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} $

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Answer:

Correct Answer: C

Solution:

  • 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} $



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