Question

From a disc of mass 'M' and radius 'R', a circular hole of diameter 'R' is cut whose rim passes through the center. The moment of inertia of the remaining part of the ruse about perpendicular axis passing through the center is ______.

Options

• `(7"MR"^2)/32`

• `(11"MR"^2)/32`

• `(9"MR"^2)/32`

• `(13"MR"^2)/32`

Answer 1

From a disc of mass 'M' and radius 'R', a circular hole of diameter 'R' is cut whose rim passes through the center. The moment of inertia of the remaining part of the ruse about perpendicular axis passing through the center is `underline((13"MR"^2)/32)`.

Explanation:

`"I"_"total" = "MR"^2/2`

`"I"_"removed" = "M"/4 ("R"//2)^2/2 + "M"/4 ("R"/2)^2 = (3 "MR"^2)/32   ....[because "mr"^2/2 + "mh"^2]`

`"I"_"remain" = "I"_"total" - "I"_"removed"`

`= "MR"^2/2 - 3/32 "MR"^2`

`= 13/32 "MR"^2`