Question

The moment of inertia of a uniform rod of mass $m$ and length $l$ is $\alpha$ when rotated about an axis passing through centre and perpendicular to the length. If the rod is broken into equal halves and arranged as shown, then the moment of inertia about the given axis is

• A. $2\alpha$
• B. $\frac{\alpha}{2}$
• C. $4\alpha$
• D. $\frac{\alpha}{4}$

Answer 1

$\alpha$

Answer 2

We start by noting that for a uniform rod of mass m and length l, the moment of inertia about an axis perpendicular to its length through its centre is

$$I_{\rm rod}=\frac{1}{12}\,m\,l^2=\alpha.$$

Now the rod is broken into two equal halves. Each half has mass = m/2 and length = l/2.

For a thin rod of length L and mass M, the moment of inertia about an axis through its centre (and perpendicular to it) is $I_{\rm cm}=\frac{1}{12}\,M\,L^2.$

Thus for one half‐rod we have

$$I_{\rm cm\, (half)}=\frac{1}{12}\left(\frac{m}{2}\right)\left(\frac{l}{2}\right)^2=\frac{1}{12}\cdot\frac{m}{2}\cdot\frac{l^2}{4}=\frac{m\,l^2}{96}\,.$$

According to the diagram each half–rod is located so that its centre of mass is at a distance $d=\frac{l}{4}$ from the given (vertical) rotation axis. Using the parallel axis theorem the moment of inertia of one half–rod about the given axis is

$$I_{\rm half}=I_{\rm cm\, (half)}+\left(\frac{m}{2}\right)d^2 =\frac{m\,l^2}{96}+\frac{m}{2}\left(\frac{l}{4}\right)^2 =\frac{m\,l^2}{96}+\frac{m}{2}\cdot\frac{l^2}{16}\,.$$

Since

$$\frac{m}{2}\cdot\frac{l^2}{16}=\frac{m\,l^2}{32}\,,$$

we have

$$I_{\rm half}=\frac{m\,l^2}{96}+\frac{m\,l^2}{32} =\frac{m\,l^2}{96}+\frac{3\,m\,l^2}{96} =\frac{4\,m\,l^2}{96}=\frac{m\,l^2}{24}\,.$$

There are two such halves so the total moment of inertia is

$$I_{\rm total}=2\,I_{\rm half}=2\left(\frac{m\,l^2}{24}\right)=\frac{m\,l^2}{12}\,.$$

But by definition the original rod’s moment of inertia is

$$\alpha=\frac{m\,l^2}{12}\,.$$

Thus, we find

$$I_{\rm total}=\alpha\,.$$

The answer is that the moment of inertia of the new arrangement is the same as that of the original rod.

Explanation (minimal):

1. For a rod of mass m and length l, $I=\frac{1}{12}ml^2=\alpha$.

2. Each half–rod (mass m/2, length l/2) has centre‐of–mass inertia $=\frac{1}{12}(m/2)(l/2)^2=\frac{m\,l^2}{96}$.

3. Using the parallel axis theorem (with d = l/4): extra term $=\frac{m}{2}(l/4)^2=\frac{m\,l^2}{32}$.

4. Thus, one half–rod’s moment $=\frac{m\,l^2}{96}+\frac{m\,l^2}{32}=\frac{m\,l^2}{24}$.

5. Two halves: $2\frac{m\,l^2}{24}=\frac{m\,l^2}{12}=\alpha$.

The moment of inertia of the new arrangement equals $\alpha$.