First slide

Moment of intertia

Question

Moment of inertia of a solid disc about a diameter is I. Then moment of inertia of the disc about an axis which passes through a point on its circumference and perpendicular to the plane of the disc is

Easy

A

6 I
B

4 I
C

2.5 I
D

1.5 I

Solution

(1)
$I = \frac{1}{4} m R^{2}$
$I_{C} =$ Moment of inertia about an axis passing through the centre of the disc and perpendicular to the plane of the disc. = $2 I = \frac{1}{2} m R^{2}$
$∴$ By parallel axis Theorem.
$I^{'} = I_{C} + m R^{2} = \frac{1}{2} m R^{2} + m R^{2} = \frac{3}{2} m R^{2} = \frac{3}{2} \times 4 I = 6 I$

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