The moment of inertia of a circular disc about an axis passing through the circumference perpendicular to the plane of the disc is :

Question

The moment of inertia of a circular disc about an axis passing through the circumference perpendicular to the plane of the disc is : (A) MR2 (B) (3/2) MR2 (C) (MR2/2) (D) (5/4) MR2

Tardigrade Admin · Accepted Answer

I=Ic m+m R2=(m R2/2)+m R2 =(3 m R2/2)

Q. The moment of inertia of a circular disc about an axis passing through the circumference perpendicular to the plane of the disc is :

$M R^{2}$

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

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

$\frac{5}{4} M R^{2}$

$I = I_{c m} + m R^{2} = \frac{m R ^{2}}{2} + m R^{2}$
$= \frac{3 m R ^{2}}{2}$

Q. The moment of inertia of a circular disc about an axis passing through the circumference perpendicular to the plane of the disc is :

MR2

23​MR2

2MR2​

45​MR2

I=Icm​+mR2=2mR2​+mR2 =23mR2​

$M R^{2}$

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

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

$\frac{5}{4} M R^{2}$

$I = I_{c m} + m R^{2} = \frac{m R ^{2}}{2} + m R^{2}$
$= \frac{3 m R ^{2}}{2}$