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