Question

Q39. 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 :-
(A). MR²
(B). 1/2 MR²
(C). 3/2 MR²
(D). 7/2 MR²

Answer 1

Answer:

• The moment of Inertia is 3/2 MR²(Option- C).

Given:

1. Mass of the Disc = M.
2. Radius of the Disc = R.

Explanation:

$\rule{300}{1.5}$

Moment of Inertia of A Disc:-

The Moment of Inertia about an Axis Perpendicular to the plane & Passing through the centre of a Disc is ,

$\large{\boxed{\bold{ I = \dfrac{MR^2}{2}}}}$

$\bold{Here}\begin{cases}\sf{M = Mass \: of \: the \; Disc} \\ \sf{R = Radius \: of\: the \: Disc}\end{cases}$

Parllel Axis Theorem:-

$\large{\boxed{\bold{ I_{CD} = I_{Z} + Mh^2}}}$

$\bold{Here}\begin{cases}\sf{I_{CD} = Moment \: of \: Inertia \; at \; Edge} \\ \sf{I_{Z} = Moment \: of \: Inertia \; at \; Center} \\ \sf{h = Distance\: Between\: Two \: axis}\end{cases}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

As We Know, the Moment of Inertia about an Axis Perpendicular to the plane & Passing through the centre of a Disc is MR²/2.

From Figure (Attachment).

We Need to find Moment of Inertia about CD axis.

Applying Parllel axis theorem

$\large{\boxed{\bold{I_{CD} = I_{Z} + Mh^2}}}$

Substituting the values,

$\large{\tt \hookrightarrow I_{CD} = \dfrac{MR^2}{2} + M(R^2)}$

∵[h = R]

$\large{\tt \hookrightarrow I_{CD} = \dfrac{MR^2}{2} + MR^2}$

$\large{\tt \hookrightarrow I_{CD} = \dfrac{MR^2 + 2MR^2}{2}}$

$\huge{\boxed{\boxed{\tt I_{CD} = \dfrac{3MR^2}{2}}}}$

So, The Moment of Inertia of a Disc About axis passing from the edge of the disc and normal to the disc is 3/2 MR².

Note:-

• Here ${\sf{I_{CD}}}$ is the Axis which is passing from the edge of the disc and is normal to the disc.

$\rule{300}{1.5}$