Evaluate the moment of inertia of a thin rod with axis of rotation at the centre , if its length is 100cm and its mass is 2.3kg.
I=mL212=2.3⋅1212=0.19 kg⋅m2.I=\frac{mL^2}{12}=\frac{2.3\cdot 1^2}{12}=0.19~kg\cdot m^2.I=12mL2=122.3⋅12=0.19 kg⋅m2.
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