Question

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I₁. The same rod is bent into a ring and its moment of inertia about a diameter is I₂. If
\(\frac{I_1}{I_2}\) is \(\frac{xπ²}{3}\)
then the value of x will be ______.

Answer 1

Correct Answer: 8

To solve for the value of x, we begin by determining the moments of inertia for both configurations of the rod. Initially, with the rod as a straight line, its moment of inertia about an axis perpendicular to one end is given by:

I₁=\(\frac{1}{3}ML^2\)

where \(M\) is the mass of the rod, and \(L\) is its length.

When bent into a ring, the length of the rod becomes the circumference of the ring:

\(L=2πR\)

where \(R\) is the radius of the ring. Thus, the moment of inertia about a diameter of the ring is:

I₂=\(\frac{1}{2}MR^2\)

Since \(L=2πR\), we have:

\(R=\frac{L}{2π}\)

Substitute \(R\) in the equation for I₂:

I₂=\(\frac{1}{2}M\left(\frac{L}{2π}\right)^2\)

Simplifying gives:

I₂=\(\frac{1}{2}M\frac{L^2}{4π^2}=\frac{ML^2}{8π^2}\)

We know from the problem conditions:

\(\frac{I_1}{I_2}=\frac{xπ^2}{3}\)

Substitute I₁ and I₂:

\(\frac{\frac{1}{3}ML^2}{\frac{ML^2}{8π^2}}=\frac{xπ^2}{3}\)

Simplify the expression:

\(\frac{8π^2}{3}=xπ^2\)

Solving for x gives:

\(x=8\)

The computed value of x is 8, which falls within the expected range of 8,8. Therefore, the value of x is 8.