Question

from a uniform disc of radius R and mass 9M a small disc of radius R by 3 is removed as shown what is the moment of inertia of the remaining disc about an Axis passing through the centre of disc and perpendicular to its plane​

Answer 1

Answer:

4mr^2

As mass removed will be m

Find moment of inertia of 9m at axis - moment of inertia of m mass at the axis

Answer 2

The moment of inertia of the remaining disc is $\dfrac{40}{9}MR^2$

Explanation:

Given that,

Mass of disc = 9 M

Radius of disc = R

Radius of small disc $r'=\dfrac{R}{3}$

Initial moment of inertia of the complete disc about an axis passing through its center and perpendicular to its plane is

We need to calculate the moment of inertia of the disc

Using formula of moment of inertia

$I=\dfrac{1}{2}mr^2$

Put the value into the formula

$I=\dfrac{1}{2}\times9M\times(R)^2$

$I=\dfrac{9}{2}MR^2$

Moment of inertia of disc removed, about an axis passing through its centre and perpendicular

We need to calculate the moment of inertia of removed disc

Using formula of moment of inertia

$I'=\dfrac{1}{2}mr^2$

Put the value into the formula

$I'=\dfrac{1}{2}M\times(\dfrac{R}{3})^2$

$I'=\dfrac{MR^2}{18}$

We need to calculate the moment of inertia of the remaining disc

Using formula for moment of inertia

$I''=I-I'$

Put the value into the formula

$I''=\dfrac{9}{2}MR^2-\dfrac{MR^2}{18}$

$I''=\dfrac{40}{9}MR^2$

Hence, The moment of inertia of the remaining disc is $\dfrac{40}{9}MR^2$

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Topic : Moment of inertia

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