Question

ans with full solutio​

Answer 1

$\Large\underline{\underline{\sf \blue{Given}:}}$

• Disc of radius = R

• Mass =9M

• Removed Radius = $\sf{\frac{R}{3}}$

$\Large\underline{\underline{\sf \blue{To\:Find}:}}$

• Moment of inertia of Remaining disc passing through the centre of the disc and perpendicular to its plane = ?

$\Large\underline{\underline{\sf \blue{Solution}:}}$

Moment of inertia of Complete disc :

$\large{\boxed{\sf I=\frac{1}{2}MR^2 }}$

$\implies{\sf I_1=\dfrac{1}{2}×9M×R^2}$

$\implies{\sf I_1=\dfrac{9}{2}MR^2}$

Mass of cut out disc :

$\implies{\sf m= \dfrac{9M}{πR^2}π\dfrac{R}{3}}$

m = M

Moment of Inertia of cut out disc :

$\implies{\sf I_2=\dfrac{1}{2}M\left(\dfrac{R}{3}\right)^2 +M\left(\dfrac{2R}{3}\right)^2}$

$\implies{\sf \dfrac{1}{2}MR^2 }$

Moment of Inertia of residue disc :

$\large{\boxed{\sf I=I_1-I_2 }}$

$\implies{\sf \dfrac{9}{2}MR^2-\dfrac{1}{2}MR^2 }$

$\implies{\sf 4MR^2}$

$\Large\underline{\underline{\sf \blue{Answer}:}}$

•°• Moment of inertia of Remaining disc passing through the centre of the disc and perpendicular to its plane is 4MR²