Find the radius of gyration and the moment of inertia of a rod of mass 200 g and length 200cm about an axis through it's centre and perpendicular to it's length.

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Solution

i) Here M = 200 = 0.2kg,
l = 200cm = 1m, K = ?, I = ?
M.I. of the rod about an axis through its center and perpendicular to its length is
I=Ml^2/12=MK^2 or K^2=l^2/12
Therefore K=l/√12
=0.2/3.464=0.0571m
2:- I=Ml^2/12=0.222/12
=0.0668kgm^2

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Find the radius of gyration and the moment of inertia of a rod of mass 200 g and length 200cm about an axis through it's centre and perpendicular to it's length.

i) Here M = 200 = 0.2kg, l = 200cm = 1m, K = ?, I = ? M.I. of the rod about an axis through its center and perpendicular to its length is I=Ml^2/12=MK^2 or K^2=l^2/12 Therefore K=l/√12 =0.2/3.464=0.0571m 2:- I=Ml^2/12=0.2*2*2/12 =0.0668kgm^2

i) Here M = 200 = 0.2kg,
l = 200cm = 1m, K = ?, I = ?
M.I. of the rod about an axis through its center and perpendicular to its length is
I=Ml^2/12=MK^2 or K^2=l^2/12
Therefore K=l/√12
=0.2/3.464=0.0571m
2:- I=Ml^2/12=0.222/12
=0.0668kgm^2