg m}^2 \]

### Step 5: Calculate the angular momentum before the collision
The angular momentum \( L \) of the particle about the center of the rod just before the collision is given by:
\[ L_{initial} = m_{particle} \cdot v \cdot d = 1 \cdot 2 \cdot \frac{1}{4} = \frac{1}{2} \, \text{kg m}^2/s \]

### Step 6: Calculate the angular momentum after the collision
After the collision, the angular momentum is given by:
\[ L_{final} = I \cdot \omega \]
where \( \omega \) is the angular velocity we want to find. Setting the initial angular momentum equal to the final angular momentum:
\[ \frac{1}{2} = \frac{1}{6} \cdot \omega \]

### Step 7: Solve for angular velocity \( \omega \)
Rearranging the equation gives:
\[ \omega = \frac{1/2}{1/6} = \frac{1/2 \cdot 6}{1} = 3 \, \text{rad/s} \]

### Conclusion
The angular velocity of the rod about its center just after the collision is \( \omega = 3 \, \text{rad/s} \).

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