Question

The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of the ring, is \((\frac{1}{x})MR^2\), where R is the radius and M is the mass of the semicircular ring. The value of x will be:

Answer 1

Correct Answer: 2

To find the moment of inertia of a semicircular ring about an axis passing through the center and perpendicular to its plane, we start by considering the moment of inertia for a full circular ring of mass \(M\) and radius \(R\). For a complete ring, the moment of inertia \(I_{\text{ring}}\) about the same axis is: \(I_{\text{ring}} = MR^2\).

Since a semicircular ring is half a full ring, its moment of inertia \(I_{\text{semicircle}}\) is half of a full circle’s moment of inertia. Thus, \(I_{\text{semicircle}} = \frac{1}{2}MR^2\).

The problem states that the moment of inertia for the semicircular ring is given by \((\frac{1}{x})MR^2\). Therefore, we equate this to our expression for the semicircular ring:

\(\frac{1}{x}MR^2 = \frac{1}{2}MR^2\).

This implies:

\(\frac{1}{x} = \frac{1}{2}\).

Solving for \(x\), we find:

\(x = 2\).

Thus, the value of \(x\) is 2, which fits the given range of [2, 2].