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 a uniform disc of mass \( M \) and radius \( R \) when a steady downward pull \( T \) is exerted on a cord wrapped around its rim, we can follow these steps:

### Step 1: Identify the Forces Acting on the Disc
The forces acting on the disc are:
- The tension \( T \) acting downward at the rim of the disc.
- The weight of the disc \( Mg \) acting downward at its center.

### Step 2: Determine the Torque Due to the Tension
The torque \( \tau \) caused by the tension \( T \) about the axis of rotation (center of the disc) can be calculated using the formula:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the disc.

### Step 3: Apply Newton's Second Law for Rotation
According to Newton's second law for rotation, the net torque is equal to the moment of inertia \( I \) times the angular acceleration \( \alpha \):
\[ \tau = I \cdot \alpha \]

### Step 4: Calculate the Moment of Inertia of the Disc
The moment of inertia \( I \) of a uniform disc about its center is given by:
\[ I = \frac{1}{2} M R^2 \]

### Step 5: Set Up the Equation
Substituting the expressions for torque and moment of inertia into the equation from Step 3:
\[ T \cdot R = \left(\frac{1}{2} M R^2\right) \cdot \alpha \]

### Step 6: Solve for Angular Acceleration \( \alpha \)
Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{2T}{M R} \]

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

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