Two masses are attached to a rod end to end . If torque is applied they rotate with angular acceleration `alpha`. If their distances are doubled and same torque is applied, then they move with angular acceleration.

Options

A. `4alpha`
B. `alpha`
C. `3alpha`
D. `alpha//4`

Explanation

To solve the problem step by step, we will analyze the situation using the concepts of torque and moment of inertia.

### Step 1: Define the initial conditions
Let the two masses be \( m_1 \) and \( m_2 \) attached to a rod at distances \( r \) from the axis of rotation. The torque \( \tau \) applied to the system results in an angular acceleration \( \alpha \).

### Step 2: Calculate the moment of inertia for the initial configuration
The moment of inertia \( I_1 \) for the two masses is given by the formula:
\[ I_1 = m_1 r^2 + m_2 r^2 = (m_1 + m_2) r^2 \]

### Step 3: Relate torque and angular acceleration for the initial configuration
Using the relationship between torque, moment of inertia, and angular acceleration:
\[ \tau = I_1 \alpha \]
Substituting \( I_1 \):
\[ \tau = (m_1 + m_2) r^2 \alpha \]

### Step 4: Define the new conditions after doubling the distances
Now, the distances of the masses from the axis of rotation are doubled, so they are at \( 2r \). The new moment of inertia \( I_2 \) is calculated as:
\[ I_2 = m_1 (2r)^2 + m_2 (2r)^2 = m_1 \cdot 4r^2 + m_2 \cdot 4r^2 = (m_1 + m_2) \cdot 4r^2 \]

### Step 5: Relate torque and angular acceleration for the new configuration
The same torque \( \tau \) is applied, so we have:
\[ \tau = I_2 \alpha' \]
Substituting \( I_2 \):
\[ \tau = (m_1 + m_2) \cdot 4r^2 \alpha' \]

### Step 6: Set the two torque equations equal to each other
Since the torque is the same in both cases:
\[ (m_1 + m_2) r^2 \alpha = (m_1 + m_2) \cdot 4r^2 \alpha' \]

### Step 7: Simplify the equation
We can cancel \( (m_1 + m_2) \) and \( r^2 \) from both sides (assuming they are not zero):
\[ \alpha = 4 \alpha' \]

### Step 8: Solve for the new angular acceleration
Rearranging gives:
\[ \alpha' = \frac{\alpha}{4} \]

### Conclusion
Thus, when the distances are doubled and the same torque is applied, the new angular acceleration \( \alpha' \) is:
\[ \alpha' = \frac{\alpha}{4} \]