Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as ; I 1= M.I. of thin circular ring about its diameter. I 2= M.I. of circular disc about an axis perpendicular to the disc and going through the centre, I3= M.I. of solid cylinder about its axis and I 4= M.I. of solid sphere about its diameter. Then :

Question

Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as ; I 1= M.I. of thin circular ring about its diameter. I 2= M.I. of circular disc about an axis perpendicular to the disc and going through the centre, I3= M.I. of solid cylinder about its axis and I 4= M.I. of solid sphere about its diameter. Then : (A) I 1+ I 3< I 2+ I 4 (B) I 1+ I 2= I 3+(5/2) I 4 (C) I 1= I 2= I 3> I 4 (D) I 1= I 2= I 3< I 4

Tardigrade Admin · Accepted Answer

Ring I1=(M R2/2) about diameter Disc I2=(M R2/2) Solid cylinder I3=(M R2/2) Solid sphere I4=(2/5) M R2 I1=I2=I3>I4

$I_{1} + I_{3} < I_{2} + I_{4}$

$I_{1} + I_{2} = I_{3} + \frac{5}{2} I_{4}$

$I_{1} = I_{2} = I_{3} > I_{4}$

$I_{1} = I_{2} = I_{3} < I_{4}$

Ring $I_{1} = \frac{M R ^{2}}{2}$ about diameter
Disc $I_{2} = \frac{M R ^{2}}{2}$
Solid cylinder $I_{3} = \frac{M R ^{2}}{2}$
Solid sphere $I_{4} = \frac{2}{5} M R^{2}$
$I_{1} = I_{2} = I_{3} > I_{4}$

I1​+I3​<I2​+I4​

I1​+I2​=I3​+25​I4​

I1​=I2​=I3​>I4​

I1​=I2​=I3​<I4​

Ring I1​=2MR2​ about diameter Disc I2​=2MR2​ Solid cylinder I3​=2MR2​ Solid sphere I4​=52​MR2 I1​=I2​=I3​>I4​

$I_{1} + I_{3} < I_{2} + I_{4}$

$I_{1} + I_{2} = I_{3} + \frac{5}{2} I_{4}$

$I_{1} = I_{2} = I_{3} > I_{4}$

$I_{1} = I_{2} = I_{3} < I_{4}$

Ring $I_{1} = \frac{M R ^{2}}{2}$ about diameter
Disc $I_{2} = \frac{M R ^{2}}{2}$
Solid cylinder $I_{3} = \frac{M R ^{2}}{2}$
Solid sphere $I_{4} = \frac{2}{5} M R^{2}$
$I_{1} = I_{2} = I_{3} > I_{4}$