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11 Nov 2007 21:18:27 IST
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Now the axis passes through one of the corners => it will be common to the corner of 2 rods.
So, moment of inertia of these 2 rods = 2/3ml2
Consider the third rod :
The moment of inertia of the third rod about its center is ml2/12.
the distance of the center of the rod from the corner =
3l/2
Using parallel axis theorem, MI of this rod about the corner
= ml2/12 + 3ml2/4 = 10ml2/12.
So net moment of inertia = 2/3ml2 + 10ml2/12
= 18ml2/12.
To find radius of gyration:
Net mass of the system = 3m. Let radius of gyration be K.
Now we know that Moment of inertia = MK2
so 18ml2/12 = 3mK2
which gives K = l/
2
Hope that helped!
Cheers!
So, moment of inertia of these 2 rods = 2/3ml2
Consider the third rod :
The moment of inertia of the third rod about its center is ml2/12.
the distance of the center of the rod from the corner =
3l/2Using parallel axis theorem, MI of this rod about the corner
= ml2/12 + 3ml2/4 = 10ml2/12.
So net moment of inertia = 2/3ml2 + 10ml2/12
= 18ml2/12.
To find radius of gyration:
Net mass of the system = 3m. Let radius of gyration be K.
Now we know that Moment of inertia = MK2
so 18ml2/12 = 3mK2
which gives K = l/
2Hope that helped!
Cheers!












let that point be intersectionbe'P' of rod L1,L2;;;;; L3 is rod oppsite to point
moment of inertia of L1 about P = ML^2/3
M.I of L2 about P = ML^2/3
M.I of L3 about P = ML^2/12 + M(
thus m.I about P = 2ML^2/3 + ML^2/12 + 3ML^2/4
18ML^2/12 = 3ML^2/2
k be radius of gyration
3Mk^2 = 3ML^2/2
k =
k= L/
do rate me if correct:)