Two men support a uniform rod of mass `M` and length `L` at its two ends. If one of them suddenly withdraws, find the force excerted by the rod on the other man
(`a`) immediately after withdrawl and
(`b`) when the rod makes angle `theta` with vertical.

Options

A. `(mv_(1)^(2))/(T_(1)^(2))t`
B. `(mv_(1))/(T_(1)^(2))t`
C. `((mv_(1))/(T_(1)))^(2)t`
D. `(mv_(1)^(2))/(T_(1))t^(2)`

Answer

To solve the problem, we need to analyze the forces acting on the rod in two scenarios: immediately after one man withdraws and when the rod makes an angle θ with the vertical.

### Part (a): Immediately After Withdrawal

1. **Identify the Forces**:
- When one man withdraws, the rod's weight (Mg) acts downwards at its center of mass (which is at L/2 from either end).
- The remaining man exerts a normal force (N) upwards at one end of the rod.

2. **Apply Newton's Second Law**:
- The net force acting on the rod immediately after one man withdraws can be expressed as:
\[ Mg - N = M \cdot a \]
- Here, \(a\) is the linear acceleration of the center of mass of the rod.

3. **Calculate the Angular Acceleration**:
- The torque about the center of mass due to the weight of the rod is:
\[ \text{Torque} = Mg \cdot \frac{L}{2} \]
- The moment of inertia (I) of the rod about its center of mass is:
\[ I = \frac{ML^2}{3} \]
- Using the relation for torque (\(\tau = I \alpha\)):
\[ Mg \cdot \frac{L}{2} = \frac{ML^2}{3} \cdot \alpha \]
- Solving for \(\alpha\):
\[ \alpha = \frac{3g}{2L} \]

4. **Find Linear Acceleration**:
- The linear acceleration of the center of mass is related to angular acceleration:
\[ a = \alpha \cdot \frac{L}{2} = \frac{3g}{2L} \cdot \frac{L}{2} = \frac{3g}{4} \]

5. **Substitute Back to Find N**:
- Substitute \(a\) back into the force equation:
\[ Mg - N = M \cdot \frac{3g}{4} \]
- Rearranging gives:
\[ N = Mg - \frac{3Mg}{4} = \frac{Mg}{4} \]

### Conclusion for Part (a):
The force exerted by the rod on the other man immediately after withdrawal is:
\[ N = \frac{Mg}{4} \]

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### Part (b): When the Rod Makes an Angle θ with the Vertical

1. **Identify Forces**:
- The weight of the rod (Mg) acts downwards at its center of mass.
- The components of the weight at angle θ:
- Perpendicular component: \(Mg \cos \theta\)
- Parallel component: \(Mg \sin \theta\)

2. **Apply Newton's Second Law**:
- The net force in the vertical direction gives:
\[ N + Mg \sin \theta = Mg \cos \theta \]
- Rearranging gives:
\[ N = Mg \cos \theta - Mg \sin \theta \]

3. **Torque Calculation**:
- The torque about the pivot point (the end where the man is) due to the weight is:
\[ \text{Torque} = Mg \cdot \frac{L}{2} \sin \theta \]
- Equating this to the moment of inertia times angular acceleration:
\[ Mg \cdot \frac{L}{2} \sin \theta = \frac{ML^2}{3} \cdot \alpha \]

4. **Find Angular Acceleration**:
- Solving for \(\alpha\):
\[ \alpha = \frac{3g \sin \theta}{2L} \]

5. **Find Linear Acceleration**:
- The linear acceleration of the center of mass:
\[ a = \alpha \cdot \frac{L}{2} = \frac{3g \sin \theta}{2L} \cdot \frac{L}{2} = \frac{3g \sin \theta}{4} \]

6. **Substitute Back to Find N**:
- Substitute \(a\) back into the force equation:
\[ Mg \cos \theta - N = M \cdot \frac{3g \sin \theta}{4} \]
- Rearranging gives:
\[ N = Mg \cos \theta - \frac{3Mg \sin \theta}{4} \]

### Conclusion for Part (b):
The force exerted by the rod on the other man when it makes an angle θ with the vertical is:
\[ N = Mg \cos \theta - \frac{3Mg \sin \theta}{4} \]

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