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 Torque
The torque \( \tau \) produced by the tangential force \( F \) acting at the radius \( r \) is given by the formula:
\[ \tau = r \times F \]
Since the force \( F \) is equal to \( mg \), we can substitute this into the torque equation:
\[ \tau = r \times mg \]
Thus, the torque is:
\[ \tau = mgr \]

### Step 2: Relate Torque to Angular Acceleration
According to Newton's second law for rotation, the torque is also related to the moment of inertia \( I \) and the angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
We can set the two expressions for torque equal to each other:
\[ mgr = I \alpha \]

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

### Step 4: Substitute Moment of Inertia into the Torque Equation
Now we can substitute the expression for \( I \) into the torque equation:
\[ mgr = \left(\frac{1}{2} m r^2\right) \alpha \]

### Step 5: Solve for Angular Acceleration
Rearranging the equation to solve for \( \alpha \):
\[ \alpha = \frac{mgr}{\frac{1}{2} m r^2} \]
We can simplify this equation:
\[ \alpha = \frac{mgr}{\frac{1}{2} m r^2} = \frac{2g}{r} \]

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

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