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

$MR squared$
$2 over 5 MR squared$
$3 over 2 MR squared$
$1 half MR squared$

Solution

3 over 2 MR squared

The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is
$straight I subscript CM space equals space 1 half MR squared$
where M is the mass of disc and R its radius. According to theorem of parallel axis, MI of circular disc about an axis touching the disc at it diameter and normal to the disc is
$straight I equals space straight I subscript CM space plus space MR squared$
$equals 1 half MR squared plus space MR squared equals 3 over 2 MR squared$

Some More Questions From Laws of Motion Chapter

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Laws Of Motion

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

$MR squared$
$2 over 5 MR squared$
$3 over 2 MR squared$
$1 half MR squared$

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