Moment of inertia of a cylinder in unusual position

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The moment of inertia for a thin disc with radius r and mass m, when the axis is perpendicular to the disc, is calculated as I₀ = 1/2 mr². Using the perpendicular axis theorem, if the axis is parallel to the disc and passes through the center, the moment of inertia becomes Iᵢ = 1/4 mr². For a disc positioned away from the center by a distance x, the moment of inertia can be expressed as 1/4 mr² + mx², where m must be adjusted accordingly. Integrating these values will yield the final answer for the moment of inertia in the specified position. This approach effectively addresses the calculation process for the unusual positioning of the cylinder.
24forChromium
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True or false.jpg

See image. It would be better if you can show me the calculation process.

EDIT: the axis of rotation goes right through the centre of the cylinder.

NEVER MIND! GOT IT!
 
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My opinion: for a thin disc whose radius is ##r## and mass is ##m,## with the axis perpendicular to the disc, the moment of inertia is ##I_0=\frac{1}{2}mr^2,## which may be easy to find. Then with the perpendicular axis theorem, if the axis is parallel to the disc and passes the center, the moment of inertia should be ##I_i=\frac{1}{2}I_0=\frac{1}{4}mr^2.##
With this result and the parallel axis theorem, we may be able to solve the problem. For the disc away from the center ##x##(based on your diagram), the disc's moment of inertia is ##\frac{1}{4}mr^2+mx^2##(of course ##m## should be adjusted. Then, just integrate them we may get the answer.
 

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