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Moment of inertia of uniform triangular ...

Moment of inertia of uniform triangular plate about axis passing through sides `AB,AC,BC` are `I_p,I_B` and `I_H` respectively and about an axis perpendicular to the plane and passing through point `C` is `I_C`. Then :
.

A

`I_(C)gtI_(P)gtI_(B)gtI_(H)`

B

`I_(H)gtI_(B)gtI_(C)gtI_(P)`

C

`I_(P)gtI_(H)gtI_(B)gtI_(C)`

D

`I_(H)gtI_(B)=I_(C)gtI_(P)`

Text Solution

Verified by Experts

The correct Answer is:
A
Moment of inertia is more when mass is farther from the axis. In case of axis `BC`, mass distribution is closed to it and in case of axis `AB` mass distribution is farthest . Hence
`I_(BC) lt I_(AC) lt I_(AB)`
`rArr I_(P) gt I_(B) gt I_(H)`

`I_(C)=i_(CM)+my^(2)`
`=I_(B)^(1)-mx^(2)+my^(2)`
`=I_(B)^(1)+m(y^(2)-x^(2))`
`I_(P)+I_(B)+m(y^(2)-x^(2))`
`gt I_(P)+I_(B)`
`gt I_(P)`
Here`I_(B)^(1)` is moment of inertia of the plate about an axis perpendicular to it and passing through `B`.
`:. I_(C) gt I_(P) gt I_(B) gt I_(H)`
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