Rotational Motion Question 12

Question: The moment of inertia of uniform semi-circular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is

[AIEEE 2005]

Options:

A) $ \frac{1}{4}Mr^{2} $

B) $ \frac{2}{5}Mr^{2} $

C) $ Mr^{2} $

D) $ \frac{1}{2}Mr^{2} $

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Answer:

Correct Answer: D

Solution:

  • [d] The mass of complete (circular) disc is

The moment of inertia of disc about the given axis is

$ l=\frac{2Mr^{2}}{2}=Mr^{2} $

The disc may be assumed as combination of two semi-circular parts.

Thus, the moment of inertia of a semi-circular part of disc,

$ l _1-l-l _1 $

$ \therefore l _1=\frac{l}{2}=\frac{Mr^{2}}{2} $



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