Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as; I1 = M.I. of thin circular ring about its diameter,I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre,I3 = M.I. of solid cylinder about its axis andI4 = 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;

I₁ = M.I. of thin circular ring about its diameter,

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

Then :

I₁ = I₂ = I₃ > I₄

I₁ + I₃ < I₂ + I₄

I₁ = I₂ = I₃ < I₄

I₁ + I₂ = I₃ + ${5 \over 2}$ I₄

Solution

Please login to view the detailed solution steps...

Go to DASH