Question

if l_(1) is te moment of inertia of a thin rod about an axis perpendicular to its length and passing thorugh its centre of mass and l_(2) te moment of inertia of the ring formed by the same rod about an axis passing through the centre of mass of the ring and perpendicular tot he plane of the ring. then find the ratio (l_(1))/(l_(2)).

Answer 1

Given :

• Moment of inertia of thin rod through perpendicular axis passing through its center = I₁
• Moment of inertia of ring formed by same rod about perpendicular axis passing though its center = I₂

To find :

• I₁/I₂ = ?

Solution :

• Let mass of the rod be m.

• We know that moment of inertia of a rod about axis perpendicular to its length passing through the center = $\frac{ml^{2} }{12}$
• Hence,  I₁ = $\frac{ml^{2} }{12}$ ....(1)
• Now, the rod is made into a ring. So, the length of the rod will be the circumference of the ring.
• ∴ $l = 2\pi r$
• ∴ $r = \frac{l}{2\pi }$
• We know that moment of inertia of a ring about axis perpendicular to the plane of the ring and passing through its center = $mr^{2}$
• Hence, I₂ = $m. (\frac{l}{2\pi }) ^{2}$
•                 = $\frac{ml^{2} }{4\pi ^{2} }$  ....(2)
• From equations (1) and (2) ,

        I₁/ I₂ = $\frac{4\pi ^{2} }{12}$

                = π²/3

Answer : I₁/I₂ = π²/3