Question

Moment of inertia of a disc of mass \(M\) and radius \(R\) about any of its diameters is \(\frac{MR^2}{4}\). The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, \(\frac{x}{2}MR^2\). The value of \(x\) is _________.

Hint:

Remember the perpendicular and parallel axis theorems. They are essential for calculating moments of inertia about different axes.

Answer 1

Correct Answer: 3

The moment of inertia (I) of a disc about any of its diameters is given by \(I_d = \frac{MR^2}{4}\). Using the perpendicular axis theorem, the moment of inertia about an axis perpendicular to the disc and passing through its center is:

\(I_{center} = I_x + I_y = 2 \cdot I_d = 2 \cdot \frac{MR^2}{4} = \frac{MR^2}{2}\)

For the moment of inertia about an axis normal to the disc and passing through a point on its edge, we apply the parallel axis theorem:

\(I_{edge} = I_{center} + Md^2\), where \(d = R\) (the distance from the center to the edge).

So, \(I_{edge} = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}\).

According to the problem, \(I_{edge} = \frac{x}{2}MR^2\). Equating the two expressions for \(I_{edge}\), we get:

\(\frac{x}{2}MR^2 = \frac{3MR^2}{2}\).

Solving for \(x\), we find \(x = 3\).

This value of \(x\) falls within the expected range of 3,3.