Rotational MotionHard
Question
The moment of inertia of a thin suare plate ABCD, of unifor thickness about an axis passing through the centre O and perpendicular to the plane of the plane of the plate is
where I1, I2, I3 and I4 are respectively moments of inertia about axes 1, 2, 3, and 4 which are in the plane of the plate.
where I1, I2, I3 and I4 are respectively moments of inertia about axes 1, 2, 3, and 4 which are in the plane of the plate.
Options
A.I1 + I2
B.I3 + I4
C.I1 + I3
D.I1 + I2 + I3 + I4
Solution
Since, it is square lamina
I3 = I4
and I1 = I2
From perpendicular axes theorem,
Moment of inertia about an axis perpendicular to square plate and passing from O is
I0 = I1 + I2 = I3 + I4
and I0 = 2I2 = 2I3
Hence, I2 = I3
Rather we can say I1 = I2 = I3 = I4
Therefore, I0 can be obtained by adding any two i.e.,
I0 = I1 + I2 = I1 + I3 = I1 + I4 = I2 + I3
= I2 + I4 = I3 + I4
I3 = I4
and I1 = I2
From perpendicular axes theorem,
Moment of inertia about an axis perpendicular to square plate and passing from O is
I0 = I1 + I2 = I3 + I4
and I0 = 2I2 = 2I3
Hence, I2 = I3
Rather we can say I1 = I2 = I3 = I4
Therefore, I0 can be obtained by adding any two i.e.,
I0 = I1 + I2 = I1 + I3 = I1 + I4 = I2 + I3
= I2 + I4 = I3 + I4
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