Two boys support by the ends a uniform rod of mass M and length 2L. The rod is horizontal. The two boys decided to change the ends of the rod by throwing the rod into air and catching it. The boys do not move from their position and the rod remained horizontal throughout its flight. Find the minimum impulse applied by each boy on the rod when it was thrown

Answer

To solve the problem of finding the minimum impulse applied by each boy on the rod when it is thrown, we can follow these steps:

### Step 1: Understand the System
We have a uniform rod of mass \( M \) and length \( 2L \) held horizontally by two boys at its ends. When they throw the rod into the air, they apply impulses that will give the rod both upward velocity and angular velocity.

**Hint:** Visualize the rod and the forces acting on it. Draw a diagram to help understand the setup.

### Step 2: Define Impulses
Let \( J_H \) be the horizontal component of the impulse and \( J_V \) be the vertical component of the impulse. The total impulse \( J \) can be expressed as:
\[ J^2 = J_H^2 + J_V^2 \]

**Hint:** Remember that impulses are vector quantities; thus, they can be combined using the Pythagorean theorem.

### Step 3: Relate Impulses to Motion
The vertical component \( J_V \) will provide the upward velocity \( U \) of the center of mass (CM) of the rod, and the horizontal component \( J_H \) will provide the angular speed \( \omega \) of the rod.

From the equations of motion, we can relate these components:
- The time of flight \( T \) of the rod can be expressed as:
\[ T = \frac{2U}{g} \]
- The angular displacement during this time is \( \pi \) radians, which gives:
\[ \omega T = \pi \implies \omega = \frac{\pi g}{2U} \]

**Hint:** Use the relationship between angular displacement, angular velocity, and time to find \( \omega \).

### Step 4: Calculate Impulses
Using the moment of inertia \( I_{CM} \) of the rod about its center of mass:
\[ I_{CM} = \frac{1}{3} M (2L)^2 = \frac{4ML^2}{3} \]
The horizontal impulse \( J_H \) can be related to the angular momentum:
\[ J_H = I_{CM} \omega \]
Substituting for \( \omega \):
\[ J_H = \frac{4ML^2}{3} \cdot \frac{\pi g}{2U} = \frac{2ML^2 \pi g}{3U} \]

**Hint:** Recall the formula for the moment of inertia and how it relates to angular momentum.

### Step 5: Express \( J_V \)
From the vertical impulse, we have:
\[ J_V = MU \]

**Hint:** The vertical impulse directly relates to the change in momentum of the rod.

### Step 6: Substitute and Minimize
Now substitute \( J_H \) and \( J_V \) into the equation for total impulse:
\[ J^2 = \left(\frac{2ML^2 \pi g}{3U}\right)^2 + (MU)^2 \]
To minimize \( J \), we take the derivative with respect to \( U \) and set it to zero.

**Hint:** Use calculus to find the minimum value of a function by setting its derivative to zero.

### Step 7: Solve for \( U \)
After differentiating and simplifying, we find:
\[ U^4 = \frac{L^2 \pi^2 g^2}{36} \implies U^2 = \frac{L \pi g}{6} \]

**Hint:** Keep track of your algebraic manipulations carefully to avoid mistakes.

### Step 8: Calculate Minimum Impulse
Substituting \( U^2 \) back into the impulse equation gives us:
\[ J^2_{min} = \frac{M^2 L^2 \pi^2 g^2}{144 \cdot \frac{L \pi g}{6}} + \frac{M^2 L \pi g}{4} \cdot 6 \]
After simplification, we find:
\[ J_{min} = \frac{M \pi L g}{6} \]

### Final Answer
The minimum impulse applied by each boy on the rod when it was thrown is:
\[ J_{min} = \frac{M \pi L g}{6} \]

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