IIT JEE , AIEEE Forum - Physics , Mechanics - Conceptual qns but not able to solve rates assured for help

rakesh61

The 2 rods shown are of length L and mass m the moment of inertia of the 2 rods about the bisector xy of angle betwen them is

Ans ml^2 / 12

Please give me a detailed solution rates assured for satisfactory ans

vk_j09a

when a rod is inclined at an angle

with d axis.. d moment of inertia is ml^2sin(^2)

whole divided by 12..

therefore the moment of inertia of one rod in d given is {ml^2}/24.

therefore the moment of inertia of d compound body abt d given axis is d answer u gave.

rooney

Draw another line perpendicular to xy and call it pq
Now, let a line perpendicular to the plane of the 2 rods and xy (i.e. the plane of the paper) be L

Since Moment of inertia of a rod through its center and perpendicular to its length is ml²/6, The moment of inertia (say I ) of each rod about L is ml²/6.

Now Let moment of inertia about xy be I'. This will be same as moment of inertia for either of the rod about pq. Since, pq and xy are perpendicular to each other, moment of inertia about any line perpendicular to both these lines (which is line L ) will be sum of the moment of inertias about these 2 lines which is I'+I' = 2I'

2I' is m.o.inertia about Line L which already is ml²/6

hence, 2l' = ml²/6
So, I' = ml²/12

-Rohil

karthik2007

Draw another symmetric bisector say cd on the other side of xy.

Now consider an axis perpendicular to the plane of the monitor, passing through the common centers of both the rods.

MI of system about this axis = ml²/12 + ml²/12 = ml²/6.

Using perpendicular axis theorem,ml²/6 = I_xy + I_cd

But by symmetry, I_xy = I_cd

So 2I_xy = ml²/6

Or I_xy = ml²/12

waterdemon

Gr8 answers!!!
Cheers!!!!@@!!!!

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