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Calculate the rotational inertia of a meter stick, with mass 0.342 kg, about an axis perpendicular...

Calculate the rotational inertia of a meter stick, with mass 0.342 kg, about an axis perpendicular to the stick and located at the 22.7 cm mark. (Treat the stick as a thin rod.)

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This problem can be solved by applying the parallel axis theorem and noting that the rotational inertia about the midpoint of the stick and perpendicular to the stick is (ML/12) where M is the mass of the stick and L is the length of the stick which is 1 meter here. The rotational inertia about the 22.7 cm mark comes out as 0.054 kg m^2. Detailed solution given below. If you have any doubt regarding the solution, ask me in the comment section. Feedback will be helpful

solution: (Rotational inertia is also called moment of inertia) Mass of the stick (M) = 0:342 kg Length of the stick (6 = 1.0

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