Question

023.
023. Four identical rings of radius R and mass M
are placed at the corner of a square in x-y
plane such that each ring touches the two
rings tangentially. The moment of inertia of
this system about z-axis passing through the
centre of the square is
(A) 6 MR2
(B)
4 MR2
(C) 8 MR2
(D
7 MR2
54 Km/Hr Sounds
024.​

Answer 1

Answer:

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Explanation:

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Answer 2

The passing through the center of the square is $4MR^{2}$.

Explanation:

We have,

$I_{4}=I_{3}=\frac{MR^{2}}{2}$

$I_{1}=I_{2}=\frac{MR^{2}}{2}+MR^{2}$

$I_{system}=I_{1}+I_{2}+I_{3}+I_{4}$

$=2*\frac{3}{2}MR^{2}+2*\frac{MR^{2}}{2}$

$=3MR^{2}+MR^{2}$

$=4MR^{2}$.