The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I 1 . The same rod is bent into a ring and its moment of inertia about a diameter is I 2 . If I 1 I 2

Question

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I 1 . The same rod is bent into a ring and its moment of inertia about a diameter is I 2 . If I 1 I 2 is x π 2 3 , then the value of x will be ______.

TestHub · Accepted Answer

Moment of inertia of rod rotating about an axis passing through one end is $I_{1} = \frac{m l^{2}}{3} \dots (1)$

Moment of inertia of ring rotating about one of its diameters is $I_{2} = \frac{m R^{2}}{2}$ .

Since the rod is bent into a ring, the circumference of the ring and the length of the rod are equal, $l = 2 π R$

Then, $I_{2} = \frac{m R^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

Thus, the ratio $\frac{I_{1}}{I_{2}} = \frac{8 π^{2}}{3}$ , so $x = 8$ .

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