Question

The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:

• A. $\frac{5}{2}MR^2$
• B. $MR^2$
• C. $\frac{7}{2}MR^2$
• D. $\frac{3}{2}MR^2$

Answer 1

Answer: (D)

$\frac{3}{2}MR^2$

Answer 2

Moment of inertia of the disc around an axis perpendicular to its plane and passing through its centre of mass ICM = $\frac{1}{2}$ MR2
The disc's moment of inertia about an axis perpendicular to its plane and running across its diameter may be calculated using the parallel axis theorem by using the formula I = ICM + Md2.

Therefore, I = $\frac{1}{2}$ MR2+MR2 = $\frac{3}{2}MR^2$.

Correct option is (D): $\frac{3}{2}MR^2$