One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has a length of 0.80 m and is uniform. It is hanging vertically straight downward. The end of the rod nearest the floor is given a linear speed `v_(0)`, so that the rod begins to rotate upward about the pivot. What must be the value of `v_(0)` such that the rod comes to a momentary halt in a straight-up orientation, exactly opposite to its initial orientation?

Options

A. `6.9 m//s`
B. `5.6 m//s`
C. `4.7 m//s`
D. `8.1 m//s`

Answer

To solve the problem, we need to determine the initial linear speed \( v_0 \) that must be given to the end of a uniform rod of length \( L = 0.80 \, \text{m} \) so that it rotates upward and comes to a momentary halt in a vertical position opposite to its initial orientation.

### Step-by-Step Solution:

1. **Identify the Initial and Final Positions**:
- The rod is initially hanging straight down (vertical downward position).
- The final position is vertical upward, opposite to the initial position.

2. **Define the Center of Mass**:
- The center of mass of the uniform rod is located at a distance of \( \frac{L}{2} \) from the pivot point (the end of the rod attached to the pivot).

3. **Apply Conservation of Energy**:
- Initially, when the rod is in the downward position, it has rotational kinetic energy and no potential energy (taking the reference point at the pivot).
- At the momentary halt in the upward position, the rod has potential energy and no rotational kinetic energy.

4. **Calculate Initial Rotational Kinetic Energy**:
- The moment of inertia \( I \) of the rod about the pivot point is given by:
\[ I = \frac{1}{3} m L^2 \]
- The initial rotational kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} I \omega^2 \]
- Here, \( \omega \) is the angular velocity, which can be related to the linear speed \( v_0 \) at the end of the rod:
\[ \omega = \frac{v_0}{L} \]

5. **Calculate Final Potential Energy**:
- When the rod reaches the upward position, the height of the center of mass from the pivot is \( L \) (it rises \( L \) from its initial position).
- The potential energy \( PE_f \) at this position is:
\[ PE_f = mgh = mgL \]

6. **Set Up the Energy Conservation Equation**:
- According to the conservation of energy:
\[ KE_i = PE_f \]
- Substituting the expressions for kinetic and potential energy:
\[ \frac{1}{2} \left(\frac{1}{3} m L^2\right) \left(\frac{v_0^2}{L^2}\right) = mgL \]

7. **Simplify the Equation**:
- Cancel \( m \) and \( L^2 \) from both sides:
\[ \frac{1}{6} v_0^2 = gL \]

8. **Solve for \( v_0 \)**:
- Rearranging gives:
\[ v_0^2 = 6gL \]
- Taking the square root:
\[ v_0 = \sqrt{6gL} \]

9. **Substitute the Values**:
- Using \( g = 9.8 \, \text{m/s}^2 \) and \( L = 0.80 \, \text{m} \):
\[ v_0 = \sqrt{6 \times 9.8 \times 0.80} \]
- Calculate:
\[ v_0 = \sqrt{47.04} \approx 6.86 \, \text{m/s} \]

### Final Answer:
The required initial speed \( v_0 \) is approximately \( 6.86 \, \text{m/s} \).