Moment of inertia of circular loop of radius R about the axis of rotation parallel to horizontal diameter at a distance R/2 from it is

Question

Moment of inertia of circular loop of radius R about the axis of rotation parallel to horizontal diameter at a distance R/2 from it is (A) MR2 (B) (1/2)MR2 (C) 2MR2 (D) (3/4)MR2

Tardigrade Admin · Accepted Answer

Applying theorem of parallel axis I=ICM+M((R/2))2 I=(1/2)MR2+(MR2/4) I=(3/4)MR2

Q. Moment of inertia of circular loop of radius R about the axis of rotation parallel to horizontal diameter at a distance R/2 from it is

$M R^{2}$

$\frac{1}{2} M R^{2}$

$2 M R^{2}$

$\frac{3}{4} M R^{2}$

Applying theorem of parallel axis
$I = I_{CM} + M (\frac{R}{2})^{2}$
$I = \frac{1}{2} M R^{2} + \frac{M R ^{2}}{4}$
$I = \frac{3}{4} M R^{2}$

Q. Moment of inertia of circular loop of radius R about the axis of rotation parallel to horizontal diameter at a distance R/2 from it is

MR2

21​MR2

2MR2

43​MR2

Applying theorem of parallel axis I=ICM​+M(2R​)2 I=21​MR2+4MR2​ I=43​MR2

$M R^{2}$

$\frac{1}{2} M R^{2}$

$2 M R^{2}$

$\frac{3}{4} M R^{2}$

Applying theorem of parallel axis
$I = I_{CM} + M (\frac{R}{2})^{2}$
$I = \frac{1}{2} M R^{2} + \frac{M R ^{2}}{4}$
$I = \frac{3}{4} M R^{2}$