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Rahul Karmakar's Avatar
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11 Nov 2007 21:03:45 IST
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Moment OF inertia
None

Three Identical Rods, each of the length L,are joined to form a rigid Equilateral triangle its radus of gyration about an axis passing through a corner and perpendicular to the plane of the triange is:
 
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nishant singh's Avatar

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11 Nov 2007 21:15:46 IST
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Find Moment of Inertia of the system about that point
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( 3L/2)^2   {used parallel axis theorem distance of centre of mass of L3 from P }

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 = (L^2)/2
k= L/ 2




do rate me if correct:)



Karthik M's Avatar

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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!

Krishna Gopal Singh's Avatar

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Joined: 29 Dec 2006
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11 Nov 2007 22:50:42 IST
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Perfect answers nishant and Kartik
Rahul Karmakar's Avatar

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Joined: 16 Dec 2006
Posts: 214
12 Nov 2007 13:20:38 IST
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Thanks Guys !

I was Mistaking for the third rod's MI
Anyways Thanks for the Answer.



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