Optimizing Pendulum Period with Center of Mass Distance

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To optimize the period of a physical pendulum, the distance x between the stick's center of mass and the pivot point O must be expressed in relation to the period formula. The relevant equation is Period = 2π√(Icom/mgh), where Icom represents the rotational inertia about the pivot, and h is the distance from the center of mass to the pivot. The discussion highlights that mass cancels out, simplifying the calculations. The user successfully resolved their confusion regarding the square root in the equation. Ultimately, determining the optimal distance x is crucial for minimizing the pendulum's period.
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In Fig. 15-46, a stick of length L = 1.8 m oscillates as a physical pendulum. (a) What value of distance x between the stick's center of mass and its pivot point O gives the least period? (b) What is that least period?

For this problem, I wasn't sure where to start, I would have thought that the answer would be as far away from the center of mass as possible but I wasn't sure how to do it. I know that you have to use the equation Period = 2 x pi x (square root of (Icom/m x g x h) and that mass cancels out, but I don't know where to go from there, thank you in advance for the help
 
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You have the equation for the period, now express the variables in terms of "x", the distance between center of mass and pivot point. (What you call "Icom" should be the rotational inertia of the pendulum about the pivot point; "h" is the distance from center of mass to pivot point.)
 
Yeah, I got the question right, I realized that I was forgetting to cancel out the square root. Thank you for the help
 

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