A student throws a horizontal stick of length `L` up into the air. At the moment it leaves her hand the speed of stick's closes end is zero. The stick completes `N` turns as it is caought by the student at the initial release point. Find the height `h` to which the centre of mass of the rod rises.

Explanation

To solve the problem of finding the height \( h \) to which the center of mass of a horizontal stick of length \( L \) rises when thrown into the air, we can follow these steps:

### Step 1: Understanding the Problem
The stick is thrown horizontally, and at the moment of release, the speed of the stick's closest end is zero. The stick completes \( N \) turns before being caught back at the release point. The center of mass of the stick is located at a distance of \( \frac{L}{2} \) from the point of release.

### Step 2: Relating Angular Velocity and Linear Velocity
The linear velocity \( v_{cm} \) of the center of mass of the stick can be related to its angular velocity \( \omega \) by the formula:
\[ v_{cm} = \omega \cdot r \]
where \( r \) is the distance from the pivot (the release point) to the center of mass. Here, \( r = \frac{L}{2} \), so:
\[ v_{cm} = \omega \cdot \frac{L}{2} \]

### Step 3: Finding the Time of Flight
The stick completes \( N \) revolutions before being caught. The time period \( T \) for one complete revolution is given by:
\[ T = \frac{2\pi}{\omega} \]
Thus, the total time \( t \) for \( N \) revolutions is:
\[ t = N \cdot T = N \cdot \frac{2\pi}{\omega} \]

### Step 4: Relating Time of Flight to Motion
The stick is thrown horizontally and falls under the influence of gravity. The time of flight can also be expressed in terms of the vertical motion:
\[ t = \frac{2 v_{cm}}{g} \]
Equating the two expressions for \( t \):
\[ \frac{2 v_{cm}}{g} = N \cdot \frac{2\pi}{\omega} \]
This simplifies to:
\[ v_{cm} = \frac{N \pi g}{\omega} \]

### Step 5: Finding the Height
The height \( h \) to which the center of mass rises can be calculated using the equation of motion for projectile motion:
\[ h = \frac{v_{cm}^2}{2g} \]
Substituting the expression for \( v_{cm} \):
\[ h = \frac{1}{2g} \left(\frac{N \pi g}{\omega}\right)^2 \]
This simplifies to:
\[ h = \frac{N^2 \pi^2 g}{2g^2 \omega^2} \]
Thus, we can express \( h \) as:
\[ h = \frac{N^2 \pi^2}{2g \omega^2} \]

### Step 6: Final Expression for Height
Using the earlier relation \( \omega^2 L = 2 \pi N g \), we can substitute \( \omega^2 \) in our height equation:
\[ h = \frac{N^2 \pi^2 L}{4g} \]

### Conclusion
The height \( h \) to which the center of mass of the rod rises is given by:
\[ h = \frac{N \pi L}{4} \]