Question

The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is :

• A. MR²
• B. \(\bigg(\frac{2}{5}\bigg)\)MR²
• C. \(\bigg(\frac{3}{2}\bigg)\)MR²
• D. \(\bigg(\frac{5}{6}\bigg)\)MR2

Answer 1

The Correct Option is C

To determine the moment of inertia of a uniform circular disc about an axis touching the disc at its diameter and normal to the disc, we need to apply the theorem of parallel axes.

The moment of inertia of a uniform circular disc about an axis through its center and normal to the plane of the disc is given by: \(\frac{1}{2} MR^2\)

According to the parallel axis theorem, the moment of inertia about an axis parallel to the axis through the center of mass and at a distance d is: I = I_{\text{cm}} + Md^2

For the given question, the axis is at the edge of the disc (i.e., touches the disc at its diameter), which means it is at a distance equal to the radius R from the center of the disc. Applying the parallel axis theorem:

\[ I = \frac{1}{2} MR^2 + MR^2 = \frac{1}{2} MR^2 + 1 \cdot MR^2 = \frac{3}{2} MR^2 \]

This calculation confirms that the correct answer is:

\(\bigg(\frac{3}{2}\bigg)MR^2\)

Therefore, among the options provided, the correct choice is the third option:

• \(\bigg(\frac{3}{2}\bigg)MR^2\)