Very Simple Rotation Inertia of a Thin Rod Problem

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To calculate the rotational inertia of a meter stick with a mass of 0.68 kg about an axis at the 21 cm mark, one can use the parallel axis theorem. The rotational inertia formula involves integrating the radius squared with respect to mass. Understanding the calculation process is crucial for grasping similar problems. The parallel axis theorem allows for determining the moment of inertia around a non-central axis by adding the product of mass and the square of the distance from the center of mass. This approach provides clarity on how to solve rotational inertia problems effectively.
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Homework Statement


Calculate the rotational inertia of a meter stick, with mass 0.68 kg, about an axis perpendicular to the stick and located at the 21 cm mark. (Treat the stick as a thin rod.)


Homework Equations


I = integral of radius squared with respect to mass


The Attempt at a Solution


I'm not quite sure how to explain my issue here, but I think all I need is to see this worked out, and then I could probably understand the nature of these problems. Thanks.
 
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You could use the parallel axis theoram if you know how to find the M.I about the centre of mass of the rod (the centre) :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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