Rotational Motion Question 61
Question: The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is:
[AIPMT (S) 2006]
Options:
A) $ MR^{2} $
B) $ \frac{2}{5}MR^{2} $
C) $ \frac{3}{2}MR^{2} $
D) $ \frac{1}{2}MR^{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
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The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
$ I _{Cm}=\frac{1}{2}MR^{2} $
where M is the mass of the disc and R its radius.
According to theorem of parallel axis, ML of circular disc about an axis touching the disc at its diameter and normal to the disc is
$ I=I _{CM}+MR^{2} $
$ =\frac{1}{2}MR^{2}+MR^{2} $
$ =\frac{3}{2}MR^{2} $
BETA
