A disc of radius R / 2 is cut from a uniform disc of radius R as shown in the figure. The mass of the disc (with cavity) is M. The moment of inertia of this disc about an axis X X ' which passes through the center 0 of the disc and is in the plane of the disc will be

Question

A disc of radius R / 2 is cut from a uniform disc of radius R as shown in the figure. The mass of the disc (with cavity) is M. The moment of inertia of this disc about an axis X X ' which passes through the center 0 of the disc and is in the plane of the disc will be (A) (7 MR 2/12) (B) (13 MR 2/24) (C) (13 MR 2/48) (D) (11 MR 2/36)

Tardigrade Admin · Accepted Answer

I Z =( M 1 R 3/ Z )[( M 2( R / 2)3/ Z )+(( R /2))2] Whose M 1=σ π R 2 M 2=σ π(( R /2))2 σ=( M /π R 2-π R 2 / y )

Q. A disc of radius $R /2$ is cut from a uniform disc of radius $R$ as shown in the figure. The mass of the disc (with cavity) is $M$ . The moment of inertia of this disc about an axis $XX$ ' which passes through the center $0$ of the disc and is in the plane of the disc will be

$\frac{7 M R ^{2}}{12}$

$\frac{13 M R ^{2}}{24}$

$\frac{13 M R ^{2}}{48}$

$\frac{11 M R ^{2}}{36}$

$I_{Z} = \frac{M _{1} R ^{3}}{Z} [\frac{M _{2} ( R /2 ) ^{3}}{Z} + (\frac{R}{2})^{2}]$
Whose $M_{1} = σπ R^{2}$
$M_{2} = σπ (\frac{R}{2})^{2}$
$σ = \frac{M}{π R ^{2} - π R ^{2} / y}$

Q. A disc of radius R/2 is cut from a uniform disc of radius R as shown in the figure. The mass of the disc (with cavity) is M. The moment of inertia of this disc about an axis XX ' which passes through the center 0 of the disc and is in the plane of the disc will be

127MR2​

2413MR2​

4813MR2​

3611MR2​