Two equal and opposite forces F are allplied tangentially to a uniform disc of mass M and radius R . If the disc is pivoted at its centre and free to rotate in its plane, the angular acceleration of the disc is :

Options

A. `(F)/(MR)`
B. `(2F)/(3MR)`
C. `(4F)/(MR)`
D. Zero

Answer

To solve the problem of finding the angular acceleration of a uniform disc when two equal and opposite forces \( F \) are applied tangentially, we can follow these steps:

### Step 1: Understand the Setup
We have a uniform disc of mass \( M \) and radius \( R \). Two equal and opposite forces \( F \) are applied tangentially to the disc. The disc is pivoted at its center and can rotate freely in its plane.

### Step 2: Identify the Torque
The torque \( \tau \) produced by a force applied tangentially at a distance \( R \) from the pivot (the center of the disc) is given by the formula:
\[ \tau = F \times R \]
Since there are two forces acting in opposite directions, the total torque \( \tau_{\text{total}} \) is:
\[ \tau_{\text{total}} = F \cdot R + F \cdot R = 2FR \]

### Step 3: 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 4: Relate Torque and Angular Acceleration
According to Newton's second law for rotation, the relationship between torque and angular acceleration \( \alpha \) is given by:
\[ \tau = I \cdot \alpha \]
From this, we can express angular acceleration as:
\[ \alpha = \frac{\tau}{I} \]

### Step 5: Substitute the Values
Now, substituting the values of torque and moment of inertia into the equation for angular acceleration:
\[ \alpha = \frac{2FR}{\frac{1}{2} M R^2} \]

### Step 6: Simplify the Expression
Simplifying the expression:
\[ \alpha = \frac{2FR \cdot 2}{M R^2} = \frac{4F}{M R} \]

### Final Answer
Thus, the angular acceleration \( \alpha \) of the disc is:
\[ \alpha = \frac{4F}{M R} \]