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The moments of inertia of a non-uniform ...

The moments of inertia of a non-uniform circular disc (of mass M and radius R) about four mutually perpendicular tangents AB, BC, CD, DA are `I_(1), I_(2), I_(3)` and `I_(4)`, respectively (the square ABCD circumscribes the circle). The distance of the center of mass of the disc from its geometrical center is given by -

A

`(1)/(4MR) sqrt((I_(1)- I_(3))^(2) + (I_(2)- I_(4))^(2))`

B

`(1)/(12MR) sqrt((I_(1)- I_(3))^(2) + (I_(2)- I_(4))^(2))`

C

`(1)/(3MR) sqrt((I_(1)- I_(2))^(2) + (I_(2)- I_(4))^(2))`

D

`(1)/(2MR) sqrt((I_(1)+ I_(3))^(2) + (I_(2) + I_(4))^(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`I_(1)= I_(CM) + M(R-y)^(2)`
`I_(3)= I_(CM) + M (R+ y)^(2)`
So, `I_(3)- I_(1)= M [(R+ y)^(2) - (R- y)^(2)]` ...(i)
Same as
`I_(4)- I_(2)= M [(R+x)^(2) -(R- x)^(2)]` ...(ii)
By Distance of CM from centre of disc `= x^(2) + y^(2)= (1)/(4MR) sqrt((I_(1)- I_(3))^(2) + (I_(2)- I_(4))^(2))`
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