four holes of radius R are cut from a thin square plate of side 4R and mas
s M .The moment of inertia of the remaining portion about Z axis is
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The options for this question are missing. Here are the missing options:
(a) πMR^2/12
(b) (4/3 - π/4) MR^2
(c) (4/3 - π/6) MR^2
(d) (8/3 - 10π/16) MR^2
Answer:
The moment of inertia of four circular plates = 4 x 1/2 M'R²
Where M' is the mass of each square plate
The moment of intertia of square plate about Z - axis = 1/6 (4R)²
= 8/3 MR²
M.I of remaining portion = 8/3 MR² - 2MR²
= (8/3M - 2M') R²
if there is any confusion please leave a comment below.
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