A uniform disc of mass `M` and radius `R` is mounted on an axle supported in frictionless bearings. A light cord is wrapped around the rim of the disc and a steady downward pull `T` is exerted on the cord. The angular acceleration of the disc is

Options

A. `T/(MR)`
B. `(MR)/T`
C. `(2T)/MR`
D. `(MR)/(2T)`

Explanation

To find the angular acceleration of the uniform disc when a steady downward pull \( T \) is exerted on a cord wrapped around its rim, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Disc**:
- The weight of the disc, \( Mg \), acts downwards at the center of the disc.
- The tension \( T \) in the cord acts tangentially at the rim of the disc.

2. **Calculate the Torque**:
- The torque \( \tau \) caused by the tension \( T \) about the axis of the disc is given by:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the disc.

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

4. **Apply the Relation Between Torque and Angular Acceleration**:
- According to Newton's second law for rotation, the torque is also related to the moment of inertia and angular acceleration \( \alpha \):
\[ \tau = I \cdot \alpha \]
- Substituting the expressions for torque and moment of inertia, we have:
\[ T \cdot R = \left(\frac{1}{2} M R^2\right) \cdot \alpha \]

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

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