Moment of inertia of ring about its diameter is I. Then, moment of inertia about an axis passing through centre perpendicular to its plane is

Question

Moment of inertia of ring about its diameter is I. Then, moment of inertia about an axis passing through centre perpendicular to its plane is (A) 2I (B) (I/2) (C) (3/2)I (D) I

Tardigrade Admin · Accepted Answer

From perpendicular axis theorem we have, I2=Ix+Iy Given, I2=I and about diameter I=Zy=I prime ∴ I=I prime+I prime I=2 I prime I prime=(I/2)

Q. Moment of inertia of ring about its diameter is I. Then, moment of inertia about an axis passing through centre perpendicular to its plane is

$2 I$

$\frac{I}{2}$

$\frac{3}{2} I$

$I$

From perpendicular axis theorem we have,
$< b r / > I_{2} = I_{x} + I_{y} < b r / >$
Given, $I_{2} = I$ and about diameter $I = Z_{y} = I^{'}$
$< b r / > ∴ I = I^{'} + I^{'} < b r / > I = 2 I^{'} < b r / > I^{'} = \frac{I}{2} < b r / >$

Q. Moment of inertia of ring about its diameter is I. Then, moment of inertia about an axis passing through centre perpendicular to its plane is

2I

2I​

23​I

I

From perpendicular axis theorem we have, <br/>I2​=Ix​+Iy​<br/> Given, I2​=I and about diameter I=Zy​=I′ <br/>∴I=I′+I′<br/>I=2I′<br/>I′=2I​<br/>

$2 I$

$\frac{I}{2}$

$\frac{3}{2} I$

$I$

From perpendicular axis theorem we have,
$< b r / > I_{2} = I_{x} + I_{y} < b r / >$
Given, $I_{2} = I$ and about diameter $I = Z_{y} = I^{'}$
$< b r / > ∴ I = I^{'} + I^{'} < b r / > I = 2 I^{'} < b r / > I^{'} = \frac{I}{2} < b r / >$