A uniform rod of mass m, hinged at its upper end, is released from rest from a horizontal position. When it passes through the vertical position, the force on the hinge is

Answer

To solve the problem of finding the force on the hinge when a uniform rod of mass \( m \) is released from a horizontal position and passes through the vertical position, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**:
We have a uniform rod of mass \( m \) and length \( l \) hinged at its upper end. The rod is released from a horizontal position.

2. **Determine the Forces Acting on the Rod**:
When the rod is in the vertical position, the forces acting on it are:
- The weight of the rod acting downwards, \( mg \).
- The normal force \( N \) exerted by the hinge acting upwards.

3. **Calculate the Potential Energy Loss**:
The center of mass of the rod is located at a distance of \( \frac{l}{2} \) from the hinge. When the rod is horizontal, the potential energy is maximum, and as it falls to the vertical position, the height of the center of mass decreases by \( \frac{l}{2} \).
\[ \text{Loss in Potential Energy} = mg \cdot \frac{l}{2} \]

4. **Use the Conservation of Energy**:
The loss in potential energy converts into kinetic energy. The kinetic energy of the rod when it is vertical can be expressed in terms of its angular velocity \( \omega \):
\[ \frac{1}{2} I \omega^2 = mg \cdot \frac{l}{2} \]
where \( I \) is the moment of inertia of the rod about the hinge, which is \( \frac{1}{3} ml^2 \).

5. **Substitute the Moment of Inertia**:
Substitute \( I \) into the equation:
\[ \frac{1}{2} \left(\frac{1}{3} ml^2\right) \omega^2 = mg \cdot \frac{l}{2} \]
Simplifying gives:
\[ \frac{ml^2 \omega^2}{6} = mg \cdot \frac{l}{2} \]

6. **Solve for Angular Velocity \( \omega^2 \)**:
Rearranging the equation:
\[ \omega^2 = \frac{3g}{l} \]

7. **Calculate the Centripetal Force**:
The centripetal force acting on the center of mass of the rod when it is vertical is given by:
\[ F_c = m \cdot \frac{l}{2} \cdot \omega^2 = m \cdot \frac{l}{2} \cdot \frac{3g}{l} = \frac{3mg}{2} \]

8. **Apply Newton’s Second Law**:
According to Newton’s second law, the net force acting on the rod in the vertical direction is:
\[ N - mg = F_c \]
Substituting \( F_c \) gives:
\[ N - mg = \frac{3mg}{2} \]

9. **Solve for the Normal Force \( N \)**:
Rearranging the equation to solve for \( N \):
\[ N = \frac{3mg}{2} + mg = \frac{5mg}{2} \]

### Final Answer:
The force on the hinge when the rod passes through the vertical position is:
\[ N = \frac{5mg}{2} \]