Rotational MotionEasy
Question
The moment of inertia of thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of plate is :-
(a) I1 + I2
(b) I3 + I4
(c) I1 + I3
(d) I1 + I2 + I3 + I4
Where I1, I2, I3 and I4 are respectively moment of inertia about axes 1, 2, 3 and 4 which are in the plane of plate, then which option will be correct :-
(a) I1 + I2
(b) I3 + I4
(c) I1 + I3
(d) I1 + I2 + I3 + I4
Where I1, I2, I3 and I4 are respectively moment of inertia about axes 1, 2, 3 and 4 which are in the plane of plate, then which option will be correct :-
Options
A.a
B.a, b
C.a, b, c
D.a, b, c, d
Solution
By applying theorem of perpendicular axis –
I = I1 + I2 ....... (i)
I = I3 + I4 ....... (ii)
Because of symmetry we have I1=I2 and I3=I4
Hence I = 2I1 = 2I2 = 2I3 = 2I4
or I1 = I2 = I3 = I4
In equation (i) put I3 in place of I2, so
I = I1 + I3 ....... (iii)
I = I1 + I2 ....... (i)
I = I3 + I4 ....... (ii)
Because of symmetry we have I1=I2 and I3=I4
Hence I = 2I1 = 2I2 = 2I3 = 2I4
or I1 = I2 = I3 = I4
In equation (i) put I3 in place of I2, so
I = I1 + I3 ....... (iii)
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