Three identical rings each of mass M and radius R are placed in the same plane touching each other such that their centres from vertices of an equilateral triangle the moment of inertia of the system about an axis passing through the centre of one of the Ring and perpendicular to its plane
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Answered by
75
The moment of inertia of a ring about axis passing through its centre and perpendicular to the plane of ring is given by I=MR2
Now we know that the moment of inertia of the whole system will be moment of inertia of one ring about axis passing through it's centre and for other ring the axis will be 2R distance from the centre. Now using the parallel axis theorem we can find out the moment of inertia of the other ring which will be
I2=MR2+M(2R)2=5MR2
So total moment of inertia will be
I=I1+2I2=MR2+2×5MR2=11MR2'
Now we know that the moment of inertia of the whole system will be moment of inertia of one ring about axis passing through it's centre and for other ring the axis will be 2R distance from the centre. Now using the parallel axis theorem we can find out the moment of inertia of the other ring which will be
I2=MR2+M(2R)2=5MR2
So total moment of inertia will be
I=I1+2I2=MR2+2×5MR2=11MR2'
Pranathi1874:
Can u put this symbol (^)for squares
can u explain why distance is taken as 2R
no need
kk i understood
Answered by
32
Answer:
11MR^2
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