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Suppose a sphere of radius a//4 is cut o...

Suppose a sphere of radius `a//4` is cut of the cube in previous problem. The center of the excised sphere is at the center of the cube. What is the moment of inertia `I_(e)` of the resulting object, about the same axis ?
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Text Solution

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The correct Answer is:
`Ma^(2) [(1)/(6) -(pi)/(1920)]`
Let mass of soild cube `= m`
man of sphere radius `(a//4), m' = (m)/(a^(3)) xx (4)/(3) pi ((1)/(4))^(3) = (m pi)/(48)`
`I_(e) =I_("radius") - I_("sphere") rArr I_(e) = (ma^(2))/(6) -(2)/(5) m' ((a)/(4))^(2)`
`I_(e) =(ma^(2))/(6) -(2)/(5) m xx(m pi)/(48) xx (a^(2))/(16)`
`I_(e) =ma^(2) [(1)/(6) -(pi)/(1920)]`.
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