Question

Q14. From a uniform disc of radius R and mass 9 M, a small disc of radius R/3 is removed. What is the moment of inertia of remaining disc about an axis passing through the centre of disc and perpendicular to its plane?​

Answer 1

Step-by-step explanation:

Correct option is

B

MR

2

Moment of inertia of a ring about an axis passing through its centre and perpendicular to it is MR

2

Answer 2

Answer:

★ Concept :-

Here the concept of Moment of Inertia had been used. We see that we are given the radius of the uniform dísc along with it's mass. Even we see that we have the radius of the disc removed from the initial disc. From this we can firstly find out the mass of the smaller disc. Then we can find out the moment of Inertia of both disc. Then we can subtract them to get the answer.

Let's do it !!

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★ Solution :-

Given,

» Radius of the Intial Disc = R

» Mass of the Intial Disc = 9M

» Radius of New Disc = R/3

» Shape of both discs = Circular (at surface)

• Let the mass of the smaller disc be m

• Let the moment of inertia of the whole disc be I₁

• Let the moment of inertia of the smaller disc be I₂

• Let the moment of Inertia of left part of disc be I

*General Moment of Inertia is denoted by M.I.

• For the mass of smaller disc ::

We see that the smaller disc is the part of uniform disc. This means that máss will be uniformly distributed. Then the ratio of mass of smaller disc by area of smaller disc must be equal to the ratio of mass of whole disc by area of whole disc. Then,

$\;\;\tt{\rightarrow\;\;\dfrac{m}{\pi\bigg(\dfrac{R}{3}\bigg)^{2}}\;=\;\dfrac{9M}{\pi R^{2}}}$

• Area of circlular region = πr² (where r is the radius)

$\;\;\tt{\rightarrow\;\;m\;=\;\dfrac{9M}{\pi R^{2}}\:\times\:\pi\bigg(\dfrac{R}{3}\bigg)^{2}}$

$\;\;\tt{\rightarrow\;\;m\;=\;\dfrac{9M}{\pi R^{2}}\:\times\:\dfrac{\pi R^{2}}{9}}$

Cancelling the like terms from numerator and denominator, we get

$\;\;\tt{\rightarrow\;\;m\;=\;\dfrac{9M}{9}}$

$\;\;\bf{\rightarrow\;\;m\;=\;M}$

• For the Moment of Inertia of Both Discs ::

We know that, for a circular disc where the axis of rotation is pássing through it's plane, the moment of inertia is given as ;

$\;\;\tt{\mapsto\;\;M.I.\;=\;\dfrac{1}{2}Mr^{2}}$

• r is the radius of disc

• M is the mass of disc

Then,

→] For moment of inertia Initial Disc -

$\;\;\tt{\mapsto\;\;I_{1}\;=\;\dfrac{1}{2}\:\times\:9M\:\times\:R^{2}}$

$\;\;\bf{\mapsto\;\;I_{1}\;=\;\dfrac{9MR^{2}}{2}}$

→] For moment of inertia Smaller Disc -

$\;\;\tt{\mapsto\;\;I_{2}\;=\;\dfrac{1}{2}\:\times\:M\:\times\:\bigg(\dfrac{R}{3}\bigg)^{2}\:+\:M\bigg(\dfrac{2R}{3}\bigg)^{2}}$

$\;\;\tt{\mapsto\;\;I_{2}\;=\;\dfrac{1}{2}\:\times\:\dfrac{MR^{2}}{9}\:+\:\dfrac{4MR^{2}}{9}}$

$\;\;\tt{\mapsto\;\;I_{2}\;=\;\dfrac{MR^{2}}{18}\:+\:\dfrac{4MR^{2}}{9}}$

$\;\;\tt{\mapsto\;\;I_{2}\;=\;\dfrac{MR^{2}}{18}\:+\:\dfrac{8MR^{2}}{18}}$

$\;\;\tt{\mapsto\;\;I_{2}\;=\;\dfrac{MR^{2}\:+\:8MR^{2}}{18}}$

$\;\;\tt{\mapsto\;\;I_{2}\;=\;\dfrac{9MR^{2}}{18}}$

$\;\;\bf{\mapsto\;\;I_{2}\;=\;\dfrac{MR^{2}}{2}}$

• For the Moment of Inertia of left part of disc ::

This will be given as, (by using Theorem of Perpendicular Axés)

$\;\;\tt{\Longrightarrow\;\;I\;=\;I_{1}\:-\:I_{2}}$

$\;\;\tt{\Longrightarrow\;\;I\;=\;\dfrac{9MR^{2}}{2}\:-\:\dfrac{MR^{2}}{2}}$

$\;\;\tt{\Longrightarrow\;\;I\;=\;\dfrac{9MR^{2}\:-\:MR^{2}}{2}}$

$\;\;\tt{\Longrightarrow\;\;I\;=\;\dfrac{8MR^{2}}{2}}$

$\;\;\bf{\Longrightarrow\;\;I\;=\;\red{4MR^{2}}}$

This is the required answer.

$\;\underline{\boxed{\tt{Moment\;of\;Inertia\;of\;left\;disc\;=\;\bf{\purple{4MR^{2}}}}}}$