Rotational Motion Question 15
Question 15 - 01 February - Shift 2
Moment of inertia of a dise of mass $M$ and radius ’ $R$ ’ about any of its diameter is $\frac{MR^{2}}{4}$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $\frac{x}{2} M R^{2}$. The value of $x$ is
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Answer: (3)
Solution:
Formula: Moment of inertia, Perpendicular Axis Theorem
$I=I _{cm}+Md^{2}$
$=\frac{MR^{2}}{MR^{2}}$
$=\frac{3}{2} MR^{2}$
