Rotational MotionEasy
Question
Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle θ with AB. The moment of inertia of the plate about the axis CD is then equal to
Options
A.I
B.I sin2θ
C.I cos2θ
D.I cos2(θ/2)
Solution
A′B′ ⏊ AB and C′D′ ⏊ CD
From symmetry IAB = IA′B′ and ICD = IC′D′
From theorem of perpendicular axes,
Izz = IAB + IA′B′ + IC′D′ = 2IAB = 2ICD
IAB = ICD.
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