A thin rod of mass M and length L is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is

Question

A thin rod of mass M and length L is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is (A) (ML2/2π2) (B) (ML2/4π2) (C) (ML2/8π2) (D) (ML2/π2)

Tardigrade Admin · Accepted Answer

I = (MR2/2) but 2 π R = L ⇒ R =(L/2π) ∴ I = (ML2/8 π2)

Q. A thin rod of mass M and length L is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is

$\frac{M L ^{2}}{2 π ^{2}}$

$\frac{M L ^{2}}{4 π ^{2}}$

$\frac{M L ^{2}}{8 π ^{2}}$

$\frac{M L ^{2}}{π ^{2}}$

$I = \frac{M R ^{2}}{2}$
but $2 π R = L \Rightarrow R = \frac{L}{2 π}$
$∴ I = \frac{M L ^{2}}{8 π ^{2}}$

Q. A thin rod of mass M and length L is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is

2π2ML2​

4π2ML2​

8π2ML2​

π2ML2​

I=2MR2​ but 2πR=L⇒R=2πL​ ∴I=8π2ML2​

$\frac{M L ^{2}}{2 π ^{2}}$

$\frac{M L ^{2}}{4 π ^{2}}$

$\frac{M L ^{2}}{8 π ^{2}}$

$\frac{M L ^{2}}{π ^{2}}$

$I = \frac{M R ^{2}}{2}$
but $2 π R = L \Rightarrow R = \frac{L}{2 π}$
$∴ I = \frac{M L ^{2}}{8 π ^{2}}$