Rotation dynamics, dealing with impulse and oscillation

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The discussion focuses on calculating the minimum distance "d" required for a vertically placed homogeneous rod to complete a 360° rotation after receiving an impulse at its top. Using angular momentum conservation and mechanical energy principles, the minimum value of "d" is derived as d_min = (m/J)√(2gL^3/3). Additionally, the period of oscillation for the rod in harmonic motion is explored, with references to the concept of a physical pendulum for further understanding. The calculations involve the rod's mass "m," length "l," and gravitational acceleration "g." This analysis provides a comprehensive approach to understanding rotation dynamics and oscillation in physical systems.
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A homogene rod with length "l" is placed vertically, and a nail is stabbed on the top of the rod (now the rod has an axis). And then an impulse is given on the rod with the separation between the impulse given to the rod's axis is "d". Earth gravitational acc is represented as g, the mass of the rod is m. Now, calculate the minimum value of d to make the rod rotate 360°.

Now if the condition above is complete, and the rod make a harmonic movement (oscillation) what is the period?

And what is the length of a mathematical pendulum should be to make the same period with the oscillating rod?
 
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The questions about the pendulum and oscillations can be answered with a little bit of internet research on key word "physical pendulum". The question about the minimum value of ##d## given an impulse ##J## can be answered by using angular momentum conservation to find the initial angular speed about the pivot and then mechanical energy conservation to say that all the initial kinetic energy is converted to potential energy when the stick is vertical with the center of mass at distance ##l/2## above the nail.
$$J d=\frac{1}{3}ml^2\omega~\rightarrow~\omega=\frac{3Jd}{ml^2}$$
$$mgl<\frac{1}{2}ml^2\omega^2=\frac{1}{2}\times\frac{1}{3}ml^2\left(\frac{3Jd}{ml^2}\right)^2~\rightarrow~d_{min}=\frac{m}{J}\sqrt{\frac{2gL^3}{3}}.$$
 
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