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} $
BETA
