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 will use the principle of conservation of energy. The initial kinetic energy of the rod when it is given a linear speed \( v_0 \) will be converted into gravitational potential energy when the rod reaches its highest point (momentarily halting in a vertical position).

### Step-by-Step Solution:

1. **Identify the Length of the Rod**:
The length of the rod \( L \) is given as 0.80 m.

2. **Determine the Change in Height**:
When the rod is hanging vertically downward, its center of mass is at a height of \( \frac{L}{2} \) (0.40 m) from the pivot. When the rod is vertical upwards, the center of mass will be at a height of \( L \) (0.80 m) from the pivot. Thus, the change in height \( h \) is:
\[ h = L - \frac{L}{2} = \frac{L}{2} = 0.40 \, \text{m} \]

3. **Calculate the Potential Energy Gain**:
The gravitational potential energy gained when the rod moves from the downward position to the upward position is given by:
\[ \Delta PE = mgh = mg \cdot 0.40 \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

4. **Calculate the Initial Kinetic Energy**:
The initial kinetic energy of the rod when it is given a linear speed \( v_0 \) is:
\[ KE = \frac{1}{2} I \omega^2 \]
For a thin rod rotating about one end, the moment of inertia \( I \) is:
\[ I = \frac{1}{3} m L^2 \]
The angular velocity \( \omega \) can be related to the linear speed \( v_0 \) by:
\[ \omega = \frac{v_0}{\frac{L}{2}} = \frac{2v_0}{L} \]
Therefore, the kinetic energy can be rewritten as:
\[ KE = \frac{1}{2} \left(\frac{1}{3} m L^2\right) \left(\frac{2v_0}{L}\right)^2 = \frac{1}{2} \cdot \frac{1}{3} m L^2 \cdot \frac{4v_0^2}{L^2} = \frac{2}{3} mv_0^2 \]

5. **Set Up the Conservation of Energy Equation**:
According to the conservation of energy:
\[ \text{Initial Kinetic Energy} = \text{Potential Energy Gain} \]
\[ \frac{2}{3} mv_0^2 = mg \cdot 0.40 \]

6. **Cancel Mass \( m \) from Both Sides**:
\[ \frac{2}{3} v_0^2 = g \cdot 0.40 \]

7. **Solve for \( v_0^2 \)**:
\[ v_0^2 = \frac{3 \cdot g \cdot 0.40}{2} \]

8. **Substitute \( g = 9.81 \, \text{m/s}^2 \)**:
\[ v_0^2 = \frac{3 \cdot 9.81 \cdot 0.40}{2} = \frac{11.772}{2} = 5.886 \]

9. **Calculate \( v_0 \)**:
\[ v_0 = \sqrt{5.886} \approx 2.43 \, \text{m/s} \]

### Final Answer:
The required linear speed \( v_0 \) is approximately \( 2.43 \, \text{m/s} \).