Rotation dynamics, dealing with impulse and oscillation

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SUMMARY

The discussion focuses on the dynamics of a homogeneous rod subjected to an impulse, specifically calculating the minimum distance "d" required for the rod to complete a 360° rotation. The impulse "J" applied at a distance "d" from the rod's axis leads to an angular speed determined by the equation ω = (3Jd)/(ml²). The minimum value of "d" is derived as d_min = (m/J)√(2gL³/3). Additionally, the discussion touches on the oscillation period of the rod and its equivalence to that of a mathematical pendulum, emphasizing the relevance of the term "physical pendulum" for further research.

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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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