A uniform disc of mass M, radius R is attached to a frictionless horizontal axis. Its rim is wound with a light string and a tension T is applied to it, the angular acceleration will be?

Options

A. 2T/M
B. 2T/MR
C. T/MR
D. T/M

Answer

To find the angular acceleration of a uniform disc of mass \( M \) and radius \( R \) when a tension \( T \) is applied to a string wound around its rim, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Torque**:
The tension \( T \) applied to the string generates a torque \( \tau \) on the disc. The torque is given by the formula:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the disc.

**Hint**: Remember that torque is the product of the force and the perpendicular distance from the axis of rotation.

2. **Moment of Inertia of the Disc**:
The moment of inertia \( I \) of a uniform disc about its central axis is given by:
\[ I = \frac{1}{2} M R^2 \]

**Hint**: The moment of inertia depends on the mass distribution of the object relative to the axis of rotation.

3. **Relate Torque to Angular Acceleration**:
According to Newton's second law for rotation, the torque is also related to the angular acceleration \( \alpha \) by:
\[ \tau = I \cdot \alpha \]

**Hint**: This equation shows how the torque applied to an object results in angular acceleration, similar to how force results in linear acceleration.

4. **Set the Equations Equal**:
From the previous steps, we have two expressions for torque:
\[ T \cdot R = I \cdot \alpha \]
Substituting the moment of inertia \( I \):
\[ T \cdot R = \left(\frac{1}{2} M R^2\right) \cdot \alpha \]

**Hint**: Make sure to substitute the correct expression for moment of inertia when relating torque to angular acceleration.

5. **Solve for Angular Acceleration \( \alpha \)**:
Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{T \cdot R}{\frac{1}{2} M R^2} \]
Simplifying this gives:
\[ \alpha = \frac{2T}{M R} \]

**Hint**: When simplifying, cancel out common terms carefully to arrive at the final expression.

### Final Result:
The angular acceleration \( \alpha \) of the disc is given by:
\[ \alpha = \frac{2T}{M R} \]