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Sagot :
Given data
*The given mass of thin rim is m = 10 kg
*The given radius of the rim is r = 20 cm = 0.20 m
The formla for the moment of inertia of thin rim with respect to an axis perpendicular to its plane, which passes through a point on its circumference is given as
[tex]\begin{gathered} I=mr^2+mr^2 \\ =2mr^2 \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} I=2(10)(0.20)^2 \\ =0.8kg.m^2 \end{gathered}[/tex]Hence, the moment of inertia of thin rim with respect to an axis perpendicular to its plane, which passes through a point on its circumference is I = 0.8 kg.m^2
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