In the above question, the angular velocity will:

Options

A. increase
B. decrease
C. not change
D. nothing can be said

Answer

To solve the problem regarding the angular velocity when an object sticks to the rim of a rotating body, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**: We have a rotating object with an initial angular velocity \( \omega \) and a moment of inertia \( I = mr^2 \), where \( m \) is the mass of the rotating object and \( r \) is its radius.

2. **Moment of Inertia Before Collision**: The initial moment of inertia of the rotating object is given by:
\[ I = mr^2 \]

3. **Moment of Inertia After Collision**: When an additional mass \( m \) sticks to the rim of the rotating object, the new moment of inertia \( I' \) becomes:
\[ I' = mr^2 + m \cdot r^2 = (m + m)r^2 = 2mr^2 \]

4. **Conservation of Angular Momentum**: Angular momentum before and after the collision must be conserved. The initial angular momentum \( L_i \) is given by:
\[ L_i = I \cdot \omega = mr^2 \cdot \omega \]
The final angular momentum \( L_f \) after the mass sticks is:
\[ L_f = I' \cdot \omega' = 2mr^2 \cdot \omega' \]

5. **Setting Up the Equation**: According to the conservation of angular momentum:
\[ L_i = L_f \]
Substituting the expressions for angular momentum:
\[ mr^2 \cdot \omega = 2mr^2 \cdot \omega' \]

6. **Solving for Final Angular Velocity**: We can simplify the equation by dividing both sides by \( mr^2 \) (assuming \( m \neq 0 \) and \( r \neq 0 \)):
\[ \omega = 2\omega' \]
Rearranging gives us:
\[ \omega' = \frac{\omega}{2} \]

7. **Conclusion**: The angular velocity after the mass sticks to the rim will be half of the initial angular velocity:
\[ \omega' = \frac{\omega}{2} \]

### Final Answer:
The angular velocity will decrease to half of its initial value.