Rotational Motion Question 60
Question: The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:
[AIPMT (S) 2005]
Options:
A) $ \frac{1}{2}MR^{2} $
B) $ MR^{2} $
C) $ \frac{7}{2}MR^{2} $
D) $ \frac{3}{2}MR^{2} $
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Answer:
Correct Answer: D
Solution:
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Key Idea: We should use parallel axis theorem.
Moment of inertia of disc passing through its centre of gravity and perpendicular to plane is
$ I _{AB}=\frac{1}{2}MR^{2} $
Using theorem of parallel axes, we have,
$ I _{CD}=I _{AB}+MR^{2} $
$ =\frac{1}{2}MR^{2}+MR^{2} $
$ =\frac{3}{2}MR^{2} $
Note: The role of moment of inertia in the study of rotational motion is analogous to that or mass in study of linear motion.
BETA
