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:

  • 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.



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