If a tangential force mg is applied to a disc of mass m and radius r, the angular acceleration produced in it is?

Options

A. gr
B. g/r
C. `(2g)/r`
D. 2gr

Answer

To find the angular acceleration produced in a disc of mass \( m \) and radius \( r \) when a tangential force \( mg \) is applied, we can follow these steps:

### Step 1: Identify the Forces and Torque
We have a disc with a mass \( m \) and radius \( r \). A tangential force \( F = mg \) is applied at the edge of the disc. The torque \( \tau \) produced by this force about the center of the disc can be calculated using the formula:
\[ \tau = r \times F \]
Substituting the value of \( F \):
\[ \tau = r \times mg \]

### Step 2: Relate Torque to Angular Acceleration
According to the rotational dynamics, the torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \) by the equation:
\[ \tau = I \alpha \]

### Step 3: Calculate the Moment of Inertia
For a solid disc rotating about its center, the moment of inertia \( I \) is given by:
\[ I = \frac{1}{2} m r^2 \]

### Step 4: Set Up the Equation
Now we can set the two expressions for torque equal to each other:
\[ r \cdot mg = I \alpha \]
Substituting the moment of inertia:
\[ r \cdot mg = \left(\frac{1}{2} m r^2\right) \alpha \]

### Step 5: Solve for Angular Acceleration
We can now solve for \( \alpha \):
1. Rearranging the equation gives:
\[ \alpha = \frac{r \cdot mg}{\frac{1}{2} m r^2} \]
2. Simplifying this expression:
\[ \alpha = \frac{2g}{r} \]

### Final Answer
Thus, the angular acceleration produced in the disc is:
\[ \alpha = \frac{2g}{r} \]

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