A thin rod has a length of 0.25 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of `0.32 "rad"//s` and a moment of inertia of `1.1 xx 10^(-3) kg.m^(2)`. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug `("mass" = 4.2 xx 10^(-3) kg)` gets where it is going, what is the angular velocity of the rod?

Options

A. `1.32 "rad"//s`
B. `0.26 "rad"//s`
C. `0.93 "rad"//s`
D. `1.66 "rad"//s`

Answer

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum of the system (rod + bug) before the bug crawls to the end of the rod must equal the angular momentum after the bug reaches the end.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Length of the rod, \( L = 0.25 \, \text{m} \)
- Initial angular velocity of the rod, \( \omega_1 = 0.32 \, \text{rad/s} \)
- Moment of inertia of the rod, \( I_{\text{rod}} = 1.1 \times 10^{-3} \, \text{kg m}^2 \)
- Mass of the bug, \( m_{\text{bug}} = 4.2 \times 10^{-3} \, \text{kg} \)

2. **Calculate Initial Angular Momentum:**
The initial angular momentum \( L_1 \) of the system is given by:
\[ L_1 = I_{\text{rod}} \cdot \omega_1 \]
Substitute the values:
\[ L_1 = (1.1 \times 10^{-3} \, \text{kg m}^2) \cdot (0.32 \, \text{rad/s}) = 3.52 \times 10^{-4} \, \text{kg m}^2/\text{s} \]

3. **Calculate the Moment of Inertia After the Bug Crawls:**
When the bug crawls to the end of the rod, its distance from the axis of rotation is equal to the length of the rod \( L \). The moment of inertia of the bug when it is at the end of the rod is given by:
\[ I_{\text{bug}} = m_{\text{bug}} \cdot L^2 \]
Substitute the values:
\[ I_{\text{bug}} = (4.2 \times 10^{-3} \, \text{kg}) \cdot (0.25 \, \text{m})^2 = 4.2 \times 10^{-3} \cdot 0.0625 = 2.625 \times 10^{-4} \, \text{kg m}^2 \]

The total moment of inertia after the bug reaches the end is:
\[ I_{\text{total}} = I_{\text{rod}} + I_{\text{bug}} = 1.1 \times 10^{-3} + 2.625 \times 10^{-4} = 1.3625 \times 10^{-3} \, \text{kg m}^2 \]

4. **Apply Conservation of Angular Momentum:**
According to the conservation of angular momentum:
\[ L_1 = L_2 \]
Where \( L_2 = I_{\text{total}} \cdot \omega_2 \) (the angular momentum after the bug crawls). Thus:
\[ 3.52 \times 10^{-4} = (1.3625 \times 10^{-3}) \cdot \omega_2 \]

5. **Solve for Final Angular Velocity \( \omega_2 \):**
Rearranging the equation gives:
\[ \omega_2 = \frac{3.52 \times 10^{-4}}{1.3625 \times 10^{-3}} \approx 0.258 \, \text{rad/s} \]

### Final Answer:
The angular velocity of the rod after the bug reaches the end is approximately \( \omega_2 \approx 0.258 \, \text{rad/s} \).