Two blocks A and B having masses 5 kg and 10 kg respectively connected through a massless string which is passing over a pulley (disc) of mass 2 kg and radius 25 cm as shown in figure If system is. released from rest. then angular acceleration of the pulley is (String does not slip over the pulley horizontal surface is smooth and `g = 10ms ^-2` `(AAK_TST_02_NEET_PHY_E02_011_Q01)`

Answer

To find the angular acceleration of the pulley in the given system, we can follow these steps:

### Step 1: Identify the forces acting on the blocks
- For block A (mass = 5 kg), the forces are:
- Weight (downward): \( W_A = m_A \cdot g = 5 \cdot 10 = 50 \, \text{N} \)
- Tension in the string (upward): \( T_A \)

- For block B (mass = 10 kg), the forces are:
- Weight (downward): \( W_B = m_B \cdot g = 10 \cdot 10 = 100 \, \text{N} \)
- Tension in the string (upward): \( T_B \)

### Step 2: Write the equations of motion for both blocks
- For block A (moving upwards):
\[ T_A = m_A \cdot g - m_A \cdot a \quad \text{(1)} \]
Substituting \( m_A = 5 \, \text{kg} \):
\[ T_A = 50 - 5a \]

- For block B (moving downwards):
\[ m_B \cdot g - T_B = m_B \cdot a \quad \text{(2)} \]
Substituting \( m_B = 10 \, \text{kg} \):
\[ 100 - T_B = 10a \]

### Step 3: Relate the tensions to the pulley
The net torque (\( \tau \)) on the pulley can be expressed in terms of the tensions:
\[ \tau = (T_B - T_A) \cdot r \]
Where \( r = 0.25 \, \text{m} \) (25 cm converted to meters).

The moment of inertia (\( I \)) of the pulley (disc) is given by:
\[ I = \frac{1}{2} m_{pulley} r^2 = \frac{1}{2} \cdot 2 \cdot (0.25)^2 = \frac{1}{2} \cdot 2 \cdot 0.0625 = 0.0625 \, \text{kg m}^2 \]

The angular acceleration (\( \alpha \)) is related to the linear acceleration (\( a \)) by:
\[ \alpha = \frac{a}{r} \]

### Step 4: Write the equation for torque
Using \( \tau = I \cdot \alpha \):
\[ (T_B - T_A) \cdot 0.25 = 0.0625 \cdot \alpha \]
Substituting \( \alpha = \frac{a}{0.25} \):
\[ (T_B - T_A) \cdot 0.25 = 0.0625 \cdot \frac{a}{0.25} \]
Simplifying gives:
\[ (T_B - T_A) = 0.25 \cdot 0.25 \cdot a = 0.015625 a \quad \text{(3)} \]

### Step 5: Substitute equations (1) and (2) into (3)
Substituting \( T_A \) and \( T_B \) from equations (1) and (2) into equation (3):
\[ (100 - 10a) - (50 - 5a) = 0.015625 a \]
Simplifying:
\[ 100 - 10a - 50 + 5a = 0.015625 a \]
\[ 50 - 5a = 0.015625 a \]
Combining like terms:
\[ 50 = 5a + 0.015625 a \]
\[ 50 = 5.015625 a \]
\[ a = \frac{50}{5.015625} \approx 9.95 \, \text{m/s}^2 \]

### Step 6: Calculate angular acceleration
Now, substituting \( a \) back to find \( \alpha \):
\[ \alpha = \frac{a}{r} = \frac{9.95}{0.25} \approx 39.8 \, \text{rad/s}^2 \]

### Final Answer
The angular acceleration of the pulley is approximately \( 39.8 \, \text{rad/s}^2 \).

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