The moment of inertia of a circular loop of radius R, at a distance of R/2 around a rotating axis parallel to horizontal diameter of loop is

Question

The moment of inertia of a circular loop of radius R, at a distance of R/2 around a rotating axis parallel to horizontal diameter of loop is (A) MR2 (B) (1/2) MR2 (C) 2MR2 (D) (3/4) MR2

Tardigrade Admin · Accepted Answer

According to theorem of parallel axis I=I CM +M((R/2))2 I=(1/2) M R2+(M R2/4) I=(3/4) M R2

Q. The moment of inertia of a circular loop of radius $R$ , at a distance of $R /2$ around a rotating axis parallel to horizontal diameter of loop is

$M R^{2}$

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

$2 M R^{2}$

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

According to 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. The moment of inertia of a circular loop of radius R, at a distance of R/2 around a rotating axis parallel to horizontal diameter of loop is

MR2

21​MR2

2MR2

43​MR2