Rotation 1 Question 18
21. The moment of inertia of a thin square plate $A B C D$, of uniform thickness about an axis passing through the centre $O$ and perpendicular to the plane of the plate is
$(1992,2 M)$
(a) $I _1+I _2$
(b) $I _3+I _4$
(c) $I _1+I _3$
(d) $I _1+I _2+I _3+I _4$
where, $I _1, I _2, I _3$ and $I _4$ are respectively moments of inertia about axes 1, 2, 3 and 4 which are in the plane of the plate.
Integer Answer Type Questions
Show Answer
Answer:
Correct Answer: 21. (a, b, c)
Solution:
- Since, it is a square lamina
$$ \begin{aligned} & I _3=I _4 \\ & \text { and } \quad I _1=I _2 \end{aligned} $$
(by symmetry)
From perpendicular axes theorem,
Moment of inertia about an axis perpendicular to square plate and passing from $O$ is
or
$$ I _o=I _1+I _2=I _3+I _4 $$
$$ \text { Hence, } \quad I _2=I _3 $$
Rather we can say $I _1=I _2=I _3=I _4$
Therefore, $I _o$ can be obtained by adding any two i.e.
$$ \begin{aligned} I _o & =I _1+I _2=I _1+I _3 \\ & =I _1+I _4=I _2+I _3 \\ & =I _2+I _4=I _3+I _4 \end{aligned} $$
